[Solution] Invoking generator for vue-cli-plugin-nativescript-vue... ERROR Error: Cannot find module '@vue/cli-service/generator/template/src/main.js'

[Solution] Invoking generator for vue-cli-plugin-nativescript-vue... ERROR Error: Cannot find module '@vue/cli-service/generator/template/src/main.js'

If you are facing "Invoking generator for vue-cli-plugin-nativescript-vue... ERROR Error: Cannot find module '@vue/cli-service/generator/template/src/main.js" error while working with vue-cli-nativescriot-vue plugin So you are at the correct place.

I have found a solution for the same and below is the solution.

Please follow below steps to get it resolved.
Please remove npm and npm-cache folder from your computer's below location.

Windows :
C:\Users\AppData\Roaming.

Linux/Ubuntu:
/usr/local

Then Follow below steps
1) npm install -g @vue/cli@3.10.0
2) Go to code sharing project Directory's root folder location where you have package.json
3)npm install --save-dev @vue/cli-service@3.5.0
4)npm install --save-dev vue-cli-plugin-nativescript-vue@0.0.12
5)vue invoke vue-cli-plugin-nativescript-vue after that it will ask you for some user input and i provided below inputs.

? Enter a unique application identifier: org.nativescript.application
? Use HTML5 history mode? (Default: hash mode) No
? Is this a brand new project? (Default: Yes) Yes
? Dual Native AND Web development experience or a Native only? (Default: Dual) Dual Native AND Web
? What type of template do you want to start with? (Default: Simple) Simple.

So by following above steps you will be able to resolve the above error..

If anyone is still facing the same issue please comment that into the post i will happy to help them to get that resolved.

Hope you enjoyed the article and it got added something new to your knowledge.

To add something new every-time into your mind stay tuned to visionfortech.

Enjoy......!!!
Happy Learning.....!!!

Angularjs forms and models

forms and models

now into review i want to insert a review form to get review from users so lets start to insert form into review tab

Into the form of review tab we are going get how much starts user want to give for our product and what are the reviews and what author gives the review.i want all above information.

Into HTML
<form name="reviewForm>
<blockquote>
<b>Stars : {{review.stars}}</b>
{{review.body}}
<author>by : {{review.author}}</author>
</blockquote>
<select ng-model="review.stars">
<option value="1">1 Star </option>
<option value="2">2 Star </option>
<option value="3">3 Star </option>
<option value="4">4 Star </option>
<option value="5">5 Star </option>
</select>
<textarea ng-model="review.body"></textarea>
<label>by : </label>
<input ng-model="review.author" type="email" />
</form>

so in above code we can't define review variable anywhere so better coding prctise we need to define that so for that we need to intialize that and commonly intialization can be done into the controller so lets create the Controller.

App.js
app.controller("reviewController",function(){
this.review={}
});

So,now we need to bind this review Controller into our form so for that we need to do the Following.

Into HTML

<form name="reviewForm ng-controller="reviewController as reviewCtrl">
<blockquote>
<b>Stars : {{review.stars}}</b>
{{reviewCtrl.review.body}}
<author>by : {{reviewCtrl.review.author}}</author>
</blockquote>
<select ng-model="reviewCtrl.review.stars">
<option value="1">1 Star </option>
<option value="2">2 Star </option>
<option value="3">3 Star </option>
<option value="4">4 Star </option>
<option value="5">5 Star </option>
</select>
<textarea ng-model="reviewCtrl.review.body"></textarea>
<label>by : </label>
<input ng-model="reviewCtrl.review.author" type="email" />
</form>

So after filling needed information we need to submit the form and into Angular we have a ng-submit directive

ng-submit allows us to call a function when the form is submitted and after that we need to define that function into our ReviewController.

So into HTML :

<form name="reviewForm ng-controller="reviewController as reviewCtrl" ng-submit="reviewCtrl .addReview(product)">
<blockquote>
<b>Stars : {{review.stars}}</b>
{{reviewCtrl.review.body}}
<author>by : {{reviewCtrl.review.author}}</author>
</blockquote>
<select ng-model="reviewCtrl.review.stars">
<option value="1">1 Star </option>
<option value="2">2 Star </option>
<option value="3">3 Star </option>
<option value="4">4 Star </option>
<option value="5">5 Star </option>
</select>
<textarea ng-model="reviewCtrl.review.body"></textarea>
<label>by : </label>
<input ng-model="reviewCtrl.review.author" type="email" />
</form>

So now we have to define this addReview function into ReviewController.

So Into app.js

app.controller("ReviewController",function(){

this.review={};

this.addReview=function(product)
{
product.review.push(this.review);
};
});

But when we went to browser and input our review and after submitting the form container which has all the input controller are not getring blank or we can say resetted so for that we just need to add one line into the addReview function of reviewController.

So Into app.js

app.controller("ReviewController",function(){

this.review={};

this.addReview=function(product)
{
product.review.push(this.review);
this.review={};
};
});

So what it can do is when this form is getting submitted and when this function got called so after pushing review into array we can assign a new javascript object to the review object which is blank.So after getting submitted the element's values will get reset.

form
1)You need to create multiple li's for each review that exists.
2)Set the blockquote stars ( strong tag) to the review stars.
3)Add the review body between the strong & cite tags.
4)Set the cite to the review author.
5)Use ng-model for review.stars select form field.
6)Use ng-model for review.body for the forms textarea.
7)Use ng-model for review.author – the email field.
8)In the strong tag, before Stars, put in the right expression to display review.stars. Keep the space prior to Stars
9)On the blank line with no tags put in the right expression to display review.body
10)In the cite tag, immediately after the -, put in the right expression to display review.author. 

Javascript JSON to string

Javascript JSON to string

Into this post we are going to discuss how to convert Javascript JSON ( Javascript Object Notation ) to string or javascript json to string.

Into this kind real IT world we can say Javascript is everywhere like front end and BackEnd both so we can say Javascript is going to rule everywhere.

So in today's article we are going to discuss JSON which generally uses to pass data from one end to another end.

Let start understanding of that using below simple examples.

So into Javascript we can do that easily using JSON.stringify().

First let's take and example of Object

let json_to_string = JSON.stringify({ javascript_json: 5, json_to_string: 6 })

console.log(json_to_string);

//output: "{"javascript_json":5,"json_to_string":6}"

Now let's take an example of run time creation of Array of Object

let json_to_string = JSON.stringify([new String('visiofortech'),new Boolean(false),new Number(7)])

console.log(json_to_string);

//output: "["visiofortech",false,7}"

Now let's take an example of Date object

let json_to_string = JSON.stringify(new Date(2018, 10, 16, 15, 4, 5))

console.log(json_to_string);

//output: ""2018-11-16T09:34:05.000Z""

Hope you enjoyed the article and it got added something new to your knowledge.

To add something new every-time into your mind stay tuned to visionfortech.

To Learn more about Javascript Javascript Treasure.

Please comment below if you have something doubts or some questions.

Enjoy...!!!

Happy Learning....!!!

Javascript Global and Local Variables

Javascript Global and Local Variables

Into this post we are going to discuss difference between how to declare javascript global variable and javascript local variable and its scope

we are going to discuss how can we declare global variable and local varible into Javascript with example.

Javascript global variable vs local variable by visionfortech

Javascript Global Variable:

If we define var and let variable same globally then let variable will not get added into global window object while var variable will get added to global window object.

So let's declare two variables at the time of window loaded

let Visionfortech_global = "Visionfortech is defined as Global variable"

var Visionfortech_local = "Visionfortech is defined as Local variable"

console.log(Visionfortech_global);
o/p : undefined

console.log(Visionfortech_local);
o/p : Visionfortech is defined as Local variable
  

Javascript Local Variable : 

let js_local_variable = 1;

if (js_local_variable === 1) {
  let js_local_variable = 2;

  console.log(js_local_variable);
  // output: 2
}

console.log(js_local_variable);
// output: 1
  

Now let see the scope example of Javascript global and Local variable.

var global_variable = 1;

var javascript_global_variable = 2;

            

if (a === 1) {

     var global_variable = 11; // the scope is global

     let javascript_global_variable = 22; // the scope is inside the if-block

            

     console.log(global_variable);  // 11

     console.log(javascript_global_variable);  // 22

}          

console.log(global_variable); // 11

console.log(javascript_global_variable); // 2  

Enjoy....!!!!

Feel free to comment below your experience with above approach and If you still find any problem  with above steps Let me know I would love to help you to resolve your  problem.

 If you want to take your Technological Knowledge to the Next Level and For More Technological information Stay tuned to Visionfortech

Javascript let vs var

Javascript Let vs Var

Into this post we are going to discuss difference between Javascript var and Let.

we are going to discuss about Javascipt variable declaration method like where user can declare variable with var and where user can declare variable with let.

Global Variable Declaration:

If we define var and let variable same globally then let variable will not get added into global window object while var variable will get added to global window object.

So let's declare two variables at the time of window loaded

let Visionfortech_let = "Visionfortech is defined as let variable"

var Visionfortech_var = "Visionfortech is defined as var variable"

console.log(Visionfortech_let);
o/p : undefined

console.log(Visionfortech_var);
o/p : Visionfortech is defined as var variable
  

Redeclare Variable:

So Here let's discuss about one more example of let vs var into Javascript.You can not declare let variable but you can redeclare var variable

'use strict';

var let_vs_var = "Javascript let vs var";

var let_vs_var = "Difference Between let and var Javascript"; 
//replaced easily



'use strict';

let let_vs_var = "Javascript let vs var";

let let_vs_var = "Difference Between let and var Javascript"; 
// Error: let_vs_var is already declared

Javascript Let Example : 

let vision = 1;

if (vision === 1) {
  let vision = 2;

  console.log(vision);
  // output: 2
}

console.log(vision);
// output: 1
  

Now let see the scope example of Javascript let vs var.

var let_vs_var = 1;

var javascript_var = 2;

            

if (a === 1) {

     var let_vs_var = 11; // the scope is global

     let javascript_var = 22; // the scope is inside the if-block

            

     console.log(let_vs_var);  // 11

     console.log(javascript_var);  // 22

}          

console.log(let_vs_var); // 11

console.log(javascript_var); // 2  

Enjoy....!!!!

Feel free to comment below your experience with above approach and If you still find any problem  with above steps Let me know I would love to help you to resolve your  problem.

 If you want to take your Technological Knowledge to the Next Level and For More Technological information Stay tuned to Visionfortech

How to prevent remote access of mognodb in windows and linux

How to prevent remote access of mognodb in windows and linux

This post is all about how to protect your mongodb server of windows or linux from ransomeware attack or we can say mongodb server is accessible only to localhost or 127.0.0.1 not any other location.

Please see the below solution.Hope you will get what you want..!!

Into Linux :

To prevent access of mongodb remotelly user need to do some changes into mongodb.conf.So to do that follow below steps.

1) Open /etc/mongodb.conf using below command
vi /etc/mongodb.conf

2) find all the statments which starts with bind_ip.Delete all the sentences.

3) Enter Below statment only into /etc/mongodb.conf
bind_ip=12.0.0.1

        If user want to give access to some machines or we can say some IP addresses for Application so user can do that by adding below line to /etc/mongod.conf

   bind_ip=<ip address>

Into windows :

1) First of all user need to turn of the firewall.

2) To turn on the firewall follow below steps.

*) Open command prompt using CMD

*) Run Below command and if it has 0.0.0.0:27017 then any user can access your database remotely.
netstat -a

*) Open Run using start -> Run or press windows+R Key to open Run prompt.

*) Write control into run prompt

*) It will open Control Pannel.

*) Go to Network and sharing center

*) Go to Windows firewall

*) Go to Advance setting

          
*) Select Inbound Rules and delete all rules related to Mongodb

                *) Again please apply below command to command prompt and you won't be able to see 0.0.0.0:27017 instead of that you will see 127.0.0.1:27017

So now no outside user will be able to access your mongodb server.Because you have grant access of only localhost or 127.0.0.1 to mongodb server only.


Cheers...!!!Enjoy...!!!

Feel free to comment below your experience with above approach and If you still find any problem with above steps Let me know I would love to help you to resolve your  problem.

If you want to take your Technological Knowledge to the Next Level and For More Technological information Stay tuned to Visionfortech

[Violation] Added non-passive event listener to a scroll-blocking 'touchstart' event. Consider marking event handler as 'passive' to make the page more responsive

[Violation] Added non-passive event listener to a scroll-blocking 'touchstart' event. Consider marking event handler as 'passive' to make the page more responsive

In this post i am going to discuss about how to remove a jQuery violation "[Violation] Added non-passive event listener to a scroll-blocking 'touchstart' event. Consider marking event handler as 'passive' to make the page more responsive" from chrome console.

Basically user can remove all the basic violations from just changing the type of log from "Verbose" to "Info".

But what if  user want to remove above violation from "Verbose" log Type ?
So Here is the solution.

Problem : 

The above violation is occur because e.preventDefault() will take much time and it will not passively load the component or we can say DOM.

So to remove that please include below code on top of your javascript.

Solution :

{passive:true}

// To remove the violation jquery-1.11.1.min.js:3 [Violation] Added non-passive event listener to a scroll-blocking 'touchstart' event. Consider marking event handler as 'passive' to make the page more responsive.

jQuery.event.special.touchstart = {
setup: function( _, ns, handle ){
if ( ns.includes("noPreventDefault") ) {
 this.addEventListener("touchstart", handle, { passive: false });
} else {
this.addEventListener("touchstart", handle, { passive: true });
}
}
};

//Remove violation code End

So now run your application and above violation will not appear into console.

Cheers...!!!Enjoy...!!!

Feel free to comment below your experience with above approach and If you still find any problem with above steps Let me know I would love to help you to resolve your  problem.

I hope this article will provided you solution which you are expecting. If you want to know more about Javascript check out Javascript World

If you want to take your Technological Knowledge to the Next Level and For More Technological information Stay tuned to Visionfortech

Bootstrap search drop down list or list with [X] sign to delete element

Bootstrap search drop down list or list with [X] sign to delete element

This post is all about a bootstrap drop down list or we can say list with search facility and delete functionality as well.

In this post we are going to discuss about a component through which user can show the drop down list with search option facility and user also can delete the option by just clicking [X] sign besides every option.So let's see how we can achieve that.

You can see the live demo at live Demo

Jsfiddle URL : JSfiddle URL

The HTML file : 

<div class="dropdown" id="option_dropdown" style="margin-left:20px;margin-top:20px">

<button class="btn btn-default btn-filter" type="button" id="savedSearch"

data-toggle="dropdown" data-submenu="" aria-expanded="true" style="word-wrap: break-word;">

<span id="filter_name">Select Language</span> <span class="caret"></span>

</button>

<ul class="dropdown-menu" id="saveLanguageUl"  style="max-height: 315px;overflow-y: scroll;"></ul>

</div>

JavaScript code :

var pra="";
$(document).ready(function(){
$("#saveLanguageUl").empty();
   $("#saveLanguageUl").append('<li><input type="text" id="searchFilterInput" placeholder="Search for Language.." title="Type in to searcg language option">');
   var language = ["angular.js","backbone.js","d3.js","amcharts.js","jquery.js","bootstrap.js","bootstrap.css","mabbox.js","python","php","ajax","json"];
 
    for(var i = 0; i < language.length; i++) {
 $("#saveLanguageUl").append('<li style="height:35px" class=""><a href="#" id='+language[i]+' name='+language[i]+'>'+language[i]  +'   <span style="float:right;" class="deleteOptionBtn glyphicon glyphicon-remove-sign" title="Delete '+language[i]+' Filter" id='+language[i]+'><br></span></a></li>');
}
       
          $("#searchFilterInput").keyup(function(){
    var input, filter, ul, li, a, i;
    input = document.getElementById("searchFilterInput");
    filter = input.value.toUpperCase();
    ul = document.getElementById("saveLanguageUl");
    li = ul.getElementsByTagName("li");
    for (i = 1; i < li.length; i++) {
                  a = li[i].getElementsByTagName("a")[0];
                  if(a.name.toUpperCase().includes(filter)){
                      li[i].style.display = "";
                  } else {
                      li[i].style.display = "none";

                  }
              }
          })
$('#saveLanguageUl').on('click', 'a', function(event) {
var filterId = $(this).attr('id');
if(pra == ""){
$("#filter_name").text(filterId);
}
else{
pra = "";
}
$("#searchFilterInput").val("");
});

$('#saveLanguageUl').on('click', 'span.deleteOptionBtn', function(event) {

var filterId = $(this).attr('id');
pra = filterId;
$(this).closest("li").remove();
$("#filter_name").text($(this).attr('Select Saved Filter'));
});
});

CSS Code:

#searchFilterInput {
  background-position: 10px 12px;
  background-repeat: no-repeat;
  width: 100%;
  font-size: 16px;
  padding: 12px 20px 12px 15px;
  border: 1px solid #ddd;
  margin-bottom: 12px;
}

You can see the live demo at live Demo
Jsfiddle URL : JSfiddle URL

Enjoy...!!!

Feel free to comment below your experience with above approach and If you still find any problem with above steps Let me know I would love to help you to resolve your  problem.

If you want to take your Technological Knowledge to the Next Level and For More Technological information Stay tuned to Visionfortech

Create E commerce website for free in 5 minutes | (with images)

Create E commerce website for free in 5 minutes | (with images)

Into this post i am going to discuss some points which are:
How to create  E commerce website in 5 minutes.
How user can create a E commerce website in which he/she can do whatever he/she want.
How easy to create a website in few minutes.

By using opencart you can create your create your own website and most important thing that it is open-source.

So by following below steps you can create a E Commerce website in few minutes.
1) Install any server(apache,tomcat) and Mysql database.
     You can install both of this by just installing one software which is XAMPP.
     You can download XAMPP from Here.

2) Download the opencart from Here.or you can download that from opencart website as well.

3) Put that downloaded files into the root directory of your server here i user XAMPP so iam going to put all the files into htdocs directory which is root directory for apache in xampp.

4) Rename config-dist.php to config.php and admin/config-dist.php to admin/config.php.

5) Open xampp control pannel and start the apache as well as mysql server.

6) Then open you browser and type http://localhost:opencart and you can see the below screen

7) Press Continue.

8) Press Continue.

Enter username and password for your admin pannel using which you can manage your shopping website and do everything what you want.

9) After that delete the install folder from your root directory of server.

Enjoy....!!!!

Feel free to comment below your experience with above approach and If you still find any problem  with above steps Let me know I would love to help you to resolve your  problem.

 If you want to take your Technological Knowledge to the Next Level and For More Technological information Stay tuned to Visionfortech

[Solution] Week 6 :Aggregation Framework : M101P: MongoDB for Developers

 [Solution] Week 5 :Aggregation Framework : M101P: MongoDB for Developers

Homework 6.1 : 

Which of the following statements are true about replication in MongoDB? Check all that apply.

Solution : 

Homework 6.2 : 

Let's suppose you have a five member replica set and want to assure that writes are committed to the journal and are acknowledged by at least 3 nodes before you proceed forward. What would be the appropriate settings for w and j?

Solution : 

Homework 6.3 :

Which of the following statements are true about choosing and using a shard key?

Solution : 

Homework 6.4 :

You have a sharded system with three shards and have sharded the collections "students" in the "school" database across those shards. The output of sh.status() when connected to mongos looks like this:

mongos> sh.status()
--- Sharding Status ---
  sharding version: {
    "_id" : 1,
    "minCompatibleVersion" : 5,
    "currentVersion" : 6,
    "clusterId" : ObjectId("5531512ac723271f602db407")
}
  shards:
    {  "_id" : "s0",  "host" : "s0/localhost:37017,localhost:37018,localhost:37019" }
    {  "_id" : "s1",  "host" : "s1/localhost:47017,localhost:47018,localhost:47019" }
    {  "_id" : "s2",  "host" : "s2/localhost:57017,localhost:57018,localhost:57019" }
  balancer:
    Currently enabled:  yes
    Currently running:  yes
        Balancer lock taken at Fri Apr 17 2015 14:32:02 GMT-0400 (EDT) by education-iMac-2.local:27017:1429295401:16807:Balancer:1622650073
    Collections with active migrations:
        school.students started at Fri Apr 17 2015 14:32:03 GMT-0400 (EDT)
    Failed balancer rounds in last 5 attempts:  0
    Migration Results for the last 24 hours:
        2 : Success
        1 : Failed with error 'migration already in progress', from s0 to s1
  databases:
    {  "_id" : "admin",  "partitioned" : false,  "primary" : "config" }
    {  "_id" : "school",  "partitioned" : true,  "primary" : "s0" }
        school.students
            shard key: { "student_id" : 1 }
            chunks:
                s0  1
                s1  3
                s2  1
            { "student_id" : { "$minKey" : 1 } } -->> { "student_id" : 0 } on : s2 Timestamp(3, 0)
            { "student_id" : 0 } -->> { "student_id" : 2 } on : s0 Timestamp(3, 1)
            { "student_id" : 2 } -->> { "student_id" : 3497 } on : s1 Timestamp(3, 2)
            { "student_id" : 3497 } -->> { "student_id" : 7778 } on : s1 Timestamp(3, 3)
            { "student_id" : 7778 } -->> { "student_id" : { "$maxKey" : 1 } } on : s1 Timestamp(3, 4)

Solution :298016

Homework 6.5 :

In this homework you will build a small replica set on your own computer. We will check that it works with validate.py, which you should download from the Download Handout link.
Create three directories for the three mongod processes. On unix, this could be done as follows:

mkdir -p /data/rs1 /data/rs2 /data/rs3

Now start three mongo instances as follows. Note that are three commands. The browser is probably wrapping them visually.

mongod --replSet m101 --logpath "1.log" --dbpath /data/rs1 --port 27017 --smallfiles --oplogSize 64 --fork

mongod --replSet m101 --logpath "2.log" --dbpath /data/rs2 --port 27018 --smallfiles --oplogSize 64 --fork

mongod --replSet m101 --logpath "3.log" --dbpath /data/rs3 --port 27019 --smallfiles --oplogSize 64 --fork

Windows users: Omit -p from mkdir. Also omit --fork and use start mongod with Windows compatible paths (i.e. backslashes "\") for the --dbpath argument (e.g; C:\data\rs1).
Now connect to a mongo shell and make sure it comes up

Now you will create the replica set. Type the following commands into the mongo shell:

config = { _id: "m101", members:[
          { _id : 0, host : "localhost:27017"},
          { _id : 1, host : "localhost:27018"},
          { _id : 2, host : "localhost:27019"} ]
};
rs.initiate(config);

At this point, the replica set should be coming up. You can type

rs.status()

to see the state of replication.
Now run validate.py to confirm that it works.

python validate.py

Validate connects to your local replica set and checks that it has three nodes. It has been tested under Pymongo 2.3 and 2.4. Type the validation code below.


Solution : kjvjkl3290mf0m20f2kjjv

Enjoy....!!!!

Feel free to comment below your experience with above approach and If you still find any problem  with above steps Let me know I would love to help you to resolve your  problem.

 If you want to take your Technological Knowledge to the Next Level and For More Technological information Stay tuned to Visionfortech

Related Post :

- Week 1 : Introduction : M101P: MongoDB for Developers

- Week 2 : CRUD : M101P: MongoDB for Developers

- Week 3 : Schema Design : M101P: MongoDB for Developers

- Week 4 : Performance : M101P: MongoDB for Developers

- Week 5 : Aggregation Framework: M101P: MongoDB for Developers

- Week 6 :Aggregation Framework : M101P: MongoDB for Developers