How do I link to a javascript code I'm working on with git?

How do I link to a javascript code I'm working on with git?

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may help you . Here's the solution I'm looking for: codepen.io.
Codepen lets you work on a JS file in-browser and in a sandbox, then use a direct link to that file that another site can use. This is perfect for me, because the site I need to embed the JS in isn't serving millions of visitors.

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Tag : javascript , By : akr
Date : March 29 2020, 07:55 AM
will be helpful for those in need You should familiarize yourself with your browser's JavaScript debugger. To open it in IE or FireFox, hit F12, in Chrome hit Ctrl+Shift+J. Here's what IE's JS debugger displays:
function loadAd(adSpecUrl, containerId)
    var xhr = new XMLHttpRequest();
    xhr.open("GET", adSpecUrl, true);
    xhr.onreadystatechange = function()
        if (xhr.readyState == 4)
            var lines = xhr.responseText.split(/\r\n|\n/);
            var src = lines[0];
            var href = lines[1];
            var w = lines[2];
            var h = lines[3];
            var container = document.getElementById(containerId);
            var link = container.appendChild(document.createElement("a"));
            link.href = href;
            link.target = "_blank";
            var img = link.appendChild(document.createElement("img"));
            img.src = src;
            img.width = w;
            img.height = h;
loadAd("/ads/sm.txt", "smusic");
loadAd("/ads/mm.txt", "mmusic");

Jquery code not working when it's use as link?

Tag : javascript , By : user180941
Date : March 29 2020, 07:55 AM
will help you I wrote a some jquery code and when i use that in tags, it works. , It solved. I replaced
var ddlSorguTalepEdenBirim = $('#<%= ddlSorguTalepEdenBirim.ClientID%>').get(0).selectedIndex;
var ddlSorguTalepEdenBirim = $('#ddlSorguTalepEdenBirim').get(0).selectedIndex;

Tag : javascript , By : Vrki
Date : March 29 2020, 07:55 AM
fixed the issue. Will look into that further I want a simple way to display a picture in html within a google maps infobox from a variable in javascipt. Currently the infobox displays the image link: , try this:
var info_text = address + '<img src="'+picture+'" alt="some alt text" />';

Tag : javascript , By : Pieter Taelman
Date : March 29 2020, 07:55 AM
like below fixes the issue It only works the first time because you are searching for elements with the test class. Then you replace the class with the ones from the function. So there are no elements with test class after the first run, and subsequently nothing is found on the following runs.
Workaround (is to keep the test class, but use it only for selecting and not for styling)
    var style = ['view view-first', 'view view-second', 'view view-third', 'view view-fourth', 'view view-fifth'];

    function changeClass(value) {

      var targetClass = "test",
          elements = document.getElementsByClassName( targetClass );
      for (var i = elements.length - 1; i >= 0; i--) {

        elements[i].className = targetClass + ' ' +  style[value];

        // elements[i] no longer exists past this point, in most browsers
<div class="test">Test #1</div>
<p> irrelevant text </p>
<div class="test">Test #2</div>
<p> irrelevant text </p>
<p> irrelevant text </p>
<div class="test">Test #3</div>
<div class="test">Test #4</div>
<div class="test">Test #5</div>
<p> irrelevant text </p>
<p> irrelevant text </p>
<p> irrelevant text </p>
<div class="test">Test #6</div>
<div class="test">Test #7</div>

<button onclick="changeClass(0);">first (red)</button>
<button onclick="changeClass(1);">second (blue)</button>
<button onclick="changeClass(2);">third (green)</button>
<button onclick="changeClass(3);">fourth (yellow)</button>
<button onclick="changeClass(4);">fifth (orange)</button>
var style = ['view view-first', 'view view-second', 'view view-third', 'view view-fourth', 'view view-fifth'];

function changeClass(value) {
    $('.test').removeClass().addClass( style[value] ).addClass( 'test' );

JAXB. Example of working code identical to broken code. (src in github link)

Tag : java , By : user94076
Date : March 29 2020, 07:55 AM
will be helpful for those in need This has to do with the fact that you're calling the Init() method for both the constructor which takes Working and the constructor which takes Broken.
Inside the Init() method you call JAXBContext.newInstance(Working.class);. When the Init() method is called through the constructor with Broken is called should be JAXBContext.newInstance(Broken.class); which it's not, and that's why it's throwing that error.
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