How to .js file $ajax code defined '$ not defined error''?

Question

var deneme = document.createElement("SCRIPT");
deneme.src = "https://ajax.googleapis.com/ajax/libs/jquery/1.12.4/jquery.min.js";
var nodeajax = document.createTextNode("function deneme() {console.log('dawdawdawd');$.ajax({ type: \"POST\", url: \"http://localhost:54371/checkbox.aspx/Check\", data: '{sayfa: 10 }', contentType: \"application/json; charset=utf-8\", dataType: \"json\", success: function (response) { console.log(response.d); alert('sdf'); }, failure: function (response) { console.log(response); } });}"););
deneme.appendChild(nodeajax);
document.body.appendChild(deneme);

How do I define ajax without using script tag in the .js file?

' $ is not defined.' I get an error.

You're appending your jQuery code (nodeajax) before you've appended the jQuery library. Trying swapping the order of your last 2 lines — Brett Gregson, Jan 27 '20 at 09:06
Either you forgot to add jQuery script in your HTML source or you are trying to call JQuery method before it was added. Keep in mind to add important dependencies on top. — Saharsh, Jan 27 '20 at 09:07
Does this answer your question? [JQuery - $ is not defined](https://stackoverflow.com/questions/2194992/jquery-is-not-defined) — Carl Binalla, Jan 27 '20 at 09:09
Why are you appending code as text like that? I have never seen this practice before — Alon Eitan, Jan 27 '20 at 09:12
@CarlBinalla I don't answer my question, because I don't use a script tag in the .js file. — simge sen, Jan 27 '20 at 09:12
@AlonEitan because I have to check the return value from web form with ajax — simge sen, Jan 27 '20 at 09:13
Try using two `script` element, instead of making the code a child of the `script` that holds the source link — Carl Binalla, Jan 27 '20 at 09:16
@simgesen Are you aware [that you don't need jQuery](https://stackoverflow.com/questions/8567114/how-to-make-an-ajax-call-without-jquery) in order to use AJAX? — Ivar, Jan 27 '20 at 09:20
Instead of appending `nodeajax` to `deneme`, create another one `var deneme_2 = document.createElement("SCRIPT");`, that's where you will append `nodeajax` — Carl Binalla, Jan 27 '20 at 09:21
@simgesen The link I added to my comment has a good explanation on how you can achieve that. — Ivar, Jan 27 '20 at 09:21

Ahmet Zeybek · Accepted Answer · 2020-01-27T11:22:52.340

0

This would be solve your problem

var jqueryScript = document.createElement('script')
var ajaxScript = document.createElement('script')
jqueryScript.src = 'https://code.jquery.com/jquery-3.4.1.min.js'
ajaxScript.innerHTML = `
window.onload = function() {
  $.ajax({
    type: "POST",
    url: "http://localhost:54371/checkbox.aspx/Check",
    data: {sayfa: 10},
    contentType: "application/json; charset=utf-8",
    dataType: "json",
    success: function(response) {
      console.log(response);
      alert('sdf');
    },
    failure: function(response) {
      console.log(response);
    }
  });
}
`
document.body.appendChild(jqueryScript)
document.body.appendChild(ajaxScript)

Here is the codepen link, it is working

edited Jan 27 '20 at 11:22

answered Jan 27 '20 at 09:22

Ahmet Zeybek

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Thanks but the function in the script does not work. It appears when the script is defined but does not work. – simge sen Jan 27 '20 at 10:11
@simgesen I updated, check again – Ahmet Zeybek Jan 27 '20 at 10:49
Thanks! I get bloked CORS access error when I apply the update. – simge sen Jan 27 '20 at 11:07
You need to set `headers` on server side, server is blocking your POST requests. Check codepen link – Ahmet Zeybek Jan 27 '20 at 11:21

How to .js file $ajax code defined '$ not defined error''?

1 Answers1