pass id from php while loop to jquery

Question

i have a form that has select boxes. this form has a table and each row has been created by php while loop from database. for example

<select name="1265483" id="1265483" class="form-control">
<option value="1">1</option>
<option value="2">2</option>
<option value="3">3</option>


<select name="5894253" id="5894253" class="form-control">
<option value="1">1</option>
<option value="2">2</option>
<option value="3">3</option>

the id of each select is that entry's id in database. i want to get those ids to make an onchange function then send that id to ajax.

Is it necessary for you to select your select elements via their ids, or would it also be possible to select them via a more general selector? — Fiona Runge
– Fiona Runge, Commented Jul 8, 2015 at 11:16
no necessity to select via their ids. i just want to send the id to ajax process. — reza
– reza, Commented Jul 8, 2015 at 11:27

Praveen Kumar Purushothaman · Accepted Answer · 2015-07-08 11:15:56Z

2

You can use something like:

$("select").change(function () {
    $this = $(this);
    $.post("some/url/dot.php", {data: $this.attr("id")}, function () {
        // code...
    });
});

answered Jul 8, 2015 at 11:15

Praveen Kumar Purushothaman

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Comments

dp4solve · Accepted Answer · 2015-07-08 11:20:19Z

0

you can do by below code

      var your_selected_value = $('#5894253 option:selected').val();

            $.ajax({
              type: "POST",
              url: "your_url",
              data: {selected: your_selected_value},
              success: function(data) {
                // Stuff
              },
              error: function(data) {
                // Stuff
              }
            });

put this code in onchange function

answered Jul 8, 2015 at 11:20

dp4solve

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2 Comments

reza Over a year ago

thanks for your answer. but that id has been randomly generated via php. i cannot write this code with specific id.

dp4solve Over a year ago

than you can use $('.form-control option:selected').val();

Vijay Porwal · Accepted Answer · 2015-07-08 11:49:24Z

0

Try using the select class name:

$("select.form-control").change(function (e) {
    var selectId = $(this).attr('id'),
    ajaxUrl = "demo_test.php?id=" + selectId;
    $.ajax({
        url: ajaxUrl,
        success: function (result) {
            if (result) {
                //success code 
            }
        }
    });
});

answered Jul 8, 2015 at 11:49

Vijay Porwal

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