Cannot Integrate Jquery Variable Into Php

I try to make a query from database to get the list of users based on their country using php/mysql and jquery. I have a mysql query that extracts the countries form database in a

Solution 1:

What you are trying to do is called AJAX. Sounds complicated, but it really isn't. See these examples for a simplistic explanation. Do not just look at them -- copy/paste to your server and make them work. Change the values. See how it works - really very simple.

A simple example

More complicated example

Populate dropdown 2 based on selection in dropdown 1

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Your code is a bit difficult for me to follow, but should be refactored something like this. (I am unsure where strTara figures in the code, but I'm sure you will be able to figure it out from here).

javascript/jQuery:

var strTara = <?phpecho _('Tara');?>

$( "#tara" ).change(function () {
    selVal = $(this).val();
    $.ajax({
        type: 'post',
         url: 'another_php_file.php',
        data: 'variabilatara=' + selVal,
        success: function(data){
            var tblHead = "
                <table class='table table-bordered table-striped'>
                    <thead><tr><th><p>strTara</p></th></tr></thead><tbody>
            ";
            $( "#testare" ).html( tblHead + data );
        }
    });
});

another_php_file.php: (your PHP AJAX processor file)

<?php$var = $_POST['variabilatara'];
    $out = '';

    $result = mysql_query("SELECT * FROM utilizatori WHERE idt='$variabilatara'") ordie(mysql_error());
    while($row=mysql_fetch_object($result)){
        $out .= "<tr>";
        $out .= "<td><p>" .$row->nume. "</p></td>"; //<== fixed$out .= "</tr>";
    }
    $out .= '</tbody></table>'; //<== fixedecho$out;
?>