Jquery show/hide element based on mysql selection

Question

I wanted to show/hide element based on MySql Value

$(document).bind('DOMNodeInserted', '#composeHeaderTable0', function(event) {
    if ($('#data_parent_type0').val() == 'Leads') {
    $("#email_template0 option[value='151ddf5a-3fe8-84bd-50db-533545893c67']").hide();
    $("#email_template0 option[value='15a6040b-bbad-1a89-d62b-537de2b73832']").show();
}
    if ($('#data_parent_type0').val() == 'Contacts') {
    $("#email_template0 option[value='1b560788-07b2-0798-6b09-5335520bd9df']").hide();
    $("#email_template0 option[value='f15bf4b4-d174-f0d6-6f09-537c8d3e9693']").show();

}
    return false;
});

Above script works, but I need to show hide based on Mysql call: I have partial php corresponding file

<?php
    mysql_connect('mysql', 'admin', '123');
    mysql_select_db('mydb');
    $Leads0emailAddress0 = mysql_real_escape_string($_POST['Leads0emailAddress0']);
    $result = mysql_query('select id from email_templates where description = 'Customers' and deleted = 0;');
...............
?>

description = 'Customers' should be description = "Customers" (with double quotes) since you are using the single quotes as surrounding quotes for query. — Mihai Iorga
– Mihai Iorga, Commented Nov 19, 2014 at 6:31

cyberrspiritt · Accepted Answer · 2014-11-19 06:02:43Z

1

Use a hidden form element, set its value to the result obtained from mysql. Assign a specific ID to that form element. From jQuery, refer that form element using the ID. That should do the trick.

answered Nov 19, 2014 at 6:02

cyberrspiritt

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2 Comments

asvasdv asdvsadvs Over a year ago

Yes, I know. I don't know how to get the MySQL output into the: $("#email_template0 option[value='value from MySQL']").hide();

cyberrspiritt Over a year ago

Try using value='<?php echo $result ?>'

Saqueib · Accepted Answer · 2014-11-19 06:29:54Z

0

Your mysql script has error

    $result = mysql_query('select id from email_templates where description = 'Customers' and deleted = 0;');

change it to

    $result = mysql_query(
              "SELECT id FROM 
                  email_templates 
                WHERE 
                  description = 'Customers' 
                AND deleted = 0"
             );

To list the result

$options = '';
while($row = mysql_fetch_assoc($result)) {
       $options .= '<option value="'.$row['id'].'">'.$row['id'].'</option>' . "\n";   
}

//display the options list in html where you want
<select id="email_template0">
   <?php echo $options; ?>
</select>

now from jQuery handle event on dropdown change

 $(function() {
 $("#email_template0").change(function() {
    alert( $('option:selected', this).val() );
    //do your hide and select here 
    });
 });

edited Nov 19, 2014 at 6:29

answered Nov 19, 2014 at 6:03

Saqueib

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4 Comments

asvasdv asdvsadvs Over a year ago

Great, thanx, that's the php part. What would be the jquery for the same? Keeping in mind there are 2 queries for Contacts and Leads.

asvasdv asdvsadvs Over a year ago

I'm still not able to get it. This is the php <?php $con = mysql_connect('mysql', 'admin', 'admin'); mysql_select_db('suitecrm'); $data_parent_type0 = mysql_real_escape_string($_POST['data_parent_type0']); $result = mysql_query('SELECT id FROM email_templates WHERE description = "'.$data_parent_type0.'" AND deleted = 0'); $options = ''; while($row = mysql_fetch_assoc($result)) { $options .= ''.$row['id'].'' . "\n"; echo $options; } mysql_close($con); ?>

asvasdv asdvsadvs Over a year ago

And this is the jquery $(document).bind('DOMNodeInserted', '#composeHeaderTable0', function(event) { $("#data_parent_type0").change(function() { var data_parent_type0 = $('#data_parent_type0').val(); $.post("etplate_dbcall.php", { data_parent_type0: data_parent_type0},'html'); }); $("#email_template0 option[value='$options']").show(); });

asvasdv asdvsadvs Over a year ago

The PHP returns the correct 'id', but the jqyuery is not re-posting

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