On select value not being passed to php

[DigiMartKe]

I have the following code for my dropbox;

 

<select name="Symptom" id="Symptomid"  onchange="LSC(this.value)"> 

 

Its options are;

 

  <option value="<?php echo $row_RsSymptom['name']?>"><?php echo $row_RsSymptom['name']?></option>

 

i have the script ;

 

<script>

function LSC(str) {

     if (str == "") {

        document.getElementById("txtHint").innerHTML = "";

        return;

    } else { 

        if (window.XMLHttpRequest) {

            // code for IE7+, Firefox, Chrome, Opera, Safari

            xmlhttp = new XMLHttpRequest();

        } else {

            // code for IE6, IE5

            xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");

        }

        xmlhttp.onreadystatechange = function() {

            if (xmlhttp.readyState == 4 && xmlhttp.status == 200) {

                document.getElementById("txtHint").innerHTML = xmlhttp.responseText;

            }

        }

        xmlhttp.open("GET","LoadSymptomDetails.php?q="+str,true);

        xmlhttp.send();

    }

}

</script>

 

 

The LoadSymptomDetails.php

 

has the following lines;

 

$sql1 = mysql_query("SELECT name,description , comments

FROM symptoms WHERE symptoms.name ='".$q."'") or die(mysql_error());

 

 

The q is not being picked. If i remove the where statement it runs and displays the table.

 

 

Kindly assist.

 

Azhar

 

 

 

[Muddy_Funster]

are you assigning a value to $q?

when passing values through form input (either by ajax or normal form submission) the value is normally passed to either the $_GET or $_POST superglobals.  To get the value out of that you would access it like any other associative array:

 

$q = $_GET['q'];

[Ch0cu3r]

Why start a new topic? You already have a topic with the same question here.

 

Please do not post duplicate topics, carry on where you left off.

 

Locked.