Passing a string value from javascipt to php

Dev Articles Community Forums Sponsor:

December 11th, 2011, 04:03 PM

siddhanta

Registered User

Join Date: Dec 2011

Posts: 1

Time spent in forums: 38 m 14 sec
Reputation Power: 0

Forms: POST & GET - Passing a string value from javascipt to php

hi,
i have two drop down list.
First one is populated from database and its working fine.
Second one will also be populated from database but as per the value selected from the first drop-down list.

Code:

<head>
<script type="text/javascript">
function xyz_list()
{

// xyz_list is the id of my first drop-down list
  var xyz_list=document.getElementById("mob_list");
  brand=xyz_list.options[mob_list.selectedIndex].text;
 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("xyz").innerHTML=xmlhttp.responseText;
    }
  }
xmlhttp.open("GET","XYZ.php?brand=" + brand,true);
xmlhttp.send();
}
</script>
</head>

<body>
<form>
<select id="xyz" onfocus="xyz_list()">
<OPTION VALUE=All>All 
<?php echo $options?> 
</SELECT> 
</form>
</body>

Below is the xyz.php script

PHP Code:


		
			
<?php

$brand=$_GET['brand'];

//alert($brand);



 //below is just inserting a blank value in drop list

 $options_mobile.="<OPTION VALUE=abc>".$brand.'</option>';

$con = mysql_connect('localhost', 'root', '');

if (!$con)

  {

  die('Could not connect: ' . mysql_error());

  }



mysql_select_db("mobile1", $con);





$sql="SELECT xyz FROM abc where pqr= '$brand'";



$result = mysql_query($sql);

$options.="<OPTION VALUE=\"$result\">".$result.'</option>';

while($row = mysql_fetch_array($result))

  {

 // $xyz=$row["xyz"]; 

  $xyz=$row["xyz"];

   $options.="<OPTION VALUE=\"$xyz\">".$xyz.'</option>';

 }

mysql_close($con);

?>

I am new to both of the languages, but i know wat my code wants to do. i googled a lot to find the error, But i cant figure out where i am wrong.
Help will be highly appreciated.

Thanks

August 27th, 2012, 07:31 AM

ganesan

Registered User

Join Date: Aug 2012

Posts: 1

Time spent in forums: 22 m 50 sec
Reputation Power: 0

PHP Development India

forum posting which is you are giving is really very nice.

August 28th, 2012, 04:35 AM

Frank451

Registered User

Join Date: Aug 2012

Posts: 3

Time spent in forums: 6 m 4 sec
Reputation Power: 0

Have you tried to import the data inside a new database file? If not then try to do it. It could solve your problem.URLURLURL

August 29th, 2012, 03:25 AM

roars1111

Registered User

Join Date: Aug 2012

Posts: 4

Time spent in forums: 43 m 48 sec
Reputation Power: 0

<?php
$appData = array();
if(!empty($signedRequest)&&!empty($signedRequest['app_data'])){
$appData = json_decode($signedRequest['app_data'],true);
}

echo '<pre>'. print_r($appData).'</pre>';
//prints Array ( [lat] => 123 [lon] => 456 ) when fer.php reloads

$params = array(
'lat'=>'123',
'lon'=>'456'
);

$encodedParams = urlencode(json_encode($params));

$tabUrl ==app_433576149993619';
//$tabUrl will open fer.php

$linkUrl = $tabUrl .'&app_data='. $encodedParams;
?>
...
function loopPage()
{
top.location = "<?= $linkUrl ?>";
//reloads fer.php
}

November 19th, 2012, 02:19 AM

lugocottewd

Registered User

Join Date: Nov 2012

Posts: 6

Time spent in forums: 18 m 31 sec
Reputation Power: 0

I know exactly how you can update multiple columns in a row.

Forms: POST & GET - Passing a string value from javascipt to php

Developer Shed Advertisers and Affiliates