peter_coop
asked on
error in ajax function
hi
when i run the following code i get an error of:
$("#addform").ajaxForm is not a function. because i am only just learning this language could someone show me why this is causing error. is it because i am not calling the form id directly but the div? many thanks
when i run the following code i get an error of:
$("#addform").ajaxForm is not a function. because i am only just learning this language could someone show me why this is causing error. is it because i am not calling the form id directly but the div? many thanks
<script type="text/javascript">
function add() {
// prepare the form when the DOM is ready
$(document).ready(function() {
// bind form using ajaxForm
$('#addform').ajaxForm({
// target identifies the element(s) to update with the server response
target: '#htmlExampleTarget',
// success identifies the function to invoke when the server response
// has been received; here we apply a fade-in effect to the new content
success: function() {
$('#htmlExampleTarget').fadeIn('slow');
}
});
});}
form
=================================================
<div id="addform" style="display:none;">
<form action="filesadd.php" method="post" class="webform">
<fieldset>
<legend><span>Enter a File</span></legend>
<label for="box">Select a Box</label>
<select name="box">
<option SELECTED VALUE="">Select a Box</option>
<?php
do {
?>
<option value="<?php echo $row_Recordset1['boxref']?>"><?php echo $row_Recordset1['boxref']?></option>
<?php
} while ($row_Recordset1 = mysql_fetch_assoc($Recordset1));
$rows = mysql_num_rows($Recordset1);
if($rows > 0) {
mysql_data_seek($Recordset1, 0);
$row_Recordset1 = mysql_fetch_assoc($Recordset1);
}
?>
</select>
<label for="fileno">File Number:</label>
<input id="fileno" name="fileno" class="text" type="text" />
<div id="htmlExampleTarget"></div>
<label for="authorised">Authorised By:</label>
<input name="authorised" type="text" class="text" id="authorised" value="<?php echo $_SESSION['kt_name_usr']; ?>" />
</fieldset>
<input class="submit" type="submit" name="submit" value="Add File" />
</form>
</div>
ASKER
hi leak. this is what i have and it is showing ok in firebug net.
<script type="text/javascript" src="/sample/admin/js/jque ry.form.js "></script >
<script type="text/javascript" src="/sample/admin/js/jque
Replace by the following and remove one after one to check when it stop to run
<link rel="stylesheet" href="http://ajax.googleapis.com/ajax/libs/jqueryui/1.7.2/themes/black-tie/jquery-ui.css" type="text/css">
<link rel="stylesheet" href="../themes/jquery-ui-sub-tabs-smoothness.css" type="text/css">
<link rel="stylesheet" href="../jq-new.css" type="text/css">
<style type="text/css">
form { background: #DFEFFC; border: 5px solid #c5dbec; margin: 10px 0; padding: 20px }
</style>
<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.4.2/jquery.min.js"></script>
<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jqueryui/1.7.2/jquery-ui.min.js"></script>
<script type="text/javascript" src="http://malsup.github.com/chili-1.7.pack.js"></script>
<script type="text/javascript" src="../malsup-tabs-init.js"></script>
<script type="text/javascript" src="http://github.com/malsup/form/raw/master/jquery.form.js?v2.43"></script>-->
<script type="text/javascript" src="http://jquery.malsup.com/form/download/jquery.form.js?v2.43"></script>
<!--
<script type="text/javascript" src="jquery.form.js"></script>
-->
<script type="text/javascript" src="malsup-form-demo.js"></script>
ASKER CERTIFIED SOLUTION
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ASKER
leak. got it working of sorts. however the success callback is being shown in the page that the form sends to addfile.php and not the div <div id="htmlExampleTarget"></d iv> in theform? ie,
ASKER
here is the php code for filesadd.php
$sql = "SELECT custref FROM files WHERE custref = '$fileno' ";
$result = runSQL($sql) or die(mysql_error());
if (mysql_num_rows($result)>0){
echo '<div style="background-color:#ffa; padding:20px">' . $fileno . 'Is already in the database' . '</div>';
}else{
//insert into db;
echo '<div style="background-color:#ffa; padding:20px">' . $fileno . 'Successfull' . '</div>';
ASKER
thank you
You're welcome! Thanks for the points!
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