JavaScript and client side scripting.
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cjkeane
Forum Contributor
Posts: 217 Joined: Fri Jun 11, 2010 1:17 pm
Post
by cjkeane » Mon Dec 19, 2011 9:38 pm
Hi everyone,
I need to select a companyname from a dropdownlist which in turn will populate the rest of the form fields. I'm not sure what I'm doing wrong.
Code: Select all
<script type="text/javascript">
var compInfoArray = new Array();
<?php
$funderquery = "SELECT * FROM icbcagencylookup ORDER BY FunderName";
$funderresult = mysql_query($funderquery) or die(mysql_error());
// build javascript array
while($funderrow=mysql_fetch_array($funderresult)){
echo 'compInfoArray['.$funderrow['FunderName'].'] = new Array();';
echo 'compInfoArray['.$funderrow['FunderName'].']["FunderTitle"] = "'.$funderrow['FunderTitle'].'";';
echo 'compInfoArray['.$funderrow['FunderName'].']["AgencyName"] = "'.$funderrow['AgencyName'].'";';
}
?>
function showname() {
var FunderName = document.form1.FunderName.value;
document.form1.FunderTitle.value = compInfoArray[FunderName]["FunderTitle"];
document.form1.FunderAgency.value = compInfoArray[FunderName]["AgencyName"];
}
window.onload=function() {
showname();
}
</script>
and the dropdownlist code:
Code: Select all
<?php
$result=mysql_query("SELECT FunderName FROM icbcagencylookup order by FunderName ASC");
$options = '';
while ($row = mysql_fetch_assoc($result)) {
$selected = ($row['FunderName']==$FunderName) ? ' selected="selected"' : '';
$options .= "<option value=\"{$row['FunderName']}\"{$selected}>{$row['FunderName']}</option>\n";
}
?>
<select name="FunderName" id="FunderName" onchange="showname()" >
<option value="">< select > <?php echo $options ?></option>
</select>
the dropdown list finds the companynames (or rather the funders) but nothing happens when the name is changed in the list. Can anyone offer any ideas? Thanks.
Gopesh
Forum Contributor
Posts: 143 Joined: Fri Dec 24, 2010 12:48 am
Location: India
Post
by Gopesh » Mon Dec 19, 2011 10:25 pm
HI,Pls post the code for other form fields also..
cjkeane
Forum Contributor
Posts: 217 Joined: Fri Jun 11, 2010 1:17 pm
Post
by cjkeane » Mon Dec 19, 2011 10:53 pm
here's the entire form. I intend to populate all fields once the fundername is chosen from the list.
Code: Select all
<table width="100%" border="0" cellspacing="0" cellpadding="0">
<tr>
<td width="50%"><span class="text-align">Name</span>
<?php
$result=mysql_query("SELECT FunderName FROM icbcagencylookup order by FunderName ASC");
$options = '';
while ($row = mysql_fetch_assoc($result)) {
$selected = ($row['FunderName']==$FunderName) ? ' selected="selected"' : '';
$options .= "<option value=\"{$row['FunderName']}\"{$selected}>{$row['FunderName']}</option>\n";
}
?>
<select name="FunderName" id="FunderName" onchange="showname()" >
<option value="">< select > <?php echo $options ?></option>
</select>
</td>
<td><span class="text-align">Address</span>
<input name="Funder_Address1" type="text" id="Funder_Address1" value="<?php echo $Funder_Address1; ?>" size="26" /></td>
</tr>
<tr>
<td><span class="text-align">Title</span>
<input name="FunderTitle" type="text" id="FunderTitle" size="26" />
</td>
<td><span class="text-align">City</span> <input name="Funder_City" type="text" id="Funder_City" value="<?php echo $Funder_City; ?>" size="26" />
</td>
</tr>
<tr>
<td><span class="text-align">Agency</span>
<input name="FunderAgency" type="text" id="FunderAgency" size="26" />
</td>
<td><span class="text-align">Province</span>
<input name="Funder_Province" type="text" id="Funder_Province" value="<?php echo $Funder_Province; ?>" size="26" /></td>
</tr>
<tr>
<td> </td>
<td><span class="text-align">Postal Code</span>
<input name="Funder_PostalCode" type="text" id="Funder_PostalCode" value="<?php echo $Funder_PostalCode; ?>" size="12" /></td>
</tr>
</table>
cjkeane
Forum Contributor
Posts: 217 Joined: Fri Jun 11, 2010 1:17 pm
Post
by cjkeane » Wed Dec 21, 2011 11:21 am
Thanks for the reply. I 'can' use the id, however I'd prefer to use the companyname (or rather fundername) as the key as it is unique. i've had a look at the link you provided. i think it might accomplish what i need to do. I'll let you know if i still have difficulty.
thanks.
Gopesh
Forum Contributor
Posts: 143 Joined: Fri Dec 24, 2010 12:48 am
Location: India
Post
by Gopesh » Wed Dec 21, 2011 10:30 pm
ok...Try it..Good Luck