Code: Select all
,Code: Select all
and [syntax="..."] tags where appropriate when posting code. Your post has been edited to reflect how we'd like it posted. Please read: [url=http://forums.devnetwork.net/viewtopic.php?t=21171]Posting Code in the Forums[/url] to learn how to do it too.[/color]
Can anyone see anything word with thie query, I keep getting
"Error in query: You have an error in your SQL syntax near 'WHERE....Code: Select all
foreach($HTTP_POST_VARS as $varname => $value)
$formVars[$varname]=$value;
$query = "SELECT
o.org_id,o.web_url,
f.salutation,f.name,f.surname,f.organisation,
f.email,f.address,f.address1,
f.telephonefax,f.mobile,f.mscints,
f.mscactive,f.otheractivities,f.gradjobs,
f.ugproj,f.pdev,f.bcsmem,f.bcspds,f.teach,
f.acconsult,f.person_id,f.org_id
FROM feedbackcontacts f LEFT JOIN organisations o
ON o.org_id = f.org_id
ORDER BY name ASC
WHERE name LIKE '$formVars[name]%'";
$result = mysql_query($query);feyd | Please use
Code: Select all
,Code: Select all
and [syntax="..."] tags where appropriate when posting code. Your post has been edited to reflect how we'd like it posted. Please read: [url=http://forums.devnetwork.net/viewtopic.php?t=21171]Posting Code in the Forums[/url] to learn how to do it too.[/color]