implode accepts an *array* as input, not a string.
you would need to change your $sqlUpdate to be an array, and each statement
look something like;
if($textfield12 != '') { $sqlUpdate[] = "textfield12='$textfield12'"; }
of course it would probably be easier to just do something like this;
if($image11_name != '') $sqlUpdate .= "image11='$image11_name', ";
if($image12_name != '') $sqlUpdate .= "image12='$image12_name', ";
...
if($textlink11 != '') $sqlUpdate .= "textlink11='$textlink11', ";
$sqlUpdate = substr($sqlUpdate, 0, -2);
(note the commas at the end of each string being concatenated?)
(not tested or anything, but should give you the idea)
HTH
Beau
// -----Original Message-----
// From: Jef [mailto:[EMAIL PROTECTED]]
// Sent: Tuesday, 26 November 2002 2:52 PM
// To: [EMAIL PROTECTED]; [EMAIL PROTECTED]
// Subject: [PHP-DB] implode()
//
//
// I am having difficulty with the implode function. What I am doing is
// building a string for an UPDATE SQL command. At the end of
// the building
// portion of the code, I want to insert a comma between the
// various fields
// that will be updated. However, it is not working. Here is a
// portion of my
// code...thanks, in advance for any assistance received....
//
// $sqlUpdate = "SET ";
// if($image11_name != '') $sqlUpdate .= "image11='$image11_name'";
// if($image12_name != '') $sqlUpdate .= "image12='$image12_name'";
// if($textfield11 != '') $sqlUpdate .= "textfield11='$textfield11'";
// if($textfield12 != '') $sqlUpdate .= "textfield12='$textfield12'";
// if($textlink11 != '') $sqlUpdate .= "textlink11='$textlink11'";
// if($textlink12 != '') $sqlUpdate .= "textlink12='$textlink12'";
//
// $update = implode(",", $sqlUpdate);
// $sql = "UPDATE mp " .$sqlUpdate;
//
// I get a bad parameter message on the line with the call to
// the Implode
// function.
//
//
//
// Questions? Comments? Suggestions?
//
//
//
// Jef
//
//
//
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