Re: implode()

On Tuesday 26 November 2002 14:51, Jef wrote: > 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. implode() expects an array as the second argument. You're passing a string. In fact with the code you have above I don't see why you don't just tack the comma on the end in the first place: if($image11_name != '') $sqlUpdate .= "image11='$image11_name', "; -- Jason Wong -> Gremlins Associates -> www.gremlins.biz Open Source Software Systems Integrators * Web Design & Hosting * Internet & Intranet Applications Development * /* Better tried by twelve than carried by six. -- Jeff Cooper */