Re: implode()

From: Date: Tue, 26 Nov 2002 18:57:34 +0000
Subject: Re: implode()
References: 1 2 3 4  Groups: php.general 
Request: Send a blank email to php-general+get-126147@lists.php.net to get a copy of this message
> > > 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', "; > > > > The reason being the comma at the end of the SQL command will prevent the > > command from working because SQL will expect another entry. However, your > > comment about implode expecting an array helps out. Thanks. > > Yes, but as you're building up the SQL statement manually you just leave out > the comma off the last line! > > Using implode() is appropriate if you're programatically (sp?) building an SQL > statement using an array. The problem is I don't know which will be the last element in the string. I am allowing 9 images to be changed, 9 text labels to be changed, and 9 links to be changed. I don't know which will be the last. I'm using an update sql command to change the database when these changes are being made. When I set the $sqlUpdate to be an array the printed result is ARRAY. When I print the sql command it reads "UPDATE mp SET ARRAY". What am I doing wrong? Jef

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