Re: problems with looping array

From: Date: Mon, 03 Dec 2001 09:30:44 +0000
Subject: Re: problems with looping array
References: 1 2  Groups: php.general 
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I tried but nothing different happened. I'm still playing though and I'll see if I can make it work. I'm new to all of this...so we'll see how it goes. Thanks for your help. -jordan Fred <fred.nospam.please@querymagic.com> wrote in message news:20011203090233.1958.qmail@pb1.pair.com... > Strings in your sql query must be quoted. Try this instead: > > > $result = mysql_query("SELECT * FROM ar > > LEFT JOIN company on ar.company_id=company.company_id > > LEFT JOIN scale on ar.scale_id=scale.scale_id > > WHERE item_number = \"$item[$index]\")"; > > Fred > > Jordan <newimagedesign@newimagedesign.com> wrote in message > news:20011203081812.81073.qmail@pb1.pair.com... > > here's the situation. I'm making a shopping cart, when an item is added > > it's item number is added into an array. I then want this array's items > to > > be compared agains the database. Unfortunately I haven't been able to use > a > > variable in the query string of MySQL. something along the lines of.... > > > > > > <?PHP > > > > include ('connect.php'); > > > > $result = mysql_query("SELECT * FROM ar > > LEFT JOIN company on ar.company_id=company.company_id > > LEFT JOIN scale on ar.scale_id=scale.scale_id > > WHERE item_number =" . $item[$index]); > > > > ?> > > > > where $item is the array of item numbers and $index is an auto > incrementing > > number that increments at the end of the loop. Please help if you know of > > anyway to use this variable. Also, I can post my whole script if > needed...I > > just didn't want unneeded info posted. Thanks in advance. > > > > -Jordan > > > > > >

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