Re: parse error AFTER end of included file??

From: Date: Tue, 28 Aug 2001 16:02:40 +0000
Subject: Re: parse error AFTER end of included file??
References: 1 2  Groups: php.general 
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Jaxon, do you have a line of white space after your closing tag? Anyway, this looks fishy to me: if (!isset($page_id)) { $sql="select page_id from table where fieldname = $value"; $link_id = mysql_connect($host, $usr, $pass) or die (mysql_error()); mysql_select_db($database, $link_id); //select database catalog $result = mysql_query ($sql, $link_id) or die (mysql_error()); //return result set to php if (!$result) echo "wait - no result set!"; $page_id = mysql_result($result, 0, fieldname); // Where is the closing } for if (!isset($page_id)) { ? // added it below } I think you'd benefit from using braces more often too, so your style is consistent..... so changing if (!$result) echo "wait - no result set!"; to if (!$result) { echo "wait - no result set!"; } might mean that you're able to spot things like this more easily. Just my opinion ;) James "Jaxon" <jaxon@salamander.net> wrote in message news:NDBBKFNCBCHANEGAKONKGEFGJKAA.jaxon@salamander.net... > Hi, > > Can anyone tell me why I have a parse error here? > > I'm including this file, which is 16 lines, but the error being thrown by > the including page reports a parse error in this file on line 17 ???: > <?php > if (!isset($page_type)) > { > $page_type = "foo"; > } > > if (!isset($page_id)) > { > $sql="select page_id from table where fieldname = $value"; > $link_id = mysql_connect($host, $usr, $pass) or die (mysql_error()); > mysql_select_db($database, $link_id); //select database catalog > $result = mysql_query ($sql, $link_id) or die (mysql_error()); > //return result set to php > if (!$result) echo "wait - no result set!"; > > $page_id = mysql_result($result, 0, fieldname); > ?> > > cheers, > jaxon >

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