Jump to content

[SOLVED] Parse Error


michaelk46

Recommended Posts

Can't figure this out...

 

I keep getting this error:

 

 

"Parse error: parse error, expecting `T_STRING' or `T_VARIABLE' or `T_NUM_STRING' in C:\wamp\www\Work\Storage\index.php on line 36"

 

 

Line 36 is $sql .= "'searchid' = ($_POST['searchid'])";

 

 

Anyone have any ideas why?

 

if ($_SERVER['REQUEST_METHOD'] != "POST")
{
	include "search.html.php";
}
else
{
	if(isset($_POST['searchid']) || (isset($_POST['pagename'])))
				{
					$sql = "SELECT * FROM `pages` WHERE ";
				}

			if(isset($_POST['searchid']))
				{
					$sql .= "'searchid' = ($_POST['searchid'])";
				}

			if(isset($_POST['searchid']),(isset($_POST['pagename']))
				{
					$sql .= " and ";
				}

			if(isset($_POST['pagename']))
				{
					$sql .= "'pagename' = ($_POST['pagename']) ";
				}


	else{
			$sql .= "You must enter search terms";


		}
	include "searchres.html.php";
}

Link to comment
https://forums.phpfreaks.com/topic/179548-solved-parse-error/
Share on other sites

Found 3 syntax errors (see comments)

<?php
if ($_SERVER['REQUEST_METHOD'] != "POST")
{
include "search.html.php";
}else{
if(isset($_POST['searchid']) || (isset($_POST['pagename'])))
{
	$sql = "SELECT * FROM `pages` WHERE ";
}

if(isset($_POST['searchid']))
{
	$sql .= "'searchid' = (".$_POST['searchid'].")"; //ADD ".."
}

if(isset($_POST['searchid']) && (isset($_POST['pagename']))) //ADDED )
{
	$sql .= " and ";
}

if(isset($_POST['pagename']))
{
	$sql .= "'pagename' = (".$_POST['pagename'].") ";//ADD ".."
}else{
	$sql .= "You must enter search terms";
}
include "searchres.html.php";
}
?>

 

EDIT: cleaned up formatting

Link to comment
https://forums.phpfreaks.com/topic/179548-solved-parse-error/#findComment-947449
Share on other sites

Archived

This topic is now archived and is closed to further replies.

×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.