Parse error: expecting T_VARIABLE or '$'???? 1

pdotd · Aug 2, 2004

hey, i was wondering if someone could tell me what is wrong with my script, its a simple script and i think it looks fine, but whenever i run it ket this errors msg saying:

Parse error: parse error, unexpected ';', expecting T_VARIABLE or '$' in F:\Websites\mrsbsgifthouse\cart\cart.cat.productstest.php on line 28

here is my script:

<html>
<head>
<title> Product List </title>
</head>
<body bgcolor="#ffffff">
<h1>Products</h1>
<table>
<tr>
<?

$y = 7;
$count =0;

for ($x=0; $x <=$y; $x++)
{

?>

<td>
<div align="center">
<a href="localhost">OUR PRICE $18.00</a><br /></a>
</div>
</td>
<?
$count = ++;
echo $count;
if ($y = 3)
{
?>
</tr>
<tr>
<?
$count =0;
}
}
?>
</tr>
</table>
</body>
</html>

sleipnir214 · Aug 2, 2004

Help us help you. When you are relaying a PHP error report, indicate where in your code the error occurs. I think it was this line:

$count = ++;

which should read:

$count++;

but I'm not sure.

Also, switching back and forth between PHP-interpreted mode (everyting between "<?php" and "?>") and HTML-througput mode (everything else) gives PHP a performance hit. You can use multiline print statements for better performance:

Code:

<?php
print '<html>
<head>
<title> Product List </title>
</head>
<body bgcolor="#ffffff">
<h1>Products</h1>
<table>
	<tr>';
        
        $y = 7;
        $count =0;

for ($x=0; $x <=$y; $x++)
{
	print '
		<td>
                <div align="center">
                <a href="localhost">OUR PRICE $18.00</a><br /></a>
                </div>
            </td>';

	$count++;
	echo $count;
	if ($y = 3) 
	{
		print '
                </tr>
                <tr>';
                $count =0;
	}
}

print '
    </tr>
</table>
</body>
</html>';
?>

Want the best answers? Ask the best questions!

TANSTAAFL!!

pdotd · Aug 2, 2004

thanks sleipnir214,

the code now works

i will remember next time to post the line error to better help you answer my questions

pdotd

kenrbnsn · Aug 2, 2004

You also probably want

Code:

            if ($y == 3) 
            {
                ?>
                </tr>
                <tr>
                <?
                $count =0;
            }

instead of

Code:

            if ($y = 3) 
            {
                ?>
                </tr>
                <tr>
                <?
                $count =0;
            }

Notice '==' instead of '=' in your "if" statement. The way you have it now will always be true and will assign '3' to '$y'.

Ken

pdotd · Aug 2, 2004

when i change my code to
if ($y == 3)
{
?>
</tr>
<tr>
<?
$count =0;
}
with my for loop = 3 it does not create a new row? it just skips the if statement

pdotd

sleipnir214 · Aug 2, 2004

pdotd:
Keep one thing in mind: of everyone in this discussion, you are the only one who understands what this code should do.

Want the best answers? Ask the best questions!

TANSTAAFL!!

kenrbnsn · Aug 2, 2004

Looking at your code again... $y will never equal 3. You set it to 7 before your loop. Maybe you want to test the value of $count in your if statement.

Ken

pdotd · Aug 2, 2004

you're right....sorry, that was my silly mistake for not seeing that. (i've been looking at code for too long)

thanks again

pdotd

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Parse error: expecting T_VARIABLE or '$'???? 1

pdotd

Programmer

sleipnir214

Programmer

pdotd

Programmer

kenrbnsn

Technical User

pdotd

Programmer

sleipnir214

Programmer

kenrbnsn

Technical User

pdotd

Programmer

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