T_ENCAPSED_AND_WHITESPACE

[GreenFanta]

Hey Guys!

 

I've been looking through my script to see what the problem could be, I've got a variable setup to generate a web page, I have no idea what could be the problem, I keep receiving the error:

 

Parse error: syntax error, unexpected T_ENCAPSED_AND_WHITESPACE, expecting T_STRING or T_VARIABLE or T_NUM_STRING in home/path/etc on line 8

 

But I don't see any miss-matching Quotations etc... I don't know what the problem could be, maybe you could help me?

 

Code:

1 <?php
2 $content = "<?php
3 $productid = 1;
4 $sql = 'SELECT * FROM table1 WHERE productid =' . $productid . '';
5 if ($result = mysql_query($sql)) {
6 if (mysql_num_rows($result)) { 
7 $row = mysql_fetch_assoc($result);
8 $leader = $row['leader'];
9 	 }
10  }
11 ?>";
12 echo($content);
13 ?>

 

Thanks!

 

Regards

Matthew

 

[kenrbnsn]

What are you trying to do?

 

 

Since you enclosed the string in double quotes, PHP is trying to do variable substitution withing the string and you can't put a quoted array reference inside a double quoted string. Either enclose the string in single quotes or quote all the $ so the variable names aren't seen as variable names:

<?php
$content = '<?php
$productid = 1;
$sql = "SELECT * FROM table1 WHERE productid =\'" . $productid . "\'";
if ($result = mysql_query($sql)) {
if (mysql_num_rows($result)) { 
$row = mysql_fetch_assoc($result);
$leader = $row["leader"];
	 }
  }
?>';
echo($content);
?>

 

or

<?php
$content = "<?php
\$productid = 1;
\$sql = 'SELECT * FROM table1 WHERE productid ='\" . $productid . \"'\";
if (\$result = mysql_query(\$sql)) {
if (mysql_num_rows(\$result)) { 
\$row = mysql_fetch_assoc(\$result);
\$leader = \$row['leader'];
	 }
  }
?>";
echo($content);
?>

 

Ken

[GreenFanta]

Thank you very much!