eval function

[Anon]

Hello friends,

I have one function which takes two parameters. I have passed the parameters like this.

$title = "Search";
$content = "echo <table border='0'><td bgcolor='red'>Test</td></table>";

testFunction($title,eval($content));

The above gives error.

Can anybody tell what could be the problem.
In the $content variable i will have the php coding part.

[znouza]

oops, you didn't read eval() function properly.

in $content you have to have VALID php statement, so valid $content will be
$content = "echo \"<table...>\";"; (don't miss the semicolon);

then eval($content) will give correct result.