Variable Scope Question PHP

Question

In the following code the variable does not seem to be getting set. Seems simple enough but for some reason this is vexing me.

function teasertext($string){
    $tstring = "";
    if (strlen($string)>9){
        $tstring .= substr($string,0,9) . "....";
    }
    else
    {
        $tstring .= $string;
    }
}
print $tstring;
return $tstring;

Rule #1, always indent consistently. It will show you why some things don't work (in your case, the print and return statements are outside of the function block)... — ircmaxell
– ircmaxell, Commented Aug 23, 2010 at 16:16

Buggabill · Accepted Answer · 2010-08-23 16:15:47Z

2

print $tstring;
return $tstring;

is outside of the function block.

function teasertext($string){
    $tstring = "";
    if (strlen($string)>9){
        $tstring .= substr($string,0,9) . "....";
    }
    else
    {
        $tstring .= $string;
    }
    print $tstring;
    return $tstring;
}

Should return $tstring properly.

answered Aug 23, 2010 at 16:15

Buggabill

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Comments

Joseph U. · Accepted Answer · 2010-08-23 16:15:13Z

-1

I placed the variables outside of the function. Silly mistake.

answered Aug 23, 2010 at 16:15

Joseph U.

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