PHP Variable help?

crashovaride · 11-06-2007, 10:04 AM

[Solved] by darkhack

Darkhack · 11-06-2007, 12:18 PM

I think the scope of $datainfo isn't set properly. You are initializing it inside a foreach loop which means that it is being recreated on each loop. Try this instead.

Code:

// Initialize variable outside foreach scope.
$datainfo = "";
foreach ($_POST as $field => $value)
{
    echo "<font color='white'>$field = $value<br></font>";
    /* Notice the dot equals (.=) sign which appends data to a variable.
     * With this, we don't need to add it again to the end of the statement. */
    $datainfo .= $field => $value . "<br>";
}

crashovaride · 11-06-2007, 12:40 PM

Thank you for the quick reply, however, when using the code you have provided (and thank you for pointing this out to me.) It seems to error on the line with the following code:unexpected T_DOUBLE_ARROW I apologize for being a bother. This is my first time writing php scripts. The code i'm using is:

$datainfo = "";

foreach ($_POST as $field => $value)
{
echo "<font color='white'>$field = $value<br></font>";
/* Notice the dot equals (.=) sign which appends data to a variable.
* With this, we don't need to add it again to the end of the statement. */
$datainfo .= $field=>$value . "<br>"; */it's erroring on this line. Any suggestions?
}

Thank you for the help.

brazilnut · 11-06-2007, 12:43 PM

Code:

$datainfo .= $field.": ".$value . "<br>";

It's already been split into it's constituent parts

crashovaride · 11-06-2007, 12:47 PM

Thank you for the code Brazilnut it's working great now. Also, thank you DarkHack for pointing me in the right direction. Much appreciated.