Line 30 - 33:
if ($isthis2) {
echo 'User Details: (' . view_cell($isthis2, "Users", "Txn_ID") . ') - Active: ' . view_cell($isthis2, "Users", "Active") . '<hr />';
echo '(' . view_cell($isthis2, "Users", "Type") . ')ID: ' . view_cell($isthis2, "Users", "ID") . ' Refer Code: ' . view_cell($isthis2, "Users", "Refer_ID") . ' Referrer: ' . . view_cell($isthis2, "Users", "Referrer") . '<br />';
echo 'E-Mail: ' . view_cell($isthis2, "Users", "EMail") . ' Paypal: ' . view_cell($isthis2, "Users", "PP_EMail") . '<br />';


Parse error: syntax error, unexpected '.' on line 32


Whut? I have tons of lines like these... I expected them to work =/. Why no such luck?


Nevermind. Lol. Sometimes just posting helps, ya know?

I don't get any parse errors for that line, though I get Fatal error: Call to undefined function view_cell() in G:\xampp\htdocs\spread\new_spread\tests\trash2.php on line 2 which is obvious.

The error might be on a line near to it.

view_cell works fine, I believe...



function view_cell($id, $table, $field) {

db_connect();

$sql = "SELECT `$field` FROM `$table` WHERE `ID`='$id'";
$result = mysql_query($sql) or die('The error was: ' . mysql_error() . '<br />The query was: ' . $sql);
$row = mysql_fetch_assoc($result);
echo $row["$field"];

}






Never mind, fixed it >.<. Thanks :P. I was looking at the wrong line, bloody wrap around lines are hard to distinguish.