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  1. #1
    Regular Coder Deacon Frost's Avatar
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    Functions in Echo....

    Line 30 - 33:
    PHP Code:
     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: ' [B][U].[/U][/B] . 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?
    Last edited by Deacon Frost; 11-26-2009 at 10:12 AM.

  • #2
    Supreme Master coder! abduraooft's Avatar
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    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.
    The Dream is not what you see in sleep; Dream is the thing which doesn't let you sleep. --(Dr. APJ. Abdul Kalam)

  • Users who have thanked abduraooft for this post:

    Deacon Frost (11-26-2009)

  • #3
    Regular Coder Deacon Frost's Avatar
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    view_cell works fine, I believe...


    PHP Code:
    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.
    Last edited by Deacon Frost; 11-26-2009 at 10:11 AM.


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