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YaBB SE Community  |  German User Help  |  Hilfe zu YaBB SE  |  2: Supplied argument is not a valid Mysql result resource « previous next »
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Author Topic: 2: Supplied argument is not a valid Mysql result resource  (Read 528 times)
MatthiasB
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Posts: 13


I'm a llama!

2: Supplied argument is not a valid Mysql result resource
« on: July 09, 2003, 12:12:22 PM »
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YaBB SE Version: 1.5.3
PHP Version: 4.1.2
MySQL Version: 3.23.x
Server Plattform: Unix, Linux, or BSD
Link zum Forum:

Problem Beschreibung:
Aus heiterem Himmel kommt folgende Fehlermeldung:

Forum-Home:

Ein Fehler ist aufgetreten!

2: Supplied argument is not a valid MySQL result resource
(/var/www/web1/html/intern/forum/Sources/BoardIndex.php ln 154)


Und im Admin-Bereich zeigt er das an:

2: Supplied argument is not a valid MySQL result resource
(/var/www/web1/html/intern/forum/Sources/Recent.php ln 48)



Was stimmt hier nicht ?
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chris
Guest
Re:2: Supplied argument is not a valid Mysql result resource
« Reply #1 on: July 09, 2003, 12:35:14 PM »
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Da stimmt primär schonmal nicht, dass der Fehler bei V1.5.3 niemals auftreten können sollte (weil eigentlich IMMER eine Klartext-Fehlermeldung kommen müsste).

Poste mal die entsprechende Zeile (+ein paar davor und ein paar danach)

bei mir wäre das:

   $result_boards = mysql_query ("
      SELECT DISTINCT c.name AS catName, c.ID_CAT, b.ID_BOARD, b.name AS boardName, b.description, b.moderators, b.numPosts, b.numTopics, c.memberGroups, m.posterName, m.posterTime, m.subject, t.ID_TOPIC, t.numReplies, IFNULL(mem.realName, m.posterName) AS realName, IFNULL(lb.logTime, 0) AS boardTime, IFNULL(lmr.logTime, 0) AS markReadTime, IFNULL(mem.ID_MEMBER, -1) AS ID_MEMBER
      FROM {$db_prefix}categories AS c
         LEFT JOIN {$db_prefix}boards AS b ON (b.ID_CAT=c.ID_CAT)
         LEFT JOIN {$db_prefix}topics AS t ON (t.ID_TOPIC=b.ID_LAST_TOPIC)
         LEFT JOIN {$db_prefix}messages AS m ON (m.ID_MSG=t.ID_LAST_MSG)
         LEFT JOIN {$db_prefix}members AS mem ON (mem.ID_MEMBER=m.ID_MEMBER)
         LEFT JOIN {$db_prefix}log_boards AS lb ON (lb.ID_BOARD=b.ID_BOARD AND lb.ID_MEMBER=$ID_MEMBER)
         LEFT JOIN {$db_prefix}log_mark_read AS lmr ON (lmr.ID_BOARD=b.ID_BOARD AND lmr.ID_MEMBER=$ID_MEMBER)
      WHERE $condition
      ORDER BY c.catOrder, c.ID_CAT, b.boardOrder, b.ID_BOARD") or database_error(__FILE__, __LINE__);

   $curcat = -1;
   while ($row_board = mysql_fetch_assoc($result_boards))
   {
      // if this is a new category
      if ($row_board['ID_CAT'] != $curcat)
      {
         $curcat=$row_board['ID_CAT'];
         echo "<tr> <td colspan=\"5\" class=\"catbg\" height=\"18\">\n";
         echo "<a name=\"$row_board[ID_CAT]\"><b>$row_board[catName]</b></a></td>\n</tr>\n";


wobei

while ($row_board = mysql_fetch_assoc($result_boards))

die Zeile ist wo es knallt... wenns bei Dir die gleiche ist, füg da mal

echo mysql_error();

vor ein... dann gibt er ne Info aus, was schief gelaufen ist...
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MatthiasB
Noobie
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Posts: 13


I'm a llama!

Re:2: Supplied argument is not a valid Mysql result resource
« Reply #2 on: July 09, 2003, 12:38:07 PM »
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das verstehe ich nicht


der zeigt das jetzt fast überall an; Supplied argument is not a valid Mysql result resource


ist die Datenbank eventuell offline oder zerschossen ?
« Last Edit: July 09, 2003, 01:20:34 PM by MatthiasB » Logged
chris
Guest
Re:2: Supplied argument is not a valid Mysql result resource
« Reply #3 on: July 09, 2003, 03:25:01 PM »
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Wäre sie offline würde garnichts gehen...

Also ist sie vermutlich kaputt/voll/was auch immer...
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