WBB User Post´s in 4images User Profile

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  • WBB User Post´s in 4images User Profile

    Hi,
    I would like to spend the written contributions from the Forum in the 4images user profile.
    The users in the Forum and 4images always the same ID

    --
    Hallo,
    ich würde gerne die geschrieben Beiträge aus dem Forum in dem 4images-User-Profil ausgeben.
    Die User haben in 4images als auch im Forum immer die gleiche ID

    Beide Systeme (4images) und das WBB 3.1.4 liegen in der gleichen DB

    member.php

    Quellcode

    1. //-----------------------------------------------------
    2. //--- Show Profile ------------------------------------
    3. //-----------------------------------------------------
    5. // WBB Datenbank auslesen
    6. $sql = "SELECT a.userID, a.posts
    7. FROM ".USERS_TABLE." u ON (".get_user_table_field("u.", "user_id")." = c.user_id)
    8. LEFT JOIN wbb1_1_user a ON (a.userID = ".get_user_table_field("u.", "user_id").")
    9. WHERE c.user_id = $user_id
    10. ORDER BY c.user_name ASC";
    11. $result = $site_db->query($sql);
    Alles anzeigen


    and

    Quellcode

    1. "lang_icq" => $lang['icq'],
    2. "wbb_post" => $result['posts'],


    in member_profile.html
    {wbb_post}

    But that does not work. Did I miss something?
    Aber das funktioniert nicht. Habe ich etwas übersehen?

    Quellcode

    1. DB Error: Bad SQL Query: SELECT a.userID, a.posts FROM 4images_users u ON (u.user_id = c.user_id) LEFT JOIN wbb1_1_user a ON (a.userID = u.user_id) WHERE c.user_id = ORDER BY c.user_name ASC
    2. You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'ON (u.user_id = c.user_id) LEFT JOIN wbb1_1_user a ON (a.userID = u.us' at line 2


    This Works:

    Quellcode

    1. $sql = "SELECT posts
    2. FROM (wbb1_1_user w, ".USERS_TABLE." u)
    3. WHERE w.userID = u.user_id";
    4. $wbb_row = $site_db->query_firstrow($sql);

    But is this Correct?

    Edit 2:
    All dont work, and this dont work too

    Quellcode

    1. $sql = "SELECT w.posts, u.user_id
    2. FROM ".USERS_TABLE." u
    3. LEFT JOIN wbb1_1_user w ON (w.userID = ".get_user_table_field("u.", "user_id").")
    4. WHERE u.user_id = $user_id";
    5. $wbb_row = $site_db->query_firstrow($sql);

    With this code, I spent every user the same ...

    Quellcode

    1. $sql = "SELECT posts
    2. FROM (wbb1_1_user w, ".USERS_TABLE." u)
    3. WHERE w.userID = u.user_id";
    4. $wbb_row = $site_db->query_firstrow($sql);

    As I said, that is USER ID in the Forum and 4images the same.

    Dont work:

    Quellcode

    1. $sql = "SELECT posts
    2. FROM wbb1_1_user
    3. WHERE userID = '".get_user_info($user_id)."'";
    4. $wbb_row = $site_db->query_firstrow($sql);


    Dont work

    Quellcode

    1. $sql = "SELECT posts
    2. FROM wbb1_1_user
    3. WHERE userID = '$user_id'";
    4. $wbb_row = $site_db->query_firstrow($sql);


    Dont work

    Quellcode

    1. $sql = "SELECT posts
    2. FROM (wbb1_1_user w, ".USERS_TABLE." u)
    3. WHERE ".$user_table_fields['user_id']." = w.userID";
    4. $wbb_row = $site_db->query_firstrow($sql);


    Quellcode

    1. $sql = "SELECT posts
    2. FROM (wbb1_1_user w, ".USERS_TABLE." u)
    3. WHERE w.userID = ".$user_table_fields['user_id']."";
    4. $wbb_row = $site_db->query_firstrow($sql);


    Quellcode

    1. $sql = "SELECT posts
    2. FROM (wbb1_1_user w, ".USERS_TABLE." u)
    3. WHERE w.userID = ".$user_row['user_id']."";
    4. $wbb_row = $site_db->query_firstrow($sql);

    Quellcode

    1. $sql = "SELECT posts
    2. FROM (wbb1_1_user w, ".USERS_TABLE." u)
    3. WHERE w.userID = '$user_id'";
    4. $wbb_row = $site_db->query_firstrow($sql);


    .... i dont no ...

    Dieser Beitrag wurde bereits 1 mal editiert, zuletzt von Sumale.nin ()