php
  • mysql
  • 2011-12-12 59 views 0 likes 
    0

    我想從php腳本更新記錄,如果複選框pendinginput_c被選中。我無法從腳本中獲取更新記錄。但是,如果我在mysql控制檯中輸入下面的確切命令,則記錄正確更新。試圖從php腳本更新記錄

       if($bean->pendinginput_c == 'on') 
          { 
          $sql = "UPDATE cases SET status='Pending Input' WHERE id='958ca4e5-51'"; 
          } 
    

    進一步在腳本中你會看到有其他的SQL語句,目前正在工作。沒有更新,只是選擇。

    $sql_nmb = "SELECT case_number FROM cases where id='".$bean->parent_id."'"; 
          $result = $bean->db->query($sql_nmb, true); 
          $case = $bean->db->fetchByAssoc($result);    
    
          $sql = 'SELECT c.assigned_user_id , c.id , cc.contact_id as contact_id, co.first_name , co.last_name , ear.email_address_id ,ea.email_address FROM cases as c 
             LEFT JOIN contacts_cases as cc on c.id = cc.case_id 
             LEFT JOIN contacts as co on co.id = contact_id 
             LEFT JOIN email_addr_bean_rel ear ON ear.bean_id = contact_id 
             LEFT JOIN email_addresses as ea on ea.id = ear.email_address_id 
             WHERE c.id = "'.$bean->parent_id.'" and cc.deleted = 0 and ear.deleted = 0 and ea.deleted = 0'; 
          $resp = $bean->db->query($sql, true); 
    
    +1

    你的mysql_query()方法在哪裏? – amrfaissal

    +1

    SQL查詢不會因爲您將它們放置在代碼中某處而被執行。 –

    +0

    我編輯了我原來的帖子,向您展示了腳本的更多內容 – ipengineer

    回答

    0

    試試這個:

    if($bean->pendinginput_c == 'on'){ 
        $sql = "UPDATE cases SET status='Pending Input' WHERE id='958ca4e5-51'"; 
        $bean->db->query($sql, true); 
    } 
    

    此外檢查$ bean-> pendinginput_c真的等於 「上」。在if語句上方回顯它以進行仔細檢查。

    祝你好運!

    0

    您是否嘗試過在這樣的查詢逃避'的:\'

    編輯:當然假設你實際上正在查詢mysql_query($sql) ......或者其他一些查詢功能。

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