UDATE查询与PHP和MySQL

Question

嵌套CASE语句我有测验例如嵌套的SQL查询来更新多个表。它包含两个case语句，该语句在对数据库执行查询时检查特定条件。下面是代码：

// form data 
$edit_qid = isset($_POST['editqid']); 
$edit_question = isset($_POST['editquestion']); 
$edit_ans1 = isset($_POST['editanswer1']); 
$edit_ans2 = isset($_POST['editanswer2']); 
$edit_ans3 = isset($_POST['editanswer3']); 
$edit_correct = isset($_POST['editcorrect']); 

// answer ids (comes from another query and it's valid) 
while ($rows = mysqli_fetch_assoc($result_aid)) { 
    $aid[] = $rows['aid']; 
} 

// update of chosen question and answer options to DB 
$upd_question = "UPDATE `question_bank` qtbl INNER JOIN `answer_bank` atbl 
         SET qtbl.`question`='".$edit_question."', 
          atbl.`answer`= CASE WHEN atbl.`aid`='".$aid[0]."' THEN '".$edit_ans1."' 
               WHEN atbl.`aid`='".$aid[1]."' THEN '".$edit_ans2."' 
               WHEN atbl.`aid`='".$aid[2]."' THEN '".$edit_ans3."' 
              END 
          atbl.`correct` = CASE WHEN ".$edit_correct."=='1' THEN '1' 
                WHEN ".$edit_correct."=='2' THEN '1' 
                WHEN ".$edit_correct."=='3' THEN '1' 
                ELSE '0' 
              END 
        WHERE qtbl.`qid`=atbl.`question_id` AND qtbl.`qid`='".$edit_qid."'"; 
mysqli_query($mysqli, $upd_question) or die ("<b>Update of question failed:</b> " . mysqli_error($mysqli)); 

可变$edit_correct包含任一值1，2或3相对于回答字符串和根据上述查询到DB那里将被保存的1或0（真/假），这意味着该如果值是1，那么真正的将被保存为第1个答案选项等这段代码的

跑步给了我以下错误：

Update of question failed: 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 'atbl.`correct` = CASE WHEN 2=='1' THEN '1' WHEN 2=='2' THEN '1' ' at line 7 

任何想法来解决这个问题？提前致谢。

UPDATE

而且，我有以下条件检查：

,atbl.`correct` = CASE WHEN '". if($edit_correct=='1') {echo 1;} ."' THEN '1' 
         WHEN '". if($edit_correct=='2') {echo 1;} ."' THEN '1' 
         WHEN '". if($edit_correct=='3') {echo 1;} ."' THEN '1' 
         ELSE '0' 
       END 

，并得到这个错误：Parse error: syntax error, unexpected 'if' (T_IF)

Answer 1

你必须把逗号介乎两列，试试这个代码：

$upd_question = "UPDATE `question_bank` qtbl INNER JOIN `answer_bank` atbl 
         SET qtbl.`question`='".$edit_question."', 
          atbl.`answer`= CASE WHEN atbl.`aid`='".$aid[0]."' THEN '".$edit_ans1."' 
               WHEN atbl.`aid`='".$aid[1]."' THEN '".$edit_ans2."' 
               WHEN atbl.`aid`='".$aid[2]."' THEN '".$edit_ans3."' 
              END 
          ,atbl.`correct` = CASE WHEN ".$edit_correct."=='1' THEN '1' 
                WHEN ".$edit_correct."=='2' THEN '1' 
                WHEN ".$edit_correct."=='3' THEN '1' 
                ELSE '0' 
              END 
        WHERE qtbl.`qid`=atbl.`question_id` AND qtbl.`qid`='".$edit_qid."'"; 
mysqli_query($mysqli, $upd_question) or die ("<b>Update of question failed:</b> " . mysqli_error($mysqli)); 

在第二种情况下你可以使用IF（）语句，使用IF会得到更多可读的代码。