Question

我有一个用于测试示例的嵌套SQL查询来更新多个表。它包含两个 case 语句，用于在对DB执行查询时检查特定条件。这是代码：

// 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));

变量$edit_correct包含关于答案字符串的值1,2或3，并且根据上面提到的对DB的查询，将保存1或0（真/假），这意味着如果值为1则true将保存为第一个答案选项等。

运行此代码会出现以下错误：

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

有什么想法可以解决这个问题吗？提前谢谢。

更新

另外，我检查了以下条件：

,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

并收到此错误：Parse error: syntax error, unexpected 'if' (T_IF)

Answer 1

你必须在两列之间加一个逗号，试试这段代码：

$upd_question = "UPDATE `question_bank` qtbl INNER JOIN `answer_bank` atbl
                        SET qtbl.`question`='".$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会更加可读代码。

使用PHP中的嵌套CASE语句更新查询; MySQL的

1 个答案: