多個SQL查詢

Question

我想要做的多個查詢在一個聲明中，這裏是我的代碼，它不工作。

$stmt = $mysqli->prepare("UPDATE menu SET DESCRIPTION= ('$txt1'), VISIBLE=('$ckHOME1') 
where id='1'; 
UPDATE menu SET DESCRIPTION= ('$txt1'), VISIBLE=('$ckHOME1') where id='1'; 
UPDATE menu SET DESCRIPTION= ('$txt2'), VISIBLE=('$ckHOME2') where id='2'; 
UPDATE menu SET DESCRIPTION= ('$txt3'), VISIBLE=('$ckHOME3') where id='3'; 
UPDATE menu SET DESCRIPTION= ('$txt4'), VISIBLE=('$ckHOME4') where id='4';"); 
$stmt->execute(); 
if (!mysqli_query($con,$sql)) 
{ 
die('Error: ' . mysqli_error($con)); 
} 
mysqli_close($con); 
} 

Answer 1

或者，您也可以把它作爲使用CASE一個UPDATE聲明。

UPDATE menu 
SET  
     DESCRIPTION = CASE WHEN id = 1 THEN '$txt1' 
          WHEN id = 2 THEN '$txt2' 
          WHEN id = 3 THEN '$txt3' 
          WHEN id = 4 THEN '$txt4' 
        END, 
     VISIBLE = CASE WHEN id = 1 THEN '$ckHOME1' 
          WHEN id = 2 THEN '$ckHOME2' 
          WHEN id = 3 THEN '$ckHOME3' 
          WHEN id = 4 THEN '$ckHOME4' 
        END 
where id IN (1, 2, 3, 4) 

Answer 2

您必須使用mysqli_multi_query()才能在一次調用中執行多個查詢。

但有沒有必要做多個查詢爲你在做什麼：

UPDATE menu 
SET description = 
     CASE id 
      WHEN 1 THEN '$txt1' 
      WHEN 2 THEN '$txt2' 
      WHEN 3 THEN '$txt3' 
      WHEN 4 THEN '$txt4' 
     END, 
    visible = 
     CASE id 
      WHEN 1 THEN '$ckHOME1' 
      WHEN 2 THEN '$ckHOME2' 
      WHEN 3 THEN '$ckHOME3' 
      WHEN 4 THEN '$ckHOME4' 
     END 
WHERE id IN (1, 2, 3, 4);