Mysql更新两个表不工作

Question

解决方案 - 在表单标签的 action 属性中包含 get 变量。问题是调用主键，我试过 $id = $_GET['ID'];它失败了。使用 $id = 1 并且效果很好。我试过 $id = (isset($_GET['ID']));也失败了。成功了，谢谢大家

  <?php

  include 'Includes/conDB.php';

  if ($_SERVER["REQUEST_METHOD"] == "POST") {

   $unoone = $_POST['stringone'];
   $unotwo = $_POST['stringtwo'];
   $unothree = $_POST['integerthree'];
   $unofour = $_POST['stringfour'];

   // a data path for images  
    $valuz_ins = explode(',', $valuz_ins);

   $id = $_GET['ID'];

   $result = "UPDATE T1 LEFT JOIN T2
   ON (T1.ID=T2.ID) SET T1.stringone=?, 

   T1.stringtwo=?, T1.integerthree=?, T1.stringfour=?,  
   T2.imge1=?, T2.imge2=?, T2.imge3=?, T2.imge4=?
   WHERE T1ID=?"; 

   // prepare and bind

   $stmt = mysqli_prepare($con,$result);

  mysqli_stmt_bind_param($stmt,'ssisssssi', $unoone, $unotwo, 

  $unothree, $unofour, $valuz_ins[0], $valuz_ins[1], $valuz_ins[2],             
   $valuz_ins[3], $id);

 /* execute prepared statement */
 mysqli_stmt_execute($stmt);


 printf("%d Record updated.\n", mysqli_stmt_affected_rows($stmt));


 } 
 /* close statement and connection */
 mysqli_stmt_close($stmt);

 mysqli_close($con);

 ?>

Answer 1

我想你错过了 T1.ID = T2.ID，请尝试下面的一个

UPDATE T1 LEFT JOIN T2
   ON (T1.ID=T2.ID) SET T1.stringone=?, 

   T1.stringtwo=?, T1.integerthree=?, T1.stringfour=?,  
   T2.imge1=?, T2.imge2=?, T2.imge3=?, T2.imge4=?
   WHERE T1.ID=?