Question

我有两个预准备语句（例如），其中第二个语句将在while(stmt1->fetch()){}循环中执行。但是，内部语句（stmt2）不会在第一个while循环中执行：

<?php
  $mysqli = new mysqli("localhost","root","","test");

  if(mysqli_connect_errno())
  {
    printf("connection failed: %s\n",mysqli_connect_error());
    exit();
  }

  $stmt1 = $mysqli->prepare("select id from posts");
  $stmt2 = $mysqli->prepare("select username from members where id=?"); 

  $stmt1->execute(); 
  $stmt1->bind_result($ID);

   while($stmt1->fetch())
   {
     echo $ID.' ';

      /*Inner query*/
      $stmt2->bind_param('i',$id);

      $id =$ID;    

      $stmt2->execute();

      $stmt2->bind_result($username);

      while($stmt2->fetch())
      {
        echo 'Username: '.$username;
      }     
      /*Inner query ends*/
   }  

 ?>

如果我将内部查询部分剪切粘贴到外部while循环之外，它会执行，但它没用。我该怎么做才能正确执行它？

Answer 1

为什么嵌套循环，当你可以改为 INNER JOIN 时。

<?php

  $mysqli = new mysqli("localhost","root","","test");

  if(mysqli_connect_errno())
  {
    printf("connection failed: %s\n",mysqli_connect_error());
    exit();
  }

  if($stmt1 = $mysqli->prepare("SELECT posts.id,members.username FROM posts INNER JOIN members ON posts.id=members.id")){

    $stmt1->execute(); 
    $stmt1->bind_result($id,$username);

    while($stmt1->fetch()){

      printf("ID # %d.<br>Username: %s<br><br>",$id,$username);

    }    
    $stmt1->close(); 
  }

  $mysqli->close();

 ?>

嵌套MySQLi准备好的语句

1 个答案: