具有多个选择的 php sql 不起作用

Question

以下代码给了我这个错误:

Call to undefined method mysqli_stmt::get_result() on line 21.

我不明白这个对象是如何工作的以及为什么我可以执行第一个数据库调用但不能执行第二个。

    <?php 

    header('Content-Type: application/json');
    include_once 'do_dbConnect.php';
    include_once 'functions.php';
    sec_session_start(); 

    //identify who took the last call
    $stmt =  $mysqli->stmt_init();
    if ($stmt->prepare("SELECT MAX(dateOfCall), id FROM call")) { //setup the query statement
      $stmt->execute(); //execute the statement
      $result = $stmt->get_result(); //get the results
      $row = $result->fetch_assoc(); //get the first row
      $user_id = $row['id']; //get the id column
    }


    //identify how many team members there are
    if ($stmt->prepare("SELECT id FROM teamMembers")) { //setup the query statement
      $stmt->execute(); //execute the statement
      $result = $stmt->get_result(); //get the results
      $memberCount = $result->num_rows;
    }


    //get next user
    if ($stmt = $mysqli->prepare("SELECT * FROM teamMembers WHERE id = (? + 1) % ?")) { //setup the query statement
      $stmt->bind_param('ii', $user_id, $memberCount);
      $stmt->execute(); //execute the statement
      $result = $stmt->get_result(); //get the results
      $row = $result->fetch_assoc(); //get the first row
      $next_user_id = $row['id']; //get the id column
      $next_user_name = $row['username'];
    }
    $stmt->close();
    //get the next call taker from the teamMember table

    echo json_encode($row);

    ?> 

Answer 1

请阅读此方法的用户说明:

http://php.net/manual/en/mysqli-stmt.get-result.php

它需要 mysqlnd 驱动程序...如果您的网站空间上没有安装它，您将必须使用 BIND_RESULT 和 FETCH!

http://www.php.net/manual/en/mysqli-stmt.bind-result.php

http://www.php.net/manual/en/mysqli-stmt.fetch.php