PHP SQL與多個選擇不工作

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

我不明白這是如何工作的對象，爲什麼我可以做的第一個數據庫調用而不是第二。

<?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); 

    ?>

來源

2014-07-25 b1shp

請指定哪些線是21 ??你怎麼知道第一個mysqli電話是好的？ –

我不知道你的問題的答案，但我總是發現，理解事情的工作往往是最重要的一步。看到這個評論：http://stackoverflow.com/a/8343970/1888402另外，不要害羞，這是一個公共論壇，人們喜歡得到幫助人們的積分，因此這對所有參與者都是雙贏的。 –

一個問題是您的第一個查詢**調用**是您需要的[* Mysql保留關鍵字*]（http://dev.mysql.com/doc/refman/5.6/en/reserved-words.html）用bact-ticks逃脫它 –