Question

执行ajax脚本时，控制台中出现500服务器错误，我不知道为什么。如果我稍微更改代码来实现mysqli函数，并在我的本地机器上运行它们一切正常，但是一旦我尝试使用codeigniter，我就会得到500服务器错误。

这是我的代码：

<?php

class Phonecall extends CI_Controller {

    public function index() {

        /*$con = mysqli_connect('localhost','root','root','MYDB');
        if (!$con) {
            die ('Could not connect: ' . mysqli_error($con));
        }*/
        //$operatorId = $_SESSION['Oper']['OperatorID']; 
        //$sql = "SELECT phone_number FROM incoming_calls WHERE OperatorID='${operatorId}'";
        //$result = mysqli_query($con,$sql);
        $query = $this->db->query("SELECT phone_number FROM MYDB.incoming_calls");// WHERE OperatorID='${operatorId}'");

        while ($row = $query->row_array()) {
            $number = $row['phone_number'];
        }

        /*$sql = "SELECT Username, UserID, Name
            FROM tblUsers 
            WHERE PhoneHome='999-999-9999' OR PhoneCell='999-999-9999' OR PhoneWork='999-999-9999'";
        */

        $query = $this->db->query("SELECT Username, UserID, Name
            FROM MYDB.tblUsers 
            WHERE PhoneHome='999-999-9999' OR PhoneCell='999-999-9999' OR PhoneWork='999-999-9999'");

        while ($row = $query->row_array()) {
            $userArray[] = array("name" => $row['Name'], "username" => $row['Username'], "user_id" => $row['UserID']);
        }

        if (!empty($userArray)) {
            echo json_encode($userArray);
        }

        if (isset($_POST["drop"])) {
            $query = $this->db->query("DELETE FROM MYDB.incoming_calls 
                WHERE phone_number = $number
                LIMIT 1");
            if (!$result) {
                die ('Could not drop row: ');
            }
        }

        $this->db->close();
    }
}

?>

以下是检索信息的ajax调用：

var user = new Array();
var user_id = new Array();
var name = new Array();

$.get(baseURL + 'phonecall/index', function(data) {//where baseURL is defined
    var loginInfo = jQuery.parseJSON(data);

    for (var i = 0; i < loginInfo.length; ++i) {
        name[i] = loginInfo[i].name;
        user[i] = loginInfo[i].username;
        user_id[i] = loginInfo[i].user_id;
    }
}

有谁知道为什么会这样？

500服务器错误codeigniter ajax

0 个答案: