javascript - AJAX 未输出正确的内容

Question

所以我尝试从 javascript 调用 php 方法，以便我可以查询数据库并将结果获取到我的 js 功能中。目前，我的 ajax 中的 'console.log(output)' 只是输出:

"array (size=1)
'action' => string 'getResults' (length=10)'"

不太确定为什么要这样做，它应该返回查询结果，这只是数据库中的一项。有人有什么想法吗？欢迎任何帮助!谢谢。

我的 Javascript 文件的一部分:

function callPHP() {
    $.ajax ({

        type: "GET",
        datatype: "application/json",
        url: "BaseClass.php",
        data: { action : 'getResults' },
        //error: function(err){console.log(err)},
        success: function(output) {

            console.log(output);
            //alert(output);
        }
        //error, function(err){console.log(err)}
    });


}

callPHP();  

BaseClass.php:

<?php

    error_reporting(E_ALL); ini_set('display_errors', 1);

    require("Conn.php");
    require("MySQLDao.php");

    $param=$_REQUEST['action'];

    echo var_dump($_GET);
    /*
    $handle = fopen("php://input", "rb");
    $param = '';
    while (!feof($handle)) {
        $param .= fread($handle, 8192);
    }
    fclose($handle);
    */

    if (empty($param))
    {
        $returnValue["status"] = false;
        $returnValue["title"] = "Error";
        $returnValue["message"] = "No Data Recieved paige" .$param ."...";
        echo json_encode($returnValue);
        return;
    }
    else
    {
        $dao = new MySQLDao();
        if ($dao->openConnection() == false)
        {
            $returnValue["status"] = false;
            $returnValue["title"] = "Error";
            $returnValue["message"] = "Connection Could Not Be Established Between Server And Database";
            echo json_encode($returnValue);
        }
        else
        {
            //Decodes data, dont change
            $body = json_decode($param, true);
            $recieved = $body["data"];

            //Gets the result of a query
            //$result = $dao->MySQLDaoMethodName(parameters);

            //Return the result of the query
            //echo json_encode($result);
        }
        $dao->closeConnection();
        return;
    }
?>

Conn.php - 这是所有连接信息，*出于保密原因而输出:*

<?php

    error_reporting(E_ALL); ini_set('display_errors', 1);

    class Conn
    {
        public static $dbhost = "***";
        public static $dbname = "***";
        public static $dbuser = "***";
        public static $dbpass = "***";
    }
?>

MySQLDao.php - 此文件保存查询:

<?php

    error_reporting(E_ALL); ini_set('display_errors', 1);

    //Class for holding queries
    class MySQLDao
    {
        var $dbhost = null;
        var $dbuser = null;
        var $dbpass = null;
        var $mysqli = null;
        var $dbname = null;
        var $result = null;


        //constructor
        function __construct()
        {
            $this->dbhost = Conn::$dbhost;
            $this->dbuser = Conn::$dbuser;
            $this->dbpass = Conn::$dbpass;
            $this->dbname = Conn::$dbname;
        }

        //Attempt a connection to the database
        public function openConnection()
        {   

            //Try and connect to the database
            $this->mysqli = new mysqli($this->dbhost, $this->dbuser, $this->dbpass, $this->dbname);
            //If the connection threw an error, report it
            if (mysqli_connect_errno())
            {
                return false;
            }
            else
            {
                return true;
            }
        }

        //Get method for retrieving the database conection
        public function getConnection()
        {
            return $this->mysqli;
        }

        //Close the connection to the database
        public function closeConnection()
        {
            //If there is a connection to the database then close it
            if ($this->mysqli != null)
                $this->mysqli->close();
        }

        //-----------------------------------QUERY METHODS-------------------------------------

        public function getResults($data)
        {

            $sql = "SELECT room.room_description FROM room WHERE room.room_id = 1";

            $result = $this->mysqli->query($sql);


            //if (mysql_num_rows($result) == 1) {
            //  $obj = mysql_fetch_object($result, 'obResults');

            //}

            echo json_encode($result);

            echo($result);

        }

    }
?>

Answer 1

我认为您误解了如何在 javscript 和 php 之间传输数据。

1. 如果您想发送对象或数据数组，您的 javascript 应该使用 $.post() 进行发布。
2. 您的 php 将从 javascript 接收 json。你需要使用php的json_decode函数来使其对php有用。
3. 如果您希望 php 脚本的输出对 javascript 有用，则需要使用 php 的 json_encode 函数对其进行编码，然后再将其返回到调用脚本。

http://php.net/manual/en/function.json-decode.php