javascript - 使用 json_encode() 将 PHP 数组转换为 JavaScript 数组

Question

如何将数组从 PHP 传递到 JavaScript 函数 console.log()？我正在模拟数据库。据我所知，我没有在代码中声明数组。我尝试过使用 .getJSON() 函数，但它不起作用。我应该为每个元素执行 AJAX 调用吗？我正在考虑这个问题，但必须有更好的方法。

JavaScript 代码:

$.ajax({
         method:"POST",
         url:'php.php',
         data:"",
         dataType:'json',
         success:function(data){
             var y1=data;
             console.log(data.name);
         }
      });

PHP 代码:

<?php

/* 
 * To change this license header, choose License Headers in Project Properties.
 * To change this template file, choose Tools | Templates
 * and open the template in the editor.
 */


if($_SERVER["REQUEST_METHOD"] == "POST")
{
    //assign vars after formatting to avoid a wrong login with correct 
    //username and password
    $username = trim($_POST["user"]);


   $customer_array = array();

    $customer1 = array(
        'name' => '3',
        'city' => 'Houston'
    );

    $customer_array[] = $customer1;

    $customer2 = array(
        'name' => '2',
        'city' => 'Boston'
    );

    $customer_array[] = $customer2;

    $customer3 = array(
        'name' => "1",
        'city' => "Bossier City"
    );

    echo json_encode($customer_array);
}

Answer 1

我只是将我的评论总结为答案，首先我将关闭 json 返回以进行故障排除。无论如何，您可以事后解析 json:

$.ajax({
    // Do "type" here
    type:"POST",
    url:'php.php',
    // Send something to check in the script
    // This is optional, but I like to do it...
    data: {
        "action":"get_name_city"
    },
    success:function(data){
        // Do a try, just incase your parse fails
        try {
            // Parse the json here
            var getJson = JSON.parse(data);
            console.log(getJson);
            console.log(data);
        }
        catch(Exception) {
            // Show any errors that might get thrown
            console.log(Exception.message);
        }
    }
});

PHP:

# Check the action is set and is named
if(isset($_POST["action"]) && $_POST["action"] == "get_name_city") {
    $username = trim($_POST["user"]);
    $customer_array = array();

    $customer1 = array(
        'name' => '3',
        'city' => 'Houston'
    );

    $customer2 = array(
        'name' => '2',
        'city' => 'Boston'
    );

    $customer3 = array(
        'name' => "1",
        'city' => "Bossier City"
    );

    $customer_array[] = $customer1;
    $customer_array[] = $customer2;
    $customer_array[] = $customer3;

    # Just die here
    die(json_encode($customer_array));
}
# Die with error so there is some feedback to your ajax
die(json_encode(array('error'=>'No data was sent.')));