php Ajax - 如何在成功函数中使用返回的数组
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Ajax - How to use a returned array in a success function
提问by Matt9Atkins
Hi I have a php code that returns an array. I want to be able to use this array in my ajax success function but I'm not sure how to go about doing this. I have tried the following, but no luck.
嗨，我有一个返回数组的 php 代码。我希望能够在我的 ajax 成功函数中使用这个数组，但我不知道如何去做。我已经尝试了以下，但没有运气。
php code:
php代码：
$arr = array();
$arr[0] = "Mark Reed"
$arr[1] = "34";
$arr[2] = "Australia";
exit($arr);
js code:
js代码：
$.ajax({
type: "POST",
url: "/returndetails.php",
data: 'id=' + userid,
success: function (data) {
document.getElementById("name").innerHTML = data[0];
document.getElementById("age").innerHTML = data[1];
document.getElementById("location").innerHTML = data[2];
}
});
回答by Hugo Tunius
You should return the data as JSON from the server.
您应该从服务器以 JSON 形式返回数据。
PHP
PHP
$arr = array();
$arr[0] = "Mark Reed";
$arr[1] = "34";
$arr[2] = "Australia";
echo json_encode($arr);
exit();
JS
JS
$.ajax({
type: "POST",
url: "/returndetails.php",
data: 'id=' + userid,
dataType: "json", // Set the data type so jQuery can parse it for you
success: function (data) {
document.getElementById("name").innerHTML = data[0];
document.getElementById("age").innerHTML = data[1];
document.getElementById("location").innerHTML = data[2];
}
});
回答by robby
A small mistake:
一个小错误：
Not: exit($arr);
不是： exit($arr);
replace with: echo json_encode($arr);
用。。。来代替： echo json_encode($arr);
回答by Tomas Grecio Ramirez
There a Problem , when you want display for example data[0]and data[1], it seems like a character from string. It Solves adding header("Content-Type: application/json");before apply echo json_encode($arr)
有一个问题，当你想要显示例如data[0]and 时data[1]，它看起来像是字符串中的一个字符。它解决了header("Content-Type: application/json");在申请前添加 echo json_encode($arr)
回答by Waqas Qayum
Here is solution
这是解决方案
$arr = array();
$arr[0] = "Mark Reed"
$arr[1] = "34";
$arr[2] = "Australia";
header("Content-Type: application/json");
echo json_encode($arr);
exit();
instead of
代替
$arr = array();
$arr[0] = "Mark Reed"
$arr[1] = "34";
$arr[2] = "Australia";
exit($arr);
