Question

我正在尝试编写这样的PHP脚本，但我不知道如何做到这一点

{
    "feed": [
        {
            "id": 1,
            "name": "National Geographic Channel",
            "image": "http://api.androidhive.info/feed/img/cosmos.jpg",
            "status": "\"Science is a beautiful and emotional human endeavor,\" says Brannon Braga, executive producer and director. \"And Cosmos is all about making science an experience.\"",
            "profilePic": "http://api.androidhive.info/feed/img/nat.jpg",
            "timeStamp": "1403375851930",
            "url": null
        },
        {
            "id": 2,
            "name": "TIME",
            "image": "http://api.androidhive.info/feed/img/time_best.jpg",
            "status": "30 years of Cirque du Soleil's best photos",
            "profilePic": "http://api.androidhive.info/feed/img/time.png",
            "timeStamp": "1403375851930",
            "url": "http://ti.me/1qW8MLB"
        }
]}

这是我的PHP脚本

<?php

require ('config.php');
$conn = mysqli_connect($servername,$username,$password,$db);
$query = "select * from playernews";

$result = mysqli_query($conn, $query);
$rows = array();
echo mysqli_error($conn);

while($row = mysqli_fetch_assoc($result)) {
    $rows[] = $row;
}

echo json_encode($rows);
?>

这是我的PHP代码脚本，所以如果有人可以帮助我吗？

Answer 1

首先，你不应该这样做。 JSON旨在将数据从一个系统传输到另一个系统，而不是作为调试介质或输出格式。额外的空格不会促进JSON的主要目标。

话虽如此，请查看the docs for json_encode。

您可以将一个选项 JSON_PRETTY_PRINT 传递给该功能，这可以帮助您实现所需目标：

json_encode($rows, JSON_PRETTY_PRINT);

Answer 2

<?php

require ('config.php');
$conn = mysqli_connect($servername, $username, $password, $db);
$query = "select * from playernews";

$result = mysqli_query($conn, $query);
$rows = array();
echo mysqli_error($conn);

while($row = mysqli_fetch_assoc($result)) {
    $rows[]=$row;
}

$data = array(
    'feed' => $rows
);

echo json_encode($data); // now will get you same format as you wanted
?>

如何获得这个Json表格？

2 个答案: