Question

我有这个JSON：

{
  "cache": true,
  "data": [
    {
      "unique_id": "914239",
      "description": "New Zealand 370/10 &amp;  335/10  v Australia 70/1 &amp;  505/10 *",
      "title": "New Zealand 370/10 &amp;  335/10  v Australia 70/1 &amp;  505/10 *"
    },
    {
      "unique_id": "973833",
      "description": "Helmand Province Under-17s 135/10 * v Khost Province Under-17s 286/9 ",
      "title": "Helmand Province Under-17s 135/10 * v Khost Province Under-17s 286/9 "
    },
    {
      "unique_id": "935949",
      "description": "Mid West Rhinos v Mashonaland Eagles 264/2 *",
      "title": "Mid West Rhinos v Mashonaland Eagles 264/2 *"
    },
    {
      "unique_id": "973379",
      "description": "Mountaineers 136/10  v Matabeleland Tuskers 42 *",
      "title": "Mountaineers 136/10  v Matabeleland Tuskers 42 *"
    },
    {
      "unique_id": "959221",
      "description": "Islamabad United v Quetta Gladiators",
      "title": "Islamabad United v Quetta Gladiators"
    }
  ],
  "provider": {
    "pubDate": "2016-02-23T14:01:01.467Z",
    "source": "http://www.cricinfo.com/",
    "url": "http://crm.wherrelz.com/"
  }
}

我需要在我的PHP网站上显示标题和数据。

Answer 1

非常简单：

$raw_json = file_get_contents('data.json');
$array = json_decode($raw_json);

foreach ($array->data AS $data) {
    echo $data->title;
    echo '<br />';
    echo $data->description;
    echo '<br />';
}

文件data.json当然包含你的json。

Answer 2

您可以使用json_decode解码数据，您可以在此处找到完整的文档：http://php.net/manual/en/function.json-decode.php

就代码示例而言，您可以执行以下操作：

$decoded = json_decode($json);

if (json_last_error() !== JSON_ERROR_NONE) {
    // Do something when you don't have valid json.
}

foreach ($decoded->data as $data) {
    echo $data->unique_id;
    echo $data->title;
}

json_last_error()将允许您捕获任何无效的JSON或json_decode()可能遇到的任何其他潜在错误，因此我绝对建议使用它并确保您在执行之前已经解码了有效的json任何事情。

来自JSON数组的PHP

2 个答案: