86

我从 SQL 查询中得到以下结果：

{"Coords":[
    {"Accuracy":"65","Latitude":"53.277720488429026","Longitude":"-9.012038778269686","Timestamp":"Fri Jul 05 2013 11:59:34 GMT+0100 (IST)"},
    {"Accuracy":"65","Latitude":"53.277720488429026","Longitude":"-9.012038778269686","Timestamp":"Fri Jul 05 2013 11:59:34 GMT+0100 (IST)"},
    {"Accuracy":"65","Latitude":"53.27770755361785","Longitude":"-9.011979642121824","Timestamp":"Fri Jul 05 2013 12:02:09 GMT+0100 (IST)"},
    {"Accuracy":"65","Latitude":"53.27769091555766","Longitude":"-9.012051410095722","Timestamp":"Fri Jul 05 2013 12:02:17 GMT+0100 (IST)"},
    {"Accuracy":"65","Latitude":"53.27769091555766","Longitude":"-9.012051410095722","Timestamp":"Fri Jul 05 2013 12:02:17 GMT+0100 (IST)"} 
    ]
}

它目前是 PHP 中的字符串。我知道它已经是 JSON 形式了，有没有一种简单的方法可以将它转换为 JSON 对象？

我需要它成为一个对象，这样我就可以添加一个额外的项目/元素/对象，就像“坐标”已经是一样。

4

4 回答 4

148

@deceze 说的是正确的，看来你的 JSON 格式不正确，试试这个：

{
    "Coords": [{
        "Accuracy": "30",
        "Latitude": "53.2778273",
        "Longitude": "-9.0121648",
        "Timestamp": "Fri Jun 28 2013 11:43:57 GMT+0100 (IST)"
    }, {
        "Accuracy": "30",
        "Latitude": "53.2778273",
        "Longitude": "-9.0121648",
        "Timestamp": "Fri Jun 28 2013 11:43:57 GMT+0100 (IST)"
    }, {
        "Accuracy": "30",
        "Latitude": "53.2778273",
        "Longitude": "-9.0121648",
        "Timestamp": "Fri Jun 28 2013 11:43:57 GMT+0100 (IST)"
    }, {
        "Accuracy": "30",
        "Latitude": "53.2778339",
        "Longitude": "-9.0121466",
        "Timestamp": "Fri Jun 28 2013 11:45:54 GMT+0100 (IST)"
    }, {
        "Accuracy": "30",
        "Latitude": "53.2778159",
        "Longitude": "-9.0121201",
        "Timestamp": "Fri Jun 28 2013 11:45:58 GMT+0100 (IST)"
    }]
}

用于json_decode将 String 转换为 Object ( stdClass) 或数组： http: //php.net/manual/en/function.json-decode.php

[编辑]

我不明白“官方 JSON 对象”是什么意思，但是假设您想通过 PHP 将内容添加到 json 中，然后将其直接转换回 JSON？

假设您有以下变量：

$data = '{"Coords":[{"Accuracy":"65","Latitude":"53.277720488429026","Longitude":"-9.012038778269686","Timestamp":"Fri Jul 05 2013 11:59:34 GMT+0100 (IST)"},{"Accuracy":"65","Latitude":"53.277720488429026","Longitude":"-9.012038778269686","Timestamp":"Fri Jul 05 2013 11:59:34 GMT+0100 (IST)"},{"Accuracy":"65","Latitude":"53.27770755361785","Longitude":"-9.011979642121824","Timestamp":"Fri Jul 05 2013 12:02:09 GMT+0100 (IST)"},{"Accuracy":"65","Latitude":"53.27769091555766","Longitude":"-9.012051410095722","Timestamp":"Fri Jul 05 2013 12:02:17 GMT+0100 (IST)"},{"Accuracy":"65","Latitude":"53.27769091555766","Longitude":"-9.012051410095722","Timestamp":"Fri Jul 05 2013 12:02:17 GMT+0100 (IST)"}]}';

您应该将其转换为对象（stdClass）：

$manage = json_decode($data);

但是使用stdClass比 PHP-Array 更复杂，然后试试这个（使用第二个参数true）：

$manage = json_decode($data, true);

这样你就可以使用数组函数： http: //php.net/manual/en/function.array.php

添加项目：

$manage = json_decode($data, true);

echo 'Before: <br>';
print_r($manage);

$manage['Coords'][] = Array(
    'Accuracy' => '90'
    'Latitude' => '53.277720488429026'
    'Longitude' => '-9.012038778269686'
    'Timestamp' => 'Fri Jul 05 2013 11:59:34 GMT+0100 (IST)'
);

echo '<br>After: <br>';
print_r($manage);

删除第一项：

$manage = json_decode($data, true);
echo 'Before: <br>';
print_r($manage);
array_shift($manage['Coords']);
echo '<br>After: <br>';
print_r($manage);

您想将 json 保存到数据库或文件的任何机会：

$data = '{"Coords":[{"Accuracy":"65","Latitude":"53.277720488429026","Longitude":"-9.012038778269686","Timestamp":"Fri Jul 05 2013 11:59:34 GMT+0100 (IST)"},{"Accuracy":"65","Latitude":"53.277720488429026","Longitude":"-9.012038778269686","Timestamp":"Fri Jul 05 2013 11:59:34 GMT+0100 (IST)"},{"Accuracy":"65","Latitude":"53.27770755361785","Longitude":"-9.011979642121824","Timestamp":"Fri Jul 05 2013 12:02:09 GMT+0100 (IST)"},{"Accuracy":"65","Latitude":"53.27769091555766","Longitude":"-9.012051410095722","Timestamp":"Fri Jul 05 2013 12:02:17 GMT+0100 (IST)"},{"Accuracy":"65","Latitude":"53.27769091555766","Longitude":"-9.012051410095722","Timestamp":"Fri Jul 05 2013 12:02:17 GMT+0100 (IST)"}]}';

$manage = json_decode($data, true);

$manage['Coords'][] = Array(
    'Accuracy' => '90'
    'Latitude' => '53.277720488429026'
    'Longitude' => '-9.012038778269686'
    'Timestamp' => 'Fri Jul 05 2013 11:59:34 GMT+0100 (IST)'
);

if (($id = fopen('datafile.txt', 'wb'))) {
    fwrite($id, json_encode($manage));
    fclose($id);
}

我希望我已经理解了你的问题。

祝你好运。

于 2013-07-05T12:26:17.717 回答
22

要将有效的 JSON 字符串转换回来，您可以使用该json_decode()方法。

要将其转换回对象，请使用以下方法：

$jObj = json_decode($jsonString);

并将其转换为关联数组，将第二个参数设置为true：

$jArr = json_decode($jsonString, true);

顺便说一句，将您提到的字符串转换回其中任何一个，您应该有一个有效的 JSON 字符串。要实现它，您应该执行以下操作：

  1. 在Coords数组中，删除"对象开头和结尾的两个（双引号）。
  2. 数组中的对象以逗号分隔 ( ,)，因此请在数组中的对象之间添加逗号Coords。

你将有一个有效的 JSON 字符串..

这是我转换为有效字符串的 JSON 字符串：http: //pastebin.com/R16NVerw

于 2013-07-05T12:13:56.833 回答
9

例如，您可以使用它

$array = json_decode($string,true)

但之前验证Json。您可以从http://jsonviewer.stack.hu/进行验证

于 2013-07-05T12:26:20.967 回答
-1

尝试使用json_encode().

再看看Valid JSON

于 2013-07-05T12:01:19.073 回答