Question

我从sql查询得到以下结果：

{"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，是否有一种简单的方法将其转换为JSON对象（我知道它已经是JSON形式）。

我需要它作为一个对象，所以我可以添加一个额外的项目/元素/对象，就像coords已经是

编辑：抱歉，我发了一条旧/错误的字符串！

Answer 1

@deceze说的是正确的，似乎你的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)"
    }]
}

使用json_decode将字符串转换为对象（stdClass）或数组：http://php.net/manual/en/function.json-decode.php

<强> [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然后将其转换回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 Object （stdClass）：

$manage = json_decode($data);

但是使用stdClass比PHP-Array更复杂，然后尝试这个（使用true的第二个参数）：

$manage = json_decode($data, true);

这样您可以使用数组函数：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);

您希望将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);
}

我希望我理解你的问题。

祝你好运。

Answer 2

要转换有效的JSON字符串，您可以使用json_decode()方法。

要将其转换回对象，请使用以下方法：

$jObj = json_decode($jsonString);

要将其转换为关联数组，请将第二个参数设置为true：

$jArr = json_decode($jsonString, true);

顺便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：

在Coords数组中，从对象的开头和结尾删除两个"（双引号）。
数组中的对象以逗号分隔（,），因此在Coords数组中的对象之间添加逗号..

你将拥有一个有效的JSON字符串..

这是你转换为有效的JSON字符串：http://pastebin.com/R16NVerw

Answer 3

你可以使用

$array = json_decode($sting,true)

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

进行验证

Answer 4

尝试使用json_encode()。

再看Valid JSON

将字符串转换为JSON对象php

4 个答案: