将字符串转换为JSON对象php

Question

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)"} 
    ]
}

当前是php中的字符串，有没有简单的方法可以将其转换为JSON对象（我知道它已经是JSON形式）。

我需要将其作为对象，以便可以添加额外的项目/元素/对象，例如已经存在的坐标

编辑：对不起，我粘贴了旧的/错误的字符串！

Answer 1

@deceze说的是正确的，看来您的JSON格式错误，请尝试以下操作：

{
    "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到的字符串转换成对象（stdClass）或数组：http://php.net/manual/en/function.json-
decode.php

[编辑]

我不明白您所说的 “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，然后再将其转换回JSON？

假设您具有以下变量：

$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它比使用PHP-Array更复杂，然后尝试一下（使用结合使用第二个参数true）：

$manage = json_decode($data, true);

这样，您可以使用数组函数：http :
//php.net/manual/en/function.array.php

添加一个项目：

$manage = json_decode($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);
}

希望我理解您的问题。

祝好运。