将字符串转换为JSON对象php

将字符串转换为JSON对象php

@deceze说的是正确的，看来您的JSON格式错误，请尝试以下 *** 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_depre

到的字符串转换成对象（

stdClass

）或数组：http://php.net/manual/en/function.json-
depre.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）：

$manage = json_depre($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

）：

$manage = json_depre($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：

$manage = json_depre($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：

$manage = json_depre($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_depre($data, true);$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_enpre($manage));    fclose($id);}

希望我理解您的问题。

祝好运。