@deceze说的是正确的，看来您的JSON格式不正确，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_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要复杂得多，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);}我希望我理解你的问题。
祝好运。
