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)"}
]
}
目前在 PHP 中是一个字符串。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?
我需要它是一个对象,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。
最佳答案
@deceze 说的是对的,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)"
}]
}
使用 json_decode
将 String 转换为 Object (stdClass
) 或数组:http://php.net/manual/en/function.json-decode.php
[编辑]
我不明白您所说的“官方 JSON 对象”是什么意思,但假设您想通过 PHP 将内容添加到 json,然后将其直接转换回 JSON?
假设你有以下变量:
$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)"}]}';
你应该把它转换成Object(stdClass):
$manage = json_decode($data);
但是使用 stdClass
比 PHP-Array 更复杂,那么试试这个(使用第二个参数和 true
):
$manage = json_decode($data, true);
这样你可以使用数组函数: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);
}
希望我能理解你的问题。
祝你好运。
关于php - 如何在 PHP 中将字符串转换为 JSON 对象,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/17488207/