更新时间:2022-04-10 04:39:29
@deceze说的是正确的,看来您的JSON格式不正确,请尝试以下操作:
What @deceze said is correct, it seems that your JSON is malformed, try this:
{
"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/zh/function.json-decode.php
Use json_decode
to convert String into Object (stdClass
) or array: http://php.net/manual/en/function.json-decode.php
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向json添加内容,然后将其直接转换回JSON?
I did not understand what do you mean by "an official JSON object", but suppose you want to add content to json via PHP and then converts it right back to 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)"}]}';
您应将其转换为 Object (stdClass):
You should convert it to Object (stdClass):
$manage = json_decode($data);
但是使用stdClass
的工作要比PHP-Array更复杂,然后尝试一下(使用true
的第二个参数):
But working with stdClass
is more complicated than PHP-Array, then try this (use second param with true
):
$manage = json_decode($data, true);
通过这种方式,您可以使用数组函数: http://php.net/manual/zh/function.array.php
This way you can use array functions: 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);
您想保存到json到数据库或文件的任何机会:
any chance you want to save to json to a database or a file:
$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.
I hope I have understood your question.
祝你好运.