如何在 PHP 中将字符串转换为 JSON 对象

我从一个 SQL 查询得到以下结果:

{"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)"}
]
}

它当前是 PHP 中的字符串。我知道它已经是 JSON 格式的了，有没有一种简单的方法可以将它转换成 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。

497314 次浏览
小开

Try with json_encode().

And look again Valid JSON

小开

To convert a valid JSON string back, you can use the json_decode() method.

To convert it back to an object use this method:

$jObj = json_decode($jsonString);

And to convert it to a associative array, set the second parameter to true:

$jArr = json_decode($jsonString, true);

By the way to convert your mentioned string back to either of those, you should have a valid JSON string. To achieve it, you should do the following:

  1. In the Coords array, remove the two " (double quote marks) from the start and end of the object.
  2. The objects in an array are comma seprated (,), so add commas between the objects in the Coords array..

And you will have a valid JSON String..

Here is your JSON String I converted to a valid one: http://pastebin.com/R16NVerw

小开
最佳答案

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)"
}]
}

Use json_decode to convert String into Object (stdClass) or array: http://php.net/manual/en/function.json-decode.php

[edited]

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?

assuming you have the following variable:

$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)"}]}';

You should convert it to Object (stdClass):

$manage = json_decode($data);

But working with stdClass is more complicated than PHP-Array, then try this (use second param with true):

$manage = json_decode($data, true);

This way you can use array functions: http://php.net/manual/en/function.array.php

adding an item:

$manage = json_decode($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);

remove first item:

$manage = json_decode($data, true);
echo 'Before: <br>';
print_r($manage);
array_shift($manage['Coords']);
echo '<br>After: <br>';
print_r($manage);

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);
}

I hope I have understood your question.

Good luck.

小开

you can use this for example

$array = json_decode($string,true)

but validate the Json before. You can validate from http://jsonviewer.stack.hu/

小开

If you don't pass second parameter, or pass false, json_decode() will parse JSON to stdClass object therefore you can use "->" arrow notation to access the object properties.

<?php


// Store JSON data in a PHP variable
$json = '{"email":"john@doe.com"}';


$obj = json_decode($json, false);
print $obj->email;


?>

Output

john@doe.com
小开

@Miro Markaravanes save me tons of time of this. In my case I stored the string of a JSON object similar to OP's in my database. When calling it back it had double quotes at the beginning and end of the string.

Due to this every time I tried to using json_decode it wouldn't work.

So I did the following:

    $data = substr($data, 1);
$data = substr($data, 0, -1);
$data = json_decode($data);

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