php 在 Codeigniter 中获取 JSON 响应
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Get JSON response In Codeigniter
提问by Jignesh.Raj
I am passing the Object from fiddler to the service written in codeigniter. My Object is something look like this :
我将对象从 fiddler 传递给用 codeigniter 编写的服务。我的对象是这样的：
My Response :
我的回复 ：
{
"GetResponse":{
"OrgName":"adfasd",
"OrgAdr1":"asdf",
"OrgAdr2":"NA",
"ProductList":[
{
"ProductID":8724,
"Pallets":0,
"Pieces":0,
"Description":"sdfsd"
}
]
}
}
What I want :
我想要的是 ：
I want to save the response as a JSON Object in codeigniter and also want to fetch the JSON Object or Array inside the main Object.
我想在 codeigniter 中将响应保存为 JSON 对象，并且还想在主对象中获取 JSON 对象或数组。
What I tried :
我试过的：
My service method in codeigniter is something like this :
我在 codeigniter 中的服务方法是这样的：
public function Save()
{
$Data = json_decode(file_get_contents('php://input'));
echo $Data;
}
But I am getting nothing in Response Bodyinside Fiddler.
但是我在 Fiddler中的Response Body 中什么也没得到。
If I use this code :
如果我使用此代码：
$Data = file_get_contents('php://input');
echo $Data;
then it is showing me the response but in the form of String. I want to save it as an JSON Object.
然后它以字符串的形式向我显示响应。我想将它保存为一个 JSON 对象。
Can anyone tell me what am I missing here ?
谁能告诉我我在这里错过了什么？
回答by Kvadiyatar
use this code:
使用此代码：
header('Content-type: application/json');
$Data = json_decode(file_get_contents('php://input'),true);
now, you will get $Dataas a array.
现在，您将获得$Data一个数组。
and you get value by $Data['name'].
然后你得到价值$Data['name']。
回答by Sakibul Alam
json_decode()parses a json string to a php variable in the form of an object or associative array. json_encode()does the reverse.
json_decode()将 json 字符串解析为对象或关联数组形式的 php 变量。json_encode()反之。
I think what is happening is that php://inputis already in json format. When you run it through json_decode()you turn it into a php object which when echoed should throw an error like
我认为正在发生的事情php://input已经是 json 格式。当你运行它时，json_decode()你把它变成一个 php 对象，当echoed 应该抛出一个错误时
Object of class stdClass could not be converted to string ....
类 stdClass 的对象无法转换为字符串 ....
If error reporting is suppressed the script stopped there and you get nothing echoed.
如果错误报告被抑制，脚本就会停在那里，你不会得到任何回应。
回答by Rajeev Ranjan
json_decode()return either an object or an array ,so your save() should be like this
json_decode()返回一个对象或一个数组，所以你的 save() 应该是这样的
public function Save()
{
$Data = json_decode(file_get_contents('php://input'));
echo '<pre>';
print_r( $Data);
}
output will be :-
输出将是：-
stdClass Object
(
[GetResponse] => stdClass Object
(
[OrgName] => adfasd
[OrgAdr1] => asdf
[OrgAdr2] => NA
[ProductList] => Array
(
[0] => stdClass Object
(
[ProductID] => 8724
[Pallets] => 0
[Pieces] => 0
[Description] => sdfsd
)
)
)
)
回答by Sherin Jose
Try this,
尝试这个，
public function Save()
{
$Data = json_decode(file_get_contents('php://input'));
var_dump($Data);
}
回答by Rohit Kumar Choudhary
To encode any value into JSON we must use json_encode(). what mistake you have done that you are decoding a non json object. So the result in null.
要将任何值编码为 JSON，我们必须使用json_encode(). 您在解码非 json 对象时犯了什么错误。所以结果为空。
public function Save()
{
$Data = json_encode(file_get_contents('php://input'));
echo $Data;
}
