[英]Decode Json Object from a url in php coming from a user
hey i am making an android app where user's request in the format of json . 嘿,我正在制作一个android应用,其中用户的请求格式为json。 Their request will look like this..
他们的要求将如下所示。
JSONObject j = new JSONObject(createJson());
String url ="http://codemoirai.esy.es/test.php?UserDetails="+j;
Where j = {"Email":"code@gmail.com","Username":"xyz","Password":"xyz"}
This is what I will be sending to the test.php , i want to know how i can fetch this data and display it using php. 这是我将发送到test.php的内容,我想知道如何获取此数据并使用php显示它。
<?php
require "init1.php";
$jsonObject = $_GET["UserDetails"];
$obj = json_decode($jsonObject);
$email = $obj->Email;
$password = $obj->Password;
.......
echo $email; //Everthing i fetch from jsonObject is null. Why?
?>
ThankYou, is it correct how i am fetching in php?? 谢谢,我在php中获取的方式正确吗?
In test.php you can catch the data with $_GET['UserDetails']
then decode it with json_decode($_GET['UserDetails')
在test.php中,您可以使用
$_GET['UserDetails']
捕获数据,然后使用json_decode($_GET['UserDetails')
对其进行解码
Then you can iterate it with a foreach
loop 然后,您可以使用
foreach
循环对其进行迭代
Example: 例:
if(isset($_GET['UserDetails'])) {
$details = json_decode($_GET['UserDetails']);
foreach($details as $key => $val) {
echo $key.' = '.$val;
}
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