您好,我正在使用凌空库从android发送一个json对象。我无法在php中收到此json对象。我通过回显ING我的Json数据进行检查,可以在“ OnResponse方法”中看到该对象。如果有人可以帮助我解决这个问题,那将是我的荣幸。我欠你很多债。这是我的代码->
Andoird Volley代码->
Requirement already satisfied: scrapy in c:\users\sameer\appdata\local\programs\python\python35\lib\site-packages (1.5.0)
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Requirement already satisfied: cffi>=1.7; platform_python_implementation != "PyPy" in c:\users\sameer\appdata\local\programs\python\python35\lib\site-packages (from cryptography>=2.2.1->pyOpenSSL->scrapy) (1.11.5)
Requirement already satisfied: asn1crypto>=0.21.0 in c:\users\sameer\appdata\local\programs\python\python35\lib\site-packages (from cryptography>=2.2.1->pyOpenSSL->scrapy) (0.24.0)
Requirement already satisfied: setuptools in c:\users\sameer\appdata\local\programs\python\python35\lib\site-packages (from zope.interface>=4.4.2->Twisted>=13.1.0->scrapy) (39.2.0)
Requirement already satisfied: pycparser in c:\users\sameer\appdata\local\programs\python\python35\lib\site-packages (from cffi>=1.7; platform_python_implementation != "PyPy"->cryptography>=2.2.1->pyOpenSSL->scrapy) (2.18)
Json格式为->
private void registerUser() {
JSONObject postObject = new JSONObject();
RequestQueue queue = Volley.newRequestQueue(this);
JSONObject historyObject = new JSONObject();
String url ="http://helpinghandbd.org/app/index.php";
try {
//historyObject.put("id","1");
historyObject.put("email","1234");
historyObject.put("password","1234");
postObject.put("user",historyObject);
} catch (JSONException e) {
e.printStackTrace();
}
Log.e("LoginActivityJsonObject",""+postObject);
JsonObjectRequest objRequest = new JsonObjectRequest(Request.Method.POST, url,postObject,
new Response.Listener<JSONObject>() {
@Override
public void onResponse(JSONObject response) {
Log.e("LoginActivity","OnResponse: "+response);
Toast.makeText(LoginActivity.this, String.valueOf(response), Toast.LENGTH_LONG).show();
}
}, new Response.ErrorListener() {
@Override
public void onErrorResponse(VolleyError error) {
Log.e("OnError", String.valueOf(error.getMessage()));
}
});
queue.add(objRequest);
}
最后,PHP代码是->
{ 'user':{
'email':'1234',
'password':'1234'
}
}
?>
我无法在PHP中接收Json对象。预先感谢。
答案 0 :(得分:0)
在您的php
代码中,更改
$decode = json_decode($data,true);
$email = $decode->user['email'];
$password = $decode->user['passowrd'];
到
$decode = json_decode($data,true);
$email = $decode['user']['email'];
$password = $decode['user']['passowrd'];