gpt4 book ai didi

php - 如何使用 android 中的 volley 库将数据保存到 mysql?

转载 作者:行者123 更新时间:2023-11-30 22:36:28 24 4
gpt4 key购买 nike

我尝试使用来自 localhost 的 volley 库将数据保存到 mysql 数据库中。当我尝试获取 json 并将一些数据添加到 mysql 时出现错误,不幸的是应用程序停止了。显示错误:Java.NullPointerException 在此行

Controller.getInstance().addToReqQueue(request);

这是我的代码:

public class MainActivity extends AppCompatActivity {

EditText firstname, lastname, age;
Button insert, show;
TextView result;
String insertUrl = "http://192.168.56.1/android_post_api/insertStudent.php";
String showUrl = "http://192.168.56.1/android_post_api/show.php";

@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);

firstname = (EditText) findViewById(R.id.editText);
lastname = (EditText) findViewById(R.id.editText2);
age = (EditText) findViewById(R.id.editText3);
insert = (Button) findViewById(R.id.insert);
show = (Button) findViewById(R.id.showstudents);
result = (TextView) findViewById(R.id.textView);
show.setOnClickListener(new View.OnClickListener() {
@Override
public void onClick(View v) {
System.out.println("ww");
JsonObjectRequest jsonObjectRequest = new JsonObjectRequest(Request.Method.POST,
showUrl, new Response.Listener<JSONObject>() {
@Override
public void onResponse(JSONObject response) {
System.out.println(response.toString());
try {
JSONArray students = response.getJSONArray("students");
for (int i = 0; i < students.length(); i++) {
JSONObject student = students.getJSONObject(i);

String firstname = student.getString("firstname");
String lastname = student.getString("lastname");
String age = student.getString("age");

result.append(firstname + " " + lastname + " " + age + " \n");
}
result.append("===\n");

} catch (JSONException e) {
e.printStackTrace();
}

}
}, new Response.ErrorListener() {
@Override
public void onErrorResponse(VolleyError error) {
System.out.append(error.getMessage());

}
});
Controller.getInstance().addToReqQueue(jsonObjectRequest);
}
});

insert.setOnClickListener(new View.OnClickListener(){

@Override
public void onClick(View v) {
StringRequest request = new StringRequest(Request.Method.POST, insertUrl, new Response.Listener<String>() {
@Override
public void onResponse(String response) {
System.out.println(response.toString());

}
}, new Response.ErrorListener() {

@Override
public void onErrorResponse(VolleyError error) {

}
}) {
@Override
protected Map<String, String> getParams() throws AuthFailureError {
Map<String,String> parameters = new HashMap<String, String>();
parameters.put("firstname",firstname.getText().toString());
parameters.put("lastname",lastname.getText().toString());
parameters.put("age",age.getText().toString());
return parameters;
}

};
Controller.getInstance().addToReqQueue(request);
}
});
}
}

最佳答案

你可以引用我下面的代码:

        JSONObject jsonBody;
try {
jsonBody = new JSONObject();
jsonBody.put("Title", "Android Volley Demo");
jsonBody.put("Author", "BNK");
jsonBody.put("Date", "2015/09/06");
mRequestBody = jsonBody.toString();

StringRequest stringRequest = new StringRequest(Request.Method.POST, url, new Response.Listener<String>() {
@Override
public void onResponse(String response) {
mTextView.setText(response);
}
}, new Response.ErrorListener() {
@Override
public void onErrorResponse(VolleyError error) {
mTextView.setText(error.toString());
}
}) {
@Override
public String getBodyContentType() {
return "application/json; charset=utf-8";
}

@Override
public byte[] getBody() throws AuthFailureError {
try {
return mRequestBody == null ? null : mRequestBody.getBytes("utf-8");
} catch (UnsupportedEncodingException uee) {
VolleyLog.wtf("Unsupported Encoding while trying to get the bytes of %s using %s",
mRequestBody, "utf-8");
return null;
}
}
};
MySingleton.getInstance(this).addToRequestQueue(stringRequest);
} catch (JSONException e) {
e.printStackTrace();
}

关于php - 如何使用 android 中的 volley 库将数据保存到 mysql?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/32418516/

24 4 0
Copyright 2021 - 2024 cfsdn All Rights Reserved 蜀ICP备2022000587号
广告合作:1813099741@qq.com 6ren.com