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我的 URL 在浏览器中返回 Json 但在 Postman 中不起作用

我有一个测试URL,它返回一个 Json 结果如下

{"result":[{"Users":{"id":"1","user_name":"raj","user_password":"rajesh123"}},{"Users":{"id":"2","user_name":"raj","user_password":"rajesh123"}},{"Users":{"id":"3","user_name":"raj","user_password":"rajesh123"}},{"Users":{"id":"4","user_name":"raj","user_password":"rajesh123"}},{"Users":{"id":"5","user_name":"raj","user_password":"rajesh123"}},{"Users":{"id":"6","user_name":"raj","user_password":"rajesh123"}},{"Users":{"id":"7","user_name":"raj","user_password":"rajesh123"}},{"Users":{"id":"8","user_name":"raj","user_password":"rajesh123"}},{"Users":{"id":"9","user_name":"raj","user_password":"rajesh123"}},{"Users":{"id":"10","user_name":"raj","user_password":"rajesh123"}},{"Users":{"id":"11","user_name":"raj","user_password":"rajesh123"}},{"Users":{"id":"12","user_name":"raj","user_password":"rajesh123"}}],"error_code":1}
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我的问题是当我在Chrome 浏览器上运行这个URL 时它返回正确的响应如上所述。但是邮递员的情况下它返回错误并显示以下消息

本网站需要 Javascript 才能运行,请在您的浏览器中启用 Javascript 或使用支持 Javascript 的浏览器

PS:在 localhost 它 [ http://localhost/test/get_user.php]浏览器和 PostMan中都可以正常工作

php json localhost postman server

2
推荐指数
1
解决办法
2913
查看次数

Android简单计算器崩溃

我是android开发新手,并做了一个简单的计算器项目.但是,我的应用程序不断崩溃,我不明白为什么.请帮忙

Java代码:

package com.example.zhiwen.calculator;
import android.support.v7.app.AppCompatActivity;
import android.os.Bundle;
import android.view.View;
import android.widget.Button;
import android.widget.EditText;
import android.widget.TextView;

public class MainActivity extends AppCompatActivity {

Button plus,minus,times,divide;
TextView textview3;
EditText first, second;
double no1 = 0, no2 = 0;
double answer = 0;

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

    plus = (Button) findViewById(R.id.plus);
    minus = (Button) findViewById(R.id.minus);
    times = (Button) findViewById(R.id.times);
    divide = (Button) findViewById(R.id.divide);
    textview3 = (TextView) findViewById(R.id.textview3);
    first = (EditText) findViewById(R.id.editText);
    second = (EditText) findViewById(R.id.editText2);
}

public void ClickMeButton(View …
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java android

-1
推荐指数
1
解决办法
128
查看次数

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