吐司不行,看不出为什么不行

Question

因此,我正在尝试编写一个包含两个微调器和一个按钮的活动,当选择了两个微调器并按下按钮时,它将带您进入另一个活动.除了一个组合,它应该产生一个Toast,说你不能这样做.

无论如何,这是代码:

public void onClick(View v) {

              String spinnerchoice1 = ("spinner1Value");
              String spinnerchoice2 = ("spinner2Value");

              if((spinnerchoice1.equals("Walking")) && (spinnerchoice2.equals("Hiking"))){

                  Toast.makeText(getBaseContext(), "I'm sorry, this is not possible.", Toast.LENGTH_LONG).show();

              }else{

                  Intent i = new Intent(GetDirections.this.getApplicationContext(), DirectionDisplay.class);
                  i.putExtra("spinner1Value", transportSpinner.getSelectedItem().toString()); 
                  i.putExtra("spinner2Value", locationSpinner.getSelectedItem().toString());
                  GetDirections.this.startActivity(i);

              }

          }     

谁能告诉我哪里出错了？

谢谢

Answer 1

您正在比较两个硬编码字符串,if条件永远不会执行.将代码更改为:

public void onClick(View v) {
  String transport = transportSpinner.getSelectedItem().toString();
  String location = locationSpinner.getSelectedItem().toString();

  if ("Walking".equals(transport) && "Hiking".equals(location)) {
      Toast.makeText(getBaseContext(), "I'm sorry, this is not possible.", Toast.LENGTH_LONG).show();
  } else {
      Intent i = new Intent(GetDirections.this.getApplicationContext(), DirectionDisplay.class);
      i.putExtra("spinner1Value", transport); 
      i.putExtra("spinner2Value", location);
      GetDirections.this.startActivity(i);
  }
}