我有一个点击监听器,在点击监听器中,我有一个if,else语句,如下所示
@Override
public void onClick(View v) {
postMethodRegister();
System.out.println("mSuccess is set to " + mSuccess);
if (mSuccess) {
Intent login = new Intent(Login.this, MainActivity.class);
startActivity(login);
}else{
Toast toast = Toast.makeText(getApplicationContext(), "Please Check your UserName or Password", Toast.LENGTH_LONG);
toast.setGravity(Gravity.CENTER | Gravity.CENTER_HORIZONTAL, 0, 0);
toast.show();
}
}
我还有一个布尔变量,默认情况下设置为false,private boolean mSuccess;如果onResponse被执行我将其设置为true,否则我将其设置为false并执行相应的响应,但由于某种原因,第一次登录尝试运行onResponse,但是当它到达onClick时mSuccess被设置为这是怎么回事? mSuccess是一个实例变量。
public void postMethodRegister() {
StringRequest request = new StringRequest(Request.Method.POST, loginUrl, new Response.Listener<String>() {
@Override
public void onResponse(String response) {
System.out.println("On Response was thrown in Login"+response);
mSuccess=true;
}
}, new Response.ErrorListener() {
@Override
public void onErrorResponse(VolleyError error) {
System.out.println("Error was thrown in Login" + error.getMessage()+"and the response code is ");
mSuccess=false;
}
})
答案 0 :(得分:1)
postMethodRegister()
方法发出异步请求。
因此,当您点击按钮onClick()
时,会调用postMethodRegister()
来调用postMethodRegister()
。
然后postMethodRegister()
发出异步网络请求,这意味着您的网络请求将在另一个线程中运行,然后您的控件将立即移至System.out.println("mSuccess is set to " + mSuccess);
onResponse()
之后的下一行。现在,由于您没有得到mSuccess
的回复,因此tvEvidenceID[i] = new TextView(context);
将保持不变(并且将为false)。