PHP MySQL INSERT语句语法错误(PHP MySQL INSERT statement syntax error)

 发布的问题对我来说并不是那么清楚，您可以随时切换activityForResult并在onActivityResult方法中检查活动结果，检查不同的请求代码。 这是一个代码演示，如何做到： 
 
 使用此切换活动： 
 
Intent intent = new Intent(this, yourClass.class); 
intent.putExtras(b);    // here
startActivityForResult(intent, 2); //put your code along : positive integer
 
 检查此方法的结果 
 
@Override
protected void onActivityResult(int requestCode, int resultCode, Intent data) {
    // TODO Auto-generated method stub
    super.onActivityResult(requestCode, resultCode, data);

    //could replace it with a switch
    if (requestCode == 1){
        //put your code here
            }
}
 
 被调用的活动不需要知道您的请求代码，但是，如果您想要执行类似的操作，可以通过在intent中传递请求代码来实现，如下所示： 
 
intent.putExtra("requestCode", requestCode); 
 
 因此，访问您切换到的活动类中的intent变量。 

The question posted was not so clear to me, You can always switch an activityForResult and check for activity result in onActivityResult method checking different request codes. Here is a code demonstration, how to do it:
 
Switch activity using this:
 
Intent intent = new Intent(this, yourClass.class); 
intent.putExtras(b);    // here
startActivityForResult(intent, 2); //put your code along : positive integer
 
Check for result in this method
 
@Override
protected void onActivityResult(int requestCode, int resultCode, Intent data) {
    // TODO Auto-generated method stub
    super.onActivityResult(requestCode, resultCode, data);

    //could replace it with a switch
    if (requestCode == 1){
        //put your code here
            }
}
 
The called activity does not need to know your requestcode, however, if you want to do something like that, you can do that by passing your request code in intent, like this:
 
intent.putExtra("requestCode", requestCode); 
 
Hence, access the intent variable in the activity class you switched onto..