Question

pattern = /domain\.com\/go\/(.*)3/i;

//alias always ends with "3"
url1 = "domain.com/go/firstname3";
url2 = "domain.com/go/secondname3?p=123";

console.log(url1.match(pattern)[1]); //firstname
console.log(url2.match(pattern)[1]); //secondname3?p=12

在两个例子中，我都需要返回参数，而不需要＆＃34; 3＆＃34; （＆＃34;名字＆＃34;，＆＃34;第二名＆＃34;）。这意味着我必须找到char＆＃34; 3＆＃34;的第一个条目。第一个例子正常工作

我该如何更改模式？

感谢。

Answer 1

使用GC代替\w*

.*

或

pattern = /domain\.com\/go\/(\w*)3/i;

Answer 2

这是因为您正在重复捕获组，并且所有语言实现仅保留最后一个捕获组。换句话说，如果在捕获secondname和secondname3?p=12之间有选择，Javascript将选择最后一个，即使两者同样有效。 This answer解释了问题的核心。

解决此类问题的正确方法是重写正则表达式组，使您想要检索的参数明确无误 - 通常通过更改.*的出现次数来完成（注意它是.个字符这是有问题的）更适合的事情，例如\w*。您希望避免重复捕获组。我将解释为什么\w这样做。

这里\w表示单词元字符 - 它与a-z，A-Z和0-9匹配，没有别的。在这种特殊情况下，正则表达式无法再与secondname3?p=12匹配，因为它包含=，而\w不允许package "name"; import android.app.Activity; import android.content.Context; import android.database.Cursor; import android.database.sqlite.SQLiteDatabase; import android.os.Bundle; import android.support.v4.view.PagerAdapter; import android.support.v4.view.ViewPager; import android.util.Log; import android.view.LayoutInflater; import android.view.View; import android.view.ViewGroup; import android.widget.TextView; import java.util.ArrayList; import java.util.List; /** * Created by name on 26-Mar-16. */ public class Test extends Activity { private static final String DB_NAME = ""; private static final String TABLE_NAME = ""; private SQLiteDatabase database; private List<String> testContactName = new ArrayList<>(); public MyAdapter mAdapter; @Override protected void onCreate(Bundle savedInstanceState) { super.onCreate(savedInstanceState); setContentView(R.layout.test); ViewPager mViewPager = (ViewPager) findViewById(R.id.testvp); mAdapter = new MyAdapter(); mViewPager.setAdapter(mAdapter); mViewPager.setCurrentItem(0); mViewPager.setOffscreenPageLimit(50); getAllContacts(); } public void getAllContacts() { DatabaseHelper dbOpenHelper = new DatabaseHelper(this, DB_NAME); database = dbOpenHelper.openDataBase(); String query = "select * from " + TABLE_NAME; Cursor cursor = database.rawQuery(query, null); if (cursor != null) { cursor.moveToFirst(); for (int c = 0; c < cursor.getCount(); c++) { String name = cursor.getString(cursor.getColumnIndex("content")); cursor.moveToNext(); /* String count= String.valueOf(cursor.getCount()); Toast.makeText(getApplicationContext(), count , Toast.LENGTH_LONG).show();*/ testContactName.add(name); // testContactName.notifyAll(); Log.i("Name: " + "" + "---", name); }mAdapter.notifyDataSetChanged(); } cursor.close(); } private class MyAdapter extends PagerAdapter { public MyAdapter() { super(); } @Override public int getCount() { return testContactName.size(); } @Override public boolean isViewFromObject(View collection, Object object) { return object == collection; } @Override public Object instantiateItem(ViewGroup collection, int position) { LayoutInflater inflater = (LayoutInflater) collection.getContext() .getSystemService(Context.LAYOUT_INFLATER_SERVICE); View view = inflater.inflate(R.layout.test1, null); TextView pagenumber; pagenumber = (TextView) view.findViewById(R.id.tv_test1); try { pagenumber.setText(testContactName.get(position)); } catch (Exception e1) { // TODO Auto-generated catch block e1.printStackTrace(); } ( collection).addView(view); return view; } @Override public void destroyItem(ViewGroup collection, int position, Object view) { ( collection).removeView((View) view); mAdapter.notifyDataSetChanged(); } } }。因此，为什么@Avinash Raj的解决方案有效 - 即使重复捕获组，整个事物只有一个匹配的实例。

正则表达式首次进入

2 个答案: