Android 无法调用 SQLite WHERE column LIKE '%?%' The statement has 0 parameters

Question

用这段代码

public static List<App> findByKeyword(Context context, String keyword){
   ContentResolver resolver = context.getContentResolver();
   Uri uri = getContentUri();
   String[] projection = DataColumns.ALL;
   String selection = DataColumns.NAME+" like '%?%' ";
   String[] selectionArgs = {keyword};
   Cursor cursor = resolver.query(uri, projection, selection, selectionArgs, DEFAULT_ORDER );
   return cursorToList(cursor);
}

我收到错误

Caused by: java.lang.IllegalArgumentException: Cannot bind argument at index 1 because the index is out of range. The statement has 0 parameters.

SQLite 原始查询应该是 SELECT * FROM table WHERE name LIKE '%?%'但是通过 ContentResolver 查询 API 使用参数存在问题

我也试过使用 " 引号

String selection = DataColumns.NAME+" like \"%?%\" ";

它还说

The statement has 0 parameters.

Answer 1

反过来试试:

String selection = DataColumns.NAME+" like ? ";
String[] selectionArgs = new String[]{ "%"+keyword+"%" };

希望对你有帮助