Android 无法调用 SQLite WHERE column LIKE '%?%' 该语句有 0 个参数 [英] Android fails to call SQLite WHERE column LIKE '%?%' The statement has 0 parameters

问题描述

使用此代码

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);
}

出现错误

引起:java.lang.IllegalArgumentException:无法在索引 1 处绑定参数，因为索引超出范围.该语句有 0 个参数.

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 '%?%'但是存在通过 ContentResolver 查询 API 使用参数的问题

The SQLite raw query should be SELECT * FROM table WHERE name LIKE '%?%' but there is problem of using parameter though ContentResolver query API

我也尝试使用 " 引号

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

它还说

该语句有 0 个参数.

The statement has 0 parameters.

推荐答案

换个角度试试:

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

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