SQLite:无法在索引1处绑定参数,因为索引超出范围。该语句具有0个参数


70

我收到以下错误,我不知道为什么会发生。我想知道是否还有其他人可以阐明这个问题。

12-25 22:52:50.252: E/AndroidRuntime(813): Caused by: java.lang.IllegalArgumentException: Cannot bind argument at index 1 because the index is out of range.  The statement has 0 parameters.
12-25 22:52:50.252: E/AndroidRuntime(813):  at android.database.sqlite.SQLiteProgram.bind(SQLiteProgram.java:212)
12-25 22:52:50.252: E/AndroidRuntime(813):  at android.database.sqlite.SQLiteProgram.bindString(SQLiteProgram.java:166)
12-25 22:52:50.252: E/AndroidRuntime(813):  at android.database.sqlite.SQLiteProgram.bindAllArgsAsStrings(SQLiteProgram.java:200)
12-25 22:52:50.252: E/AndroidRuntime(813):  at android.database.sqlite.SQLiteDirectCursorDriver.query(SQLiteDirectCursorDriver.java:47)
12-25 22:52:50.252: E/AndroidRuntime(813):  at android.database.sqlite.SQLiteDatabase.rawQueryWithFactory(SQLiteDatabase.java:1314)
12-25 22:52:50.252: E/AndroidRuntime(813):  at android.database.sqlite.SQLiteDatabase.queryWithFactory(SQLiteDatabase.java:1161)
12-25 22:52:50.252: E/AndroidRuntime(813):  at android.database.sqlite.SQLiteDatabase.query(SQLiteDatabase.java:1032)
12-25 22:52:50.252: E/AndroidRuntime(813):  at android.database.sqlite.SQLiteDatabase.query(SQLiteDatabase.java:1200)

代码在这里:

public Player getPlayer(String name) {
    SQLiteDatabase db = this.getReadableDatabase();

    String[] projection = {
            PlayerEntry.COLUMN_NAME_PLAYER_NAME,
            PlayerEntry.COLUMN_NAME_PLAYED_GAMES,
            };

    String selection =  PlayerEntry.COLUMN_NAME_PLAYER_NAME ;
    String[] selectionArgs = new String[1];
    selectionArgs[0] = name;

    Cursor cursor = db.query(
            PlayerEntry.TABLE_NAME,  // The table to query
            projection,                               // The columns to return
            selection,                                // The columns for the WHERE clause
            selectionArgs,                            // The values for the WHERE clause
            null,                                     // don't group the rows
            null,                                     // don't filter by row groups
            null                                 // The sort order
            );

    if (cursor != null)
        cursor.moveToFirst();

3
根据我对querymethod的阅读,似乎selection不是列名,而是可能包含?对的引用的完整WHERE子句selectionArgs。也就是说,selection应该像一样"PlayerName=?"selectionArgs是一个值的列表(如您的代码中所示)。
安东·科瓦连科

Answers:


113

selection应当是一个表达式,并且selectionArgs应具有与中的?文字占位符一样多的元素selection

selection不是表达式,也没有任何表达式,?但是您有一个元素selectionArgs

您可能想要类似的东西:

String selection =  PlayerEntry.COLUMN_NAME_PLAYER_NAME + "=?";

使它成为与玩家名称列匹配的表达式,再加上您要绑定的文字selectionArgs[0]


50
就我而言,我在这样的单引号中有问号'?'。删除单引号可以解决该错误。
theblang 2014年

4
@mattblang我认为将您的评论添加到新答案中是一个好主意,这样可以帮助更多像我一样遇到此问题的人。想要尝试一下吗?
滕腾宝

1
@theblang同样的问题导致了相当长的时间...嘿,如果我们需要在绑定的预备变量周围加上引号怎么办?例如,何时匹配字符串?
瓦伦

0

就我而言,我在单引号内有问号,例如“?”。删除单引号可以解决该错误。

摘自上述theblang的评论。

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