为什么在“WHERE”附近的查询中出现语法错误？(多桌)

debugcn 发表于 Dev

佐藤绫

我还是个新手，这是我第一次使用这些多表查询。为什么我有这样的错误？这是我的代码：

 String selectQuery =
            " select *" +
            "    from tableassign left outer join\n" +
            "         tableacc\n" +
            "         on tableassign.signeeid = tableacc.userid left outer join\n" +
            "         tableinfo\n" +
            "         on tableassign.signeeid = tableinfo.userid left outer join\n" +
                    " WHERE tableassign.signeedepid =?";
    Cursor data = db.rawQuery(selectQuery , new String[]{ signeedep });

错误：

android.database.sqlite.SQLiteException: near "WHERE": syntax error (code 1):
 , while compiling: select *  from tableassign left outer join  tableacc on
 tableassign.signeeid = tableacc.userid left outer join  tableinfo on 
tableassign.signeeid = tableinfo.userid left outer join WHERE 
tableassign.signeedepid =?

拉西尔·希兰

您似乎有复制/粘贴问题。删除left outer join之前的最后一个where：

String selectQuery =
        " select *" +
        "    from tableassign left outer join\n" +
        "         tableacc\n" +
        "         on tableassign.signeeid = tableacc.userid left outer join\n" +
        "         tableinfo\n" +
        "         on tableassign.signeeid = tableinfo.userid\n" +
                " WHERE tableassign.signeedepid =?";
Cursor data = db.rawQuery(selectQuery , new String[]{ signeedep });

我会以不同的方式格式化它，主要是为了可读性：

String selectQuery =
        " SELECT *\n" +
        " FROM tableassign\n" +
        " LEFT JOIN tableacc ON tableassign.signeeid = tableacc.userid\n" +
        " LEFT JOIN tableinfo ON tableassign.signeeid = tableinfo.userid\n" +
        " WHERE tableassign.signeedepid =?";
Cursor data = db.rawQuery(selectQuery , new String[]{ signeedep });

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