MySQL报错解决--Parameter index out of range (1 > number of parameters, which is 0)

MySQL报错解决–Parameter index out of range (1 > number of parameters, which is 0)

今天写数据库批量删除的时候,碰到了这个错误,我靠,改了我1个半小时。(发出来长个记性)

解决方案:查看?周围的单引号或者双引号。

	public int resumedeletemore(Connection con, List<String> list) throws Exception {
		int row = 0;
		String temp1 = "";
	    //开始截取
			for (int i = 0; i < list.size(); i++) {
				temp1 +="'"+list.get(i)+"',";
			}
			temp1 = temp1.substring(0, temp1.length() - 1);
		String sql ="delete from user_resume where user_id in("+temp1+");";
         System.out.println(sql);
         PreparedStatement stmt=con.prepareStatement(sql);
          row=stmt.executeUpdate();

		return row;
	}
}
posted @ 2020-05-01 02:35  别团等shy哥发育  阅读(154)  评论(0)    收藏  举报