2023年6月21日发(作者：)

c#mysql参数化like_C#参数化SQL语句中的like和in在写项⽬的时候遇到⼀个问题，sql 语句进⾏ like in 参数化，按照正常的⽅式是⽆法实现的我们⼀般的思维是：Like 参数：string strSql = "select * from s where City like '%@add%'";SqlParameter[] Parameters=new SqlParameter[1];Parameters[0] = new SqlParameter("@add", "bre");In 参数string strSql = "select * from s where AddressID in (@add)";SqlParameter[] Parameters = new SqlParameter[1];Parameters[0] = new SqlParameter("@add", "343,372,11481,11533,11535,11755,11884,12092,12093,12143");可是这样放在程序⾥⾯是⽆法执⾏的，即使不报错，也是搜索不出来结果的，去⽹上搜索也没有⼀个明确的答案，经过反复试验，终于解决这个问题正确解法如下：like 参数string strSql = "select * from s where City like '%'+ @add + '%'";SqlParameter[] Parameters=new SqlParameter[1];Parameters[0] = new SqlParameter("@add", "bre");in 参数string strSql = "exec('select * from s where AddressID in ('+@add+')')";SqlParameter[] Parameters = new SqlParameter[1];Parameters[0] = new SqlParameter("@add", "343,372,11481,11533,11535,11755,11884,12092,12093,12143");

2023年6月21日发(作者：)

c#mysql参数化like_C#参数化SQL语句中的like和in在写项⽬的时候遇到⼀个问题，sql 语句进⾏ like in 参数化，按照正常的⽅式是⽆法实现的我们⼀般的思维是：Like 参数：string strSql = "select * from s where City like '%@add%'";SqlParameter[] Parameters=new SqlParameter[1];Parameters[0] = new SqlParameter("@add", "bre");In 参数string strSql = "select * from s where AddressID in (@add)";SqlParameter[] Parameters = new SqlParameter[1];Parameters[0] = new SqlParameter("@add", "343,372,11481,11533,11535,11755,11884,12092,12093,12143");可是这样放在程序⾥⾯是⽆法执⾏的，即使不报错，也是搜索不出来结果的，去⽹上搜索也没有⼀个明确的答案，经过反复试验，终于解决这个问题正确解法如下：like 参数string strSql = "select * from s where City like '%'+ @add + '%'";SqlParameter[] Parameters=new SqlParameter[1];Parameters[0] = new SqlParameter("@add", "bre");in 参数string strSql = "exec('select * from s where AddressID in ('+@add+')')";SqlParameter[] Parameters = new SqlParameter[1];Parameters[0] = new SqlParameter("@add", "343,372,11481,11533,11535,11755,11884,12092,12093,12143");