------解决方案-------------------- select [name],password,phonenumber from UserTable where [name] like '%黄%'
------解决方案--------------------
------解决方案-------------------- 我把你改下吧 SELECT cname AS '用户名', csex AS '性别', cid AS '身份证', cphonenumber AS '手机号码', cpassword AS '密码'FROM customers WHERE cname like '%"+ @cname+"%'";
------解决方案-------------------- 然后ListView中cname的value值设为"'%"+TextBox3.Text+"%'"
------解决方案-------------------- SELECT cname AS '用户名', csex AS '性别', cid AS '身份证', cphonenumber AS '手机号码', cpassword AS '密码' FROM customers WHERE (cname like @cname) 你写的这个语句中的下面代码 你把: WHERE (cname like @cname) 改成: WHERE (cname like '%"+@cname+"%')