在前台有几个查询条件的填写。填写了姓名后按照姓名模糊查找，填写了年龄后可以按年龄范围查找，填写了成绩后可以按成绩查找，这三个条件任写其中的两个或多个都可以按其对应的条件查找出来！不分先后！
------解决方案--------------------
StringBuffer，然后用append
------解决方案--------------------
把条件理清楚了，很easy的！不相信，自己试试呗
------解决方案--------------------
------解决方案--------------------
- Java code
package com.bt.employee.dao.impl;import java.sql.Connection;import java.sql.PreparedStatement;import java.sql.ResultSet;import java.sql.SQLException;import java.util.ArrayList;import java.util.List;import com.bt.employee.dao.EmployeeDAO_Query;import com.bt.employee.entity.Employee;import com.bt.employee.util.DBUtil;public class EmployeeDAOImpl_Query1 implements EmployeeDAO_Query{ public List<Employee> specificInquiresr(String ename, String age, String grade) throws SQLException { Connection conn = DBUtil.getConnection(); List<Employee> employees = new ArrayList<Employee>(); StringBuffer sql = new StringBuffer(); sql.append("select * from 表 where 1=1"); //String ename1 = ename.trim(); if(ename.trim()!=""){ sql.append(" and ename like '%" + ename +"%'"); } if(age.equals("0")){ sql.append(" " ); }else if(age.equals("1")){ sql.append(" and age<60" ); }else if(age.equals("2")){ sql.append(" and salary between 60 and 80" ); }else if(age.equals("3")){ sql.append(" and salary between 80 and 100" ); } if(grade!=""){ sql.append(" and grade= " + grade); } String sql1 = sql.toString(); PreparedStatement ps = conn.prepareStatement(sql1); System.out.println(sql1); ResultSet rs = ps.executeQuery(); while(rs.next()){ Employee e = new Employee(); e.setId(rs.getLong("id")); e.setEname(rs.getString("ename")); e.setAge(rs.getInt("age")); e.setSalary(rs.getDouble("grade")); employees.add(e); } return employees; }}
------解决方案--------------------
ibatis的SQLmapping里面有isNotEmpty判断