Question

以下是我的PrepareStatement代码。它不会生成正确的SQL查询。它也没有超越1st println-statement。另外它在查询中说**未指明**（请参见下文）。

我们怎么解决这个问题呢？

   ps1 = con.prepareStatement(
                     "select stuId, name, relationsName, houseAddress, houseNumber from temp where "
                        + " stuId like '?%' and " 
                        + " sex = '?' and " 
                        + " name like '?%' and "
                        + " age BETWEEN ? and ? and "
                        + " relationsName like '?%' " 
                        + " order by name asc limit 0, 150000 "
               );

               System.out.println("ps1 Before : " + ps1);

输出：

ps1之前：com.mysql.jdbc.JDBC4PreparedStatement@14d55de：选择 stuId，name，relationsName，houseAddress，houseNumber from temp where stuId喜欢'？％'和sex ='？'并命名为'？％'和年龄BETWEEN ** 未指明**和**未指定**和relationsName喜欢'？％' 按名称排序asc limit 0,150000

它没有超出这一点。此外，它在查询中显示 NOT SPECIFIED （请参见结尾）。

有什么见解吗？

       ps1.setString(1, stuId);
       ps1.setString(2, gender);
       ps1.setString(3, name);
       ps1.setInt(4, startAge);
       ps1.setInt(5, endAge);
       ps1.setString(6, relationsName);

       System.out.println("ps1 After : " + ps1);

       rs = ps1.executeQuery();

Answer 1

因为占位符用单引号括起来，因此使它成为一个值而不再是一个参数。你应该摆脱它，例如

ps1 = con.prepareStatement(
                 "select stuId, name, relationsName, houseAddress, houseNumber from temp where "
                    + " stuId like ? and " 
                    + " sex = ? and " 
                    + " name like ? and "
                    + " age BETWEEN ? and ? and "
                    + " relationsName like ? " 
                    + " order by name asc limit 0, 150000 "
           );

对于LIKE语句，你应该连接java中的值，而不是sql，

ps1.setString(1, stuId + '%');

为什么PrepareStatement不能生成正确的SQL查询？

1 个答案: