[英]Why does the PrepareStatement not generate correct SQL query?
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:从temp中选择stuId,name,relationsName,houseAddress,houseNumber,其中stuId喜欢'?%'和sex ='?' 并命名为'?%'和年龄BETWEEN **未指明**和**未指明**和relationsName喜欢'?%'按名称排序asc limit 0,150000
它没有超出这一点。 此外,它说不在查询中指定 (请参阅为末)。
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();
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,你应该在java中连接值,而不是在sql中,
ps1.setString(1, stuId + '%');
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