执行SQL查询时的MySQL语法错误 | 执行SQL查询时的MySQL语法错误

本文介绍了执行SQL查询时的MySQL语法错误的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！

问题描述

当我尝试运行下面的代码时，我得到:

When I try to run the code below I am getting:

String query="Select * from DB.Admin where username = ?";
PreparedStatement st=connection.prepareStatement(query);
st.setString(1,request.getParameter("loginid"));
ResultSet rst= st.executeQuery(query);
int count=0;
while(rst.next()){
   count++;
}

请帮助我.

推荐答案

您将必须从executeQuery调用中删除query参数.如果提供该参数，则将在不绑定任何值的情况下执行查询(请参见声明以获得详细信息)-这就是为什么语法(即?)无效的原因.

You will have to remove the query argument from your executeQuery call. If you provide the parameter, the query will be executed without binding any values (see Statement for details) - this is why the syntax (i.e. the ?) is invalid.

执行如下查询:

ResultSet rst = st.executeQuery();

请注意:您应始终用 try-with-resources 块包装Connection，PreparedStatement和ResultSet，例如

As a side note: you should always wrap Connection, PreparedStatement and ResultSet with a try-with-resources block, e.g.

try (ResultSet rst = st.executeQuery()) {
    // read the results
}

这样，无论发生什么情况，您都可以确保ResultSet将被关闭.

This way you can be sure the ResultSet will be closed no matter what happens.

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