更新查询附近的语法错误'？在Eclipse中，但在MySQL工作台中工作

I'm getting this error in eclipse

我在eclipse中遇到这个错误

com.mysql.jdbc.exceptions.jdbc4.MySQLSyntaxErrorException: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '? , login = ? , pwd = ? WHERE login = 'pp'' at line 1

com.mysql.jdbc.exceptions.jdbc4.MySQLSyntaxErrorException：SQL语法中有错误;检查与您的MySQL服务器版本对应的手册，以便在'附近使用正确的语法？，登录=？，pwd =？在第1行登录='pp''

This is my query in my source code:

这是我在源代码中的查询：

String query2 = "UPDATE usuarios SET nombre = ? , login = ? , pwd = ? WHERE login = '" + login2 + "'";

And this is the whole code from my method:

这是我方法的全部代码：

private void modificar() {
        // Prints the content of the table
        String query = "SELECT * FROM usuarios";
        try {
            pst = con.Conectar().prepareStatement(query);
            rs = pst.executeQuery(query);

            // Itarate over the registries
            int i = 0;
            while (rs.next()) {
                i++;
                //print them
                System.out.println(i + "  " + rs.getString("id") + " " + rs.getString("login"));
            }
            //There are X registries
            System.out.println("Existen " + i + " usuarios actualmente");
            pst.close();

            //Wich registry do you need to modify?
            System.out.println("Ingrese el login del usuarios a modificar");
            String login2 = scanner.nextLine();

            System.out.println("Ingrese datos a modificar");
            System.out.print("Nombre: ");
            nombre = scanner.nextLine();
            System.out.print("Login: ");
            login = scanner.nextLine();
            System.out.print("Password: ");
            pwd = scanner.nextLine();

            String query2 = "UPDATE usuarios SET nombre = ? , login = ? , pwd = ? WHERE login = '" + login2 + "'";

            pst = con.Conectar().prepareStatement(query2);
            pst.setString(1, nombre);
            pst.setString(2, login);
            pst.setString(3, pwd);
            /*
             * Aqui da error de sintaxis en query2
             */
            pst.executeUpdate(query2);
            pst.close();

            String query3 = "SELECT * FROM usuarios where login =" + login;

            pst = con.Conectar().prepareStatement(query3);
            rs = pst.executeQuery(query3);
            rs.next();

            System.out.println("ahora quedo asi " + rs.getString("login"));

        } catch (SQLException e) {
            // TODO: handle exception
            e.printStackTrace();
        } finally {
            cerrarConsultas();
        }
    }

But is working fine when I use it in MySQL Workbench, this is my test in the workbench.

但是当我在MySQL Workbench中使用它时工作正常，这是我在工作台中的测试。

prepare insertar from "UPDATE usuarios SET nombre = ?, login = ?, pwd = ? WHERE login = 'pp'"; 
-- "UPDATE usuarios SET nombre = ?, login = ?, pwd = ? WHERE login = 'pablo'";
set @nombre = 'pp';
set @login = 'pp';
set @pwd = 'pp';
execute insertar using @nombre, @login, @pwd;
deallocate prepare insertar;

I've tried even with literal qoutes and still doesn't work.

我甚至尝试过文字qoutes但仍然无法正常工作。

String query2 = "UPDATE usuarios SET `nombre` = ? , `login` = ? , `pwd` = ? WHERE login = '" + login2 + "'";

Also tried:

还尝试过：

String query2 = "UPDATE usuarios SET `nombre` = ? , `login` = ? , `pwd` = ? WHERE login = "+ login2;

Same result.

结果相同。

2 个解决方案

#1

Replace pst.executeUpdate(query2); with pst.executeUpdate();

替换pst.executeUpdate（query2）;用pst.executeUpdate（）;

Otherwise you will end up ignoring the parameter binding you did with with the various pst.setString(...) hence the db engine will receive a query with ? instead of the values you meant to bind.

否则你最终会忽略你用各种pst.setString（...）进行的参数绑定，因此db引擎会收到一个查询？而不是你想绑定的值。

#2

Why not using a parameter for your WHERE clause?

为什么不为WHERE子句使用参数？

String query2 = "UPDATE usuarios SET nombre = ? , login = ? , pwd = ? WHERE login = ?";

pst = con.Conectar().prepareStatement(query2);
pst.setString(1, nombre);
pst.setString(2, login);
pst.setString(3, pwd);
pst.setString(4, login2 );

#1