小编Rub*_*num的帖子

我有一个带有两个容器的docker-compose设置:一个是php/apache服务,另一个容器是数据库(mysql).

这是我的docker-compose.yml

version: '2'
services:
   app:
     depends_on:
       - db
     links:
       - db:mysql
     build: .
     image: app
     ports:
       - "80:80"
     restart: always
     links:
       - db:db
     volumes:
       - ../:/var/www/html/

   db:
     image: mysql:latest
     restart: unless-stopped
     volumes:
       - ./db_data:/var/lib/mysql
       - ./databaseDumps:/tmp/databaseDumps
     environment:
       MYSQL_USER: "myApp"
       MYSQL_PASSWORD: "root"
       MYSQL_ROOT_PASSWORD: "root"
       MYSQL_DATABASE: "myAppDatabase"
       MYSQL_ROOT_HOST: "%"

这是我的应用程序Dockerfile:

FROM php:7-apache

COPY prefilled_files/000-default.conf /etc/apache2/sites-available/000-default.conf
RUN apt-get -qq update
RUN apt-get -qq -y install libpng-dev curl git nano vim zip unzip mysql-client libmysqlclient-dev
RUN curl -sS https://getcomposer.org/installer | php -- --install-dir=/usr/local/bin …

我正在基于摇摆的项目工作,并使用预备语句,但它给出异常,如超出范围,当我使用参数,不要把'？' 但简单？然后显示异常,如"jdbc语法错误检查附近的手册？？,？" 在第一行"我很困惑发生了什么.检查我的代码有什么问题？

private void AddActionPerformed(java.awt.event.ActionEvent evt) {
    if (((JTextField) chose.getDateEditor().getUiComponent()).getText() == null) {
        String sql = "INSERT INTO expance1 ( Breakfast,Date) VALUES (?,?)";
    } else {
        String sql = "INSERT INTO expance1 ( Breakfast,Date) VALUES (?,?)";
        try {
            pst = conn.prepareStatement(sql);
            pst.setString(1, breakfast.getText());
            pst.setString(2, ((JTextField) chose.getDateEditor().getUiComponent()).getText());
            pst.executeUpdate(sql);
            JOptionPane.showMessageDialog(null, "insert sucessfully");
        } catch (Exception e) {
            JOptionPane.showMessageDialog(null, e);
        } finally {
            try {
                rs.close();
                pst.close();

            } catch (Exception e) {
            }
        }
    }
}