Mysql 如何仅从付款和学校名称中提取金额？

与：

如果您试图将学校名称与金额匹配，则此操作将始终失败

尝试：

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我尝试过这个方法，但它没有提供任何SELECT*，从学校内部加入对学校的付款。schoolname=payments.amount；将您的条件更改为schools.paymentId=付款。paymentId@MKhalid，你是对的，谢谢它起作用了。哦，试着把你的问题和下次你已经试过的东西弄清楚，这样你就不会那么快被否决了。问题是直截了当的，我只需要两个表中的两列，即-schoolname和amount。你说SELECT s.name，p.amount…from schools s…这不起作用，因为我不懂你的首字母缩写。@mpile这更好吗？是的，谢谢，我刚刚更改了schools.name的列名以适合我的schoolname。。太好了。

-- Database: `kanda`

-- Table structure for table `payments`
--

CREATE TABLE IF NOT EXISTS `payments` (
  `paymentId` int(11) NOT NULL AUTO_INCREMENT,
  `amount` double NOT NULL,
  `date` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP ON UPDATE CURRENT_TIMESTAMP,
  PRIMARY KEY (`paymentId`)
) ENGINE=InnoDB  DEFAULT CHARSET=latin1 AUTO_INCREMENT=3 ;

--
-- Dumping data for table `payments`
--

INSERT INTO `payments` (`paymentId`, `amount`, `date`) VALUES
(1, 3000, '2015-01-13 08:47:46'),
(2, 3000, '2015-01-13 08:47:56');

-- phpMyAdmin SQL Dump
-- version 4.1.6
-- http://www.phpmyadmin.net
--
-- Host: 127.0.0.1
-- Generation Time: Jan 14, 2015 at 01:28 PM
-- Server version: 5.6.16
-- PHP Version: 5.5.9

SET SQL_MODE = "NO_AUTO_VALUE_ON_ZERO";
SET time_zone = "+00:00";


--


--
-- Table structure for table `schools`
--

CREATE TABLE IF NOT EXISTS `schools` (
  `schoolId` int(11) NOT NULL AUTO_INCREMENT,
  `paymentId` int(11) NOT NULL,
  `schoolname` varchar(20) NOT NULL,
  PRIMARY KEY (`schoolId`)
) ENGINE=InnoDB  DEFAULT CHARSET=latin1 AUTO_INCREMENT=4 ;

--
-- Dumping data for table `schools`
--

INSERT INTO `schools` (`schoolId`, `paymentId`, `schoolname`) VALUES
(1, 1, 'westlands primary'),
(2, 2, 'st.johns primary');

SELECT * FROM schools INNER JOIN payments ON schools.schoolname = payments.amount;

SELECT schools.name, payments.amount FROM schools INNER JOIN payments ON schools.paymentId = payments.paymentId;