0

我需要计算每个类别的计数（vendors.id） 。即：每个类别（不是子类别） 有多少供应商

我正在使用 cakephp 3.6 框架和 MySQL 作为数据库

我尝试了所有可能的方法，我知道但没有找到任何解决方案。任何人都可以帮助我，对我的项目非常重要

[更新] 我使用过的 sql 查询：

SELECT cat.id,cat.name ,COUNT(`vendor_id`) AS vendor_count FROM `vendor_services` JOIN `categories` ON(`vendor_services`.`category_id` = `categories`.`id`) JOIN `categories` AS cat ON(categories.category_id = cat.id) WHERE 1 GROUP BY cat.id

[更新]下面是创建对应表的sql

-- phpMyAdmin SQL Dump
-- version 4.7.0
-- https://www.phpmyadmin.net/
--
-- Host: 127.0.0.1
-- Generation Time: Dec 04, 2018 at 10:50 AM
-- Server version: 10.1.24-MariaDB
-- PHP Version: 7.1.6

SET SQL_MODE = "NO_AUTO_VALUE_ON_ZERO";
SET AUTOCOMMIT = 0;
START TRANSACTION;
SET time_zone = "+00:00";

--
-- Database: `demo123`
--

-- --------------------------------------------------------

--
-- Table structure for table `categories`
--

CREATE TABLE `categories` (
  `id` int(11) NOT NULL,
  `category_id` tinyint(4) NOT NULL,
  `name` varchar(60) NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=latin1;

--
-- Dumping data for table `categories`
--

INSERT INTO `categories` (`id`, `category_id`, `name`) VALUES
(1, 0, 'Books'),
(2, 0, 'Electronics'),
(3, 0, 'Garden'),
(4, 1, 'Novel'),
(5, 1, 'Science'),
(6, 1, 'Story'),
(7, 2, 'Mobile'),
(8, 2, 'Tablet'),
(9, 2, 'Headphone'),
(10, 3, 'Pumps'),
(11, 3, 'Pipes');

-- --------------------------------------------------------

--
-- Table structure for table `vendors`
--

CREATE TABLE `vendors` (
  `id` int(11) NOT NULL,
  `name` varchar(60) NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=latin1;

--
-- Dumping data for table `vendors`
--

INSERT INTO `vendors` (`id`, `name`) VALUES
(1, 'VR Enterprizes'),
(2, 'RR Vendors'),
(3, 'JK Suppliers');

-- --------------------------------------------------------

--
-- Table structure for table `vendor_services`
--

CREATE TABLE `vendor_services` (
  `id` int(11) NOT NULL,
  `vendor_id` int(11) NOT NULL,
  `category_id` int(11) NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=latin1;

--
-- Dumping data for table `vendor_services`
--

INSERT INTO `vendor_services` (`id`, `vendor_id`, `category_id`) VALUES
(1, 1, 4),
(2, 1, 5),
(3, 1, 6),
(4, 1, 11),
(5, 2, 7),
(6, 2, 8),
(7, 2, 9),
(8, 3, 10),
(9, 3, 11);

--
-- Indexes for dumped tables
--

--
-- Indexes for table `categories`
--
ALTER TABLE `categories`
  ADD PRIMARY KEY (`id`),
  ADD KEY `category_id` (`category_id`);

--
-- Indexes for table `vendors`
--
ALTER TABLE `vendors`
  ADD PRIMARY KEY (`id`);

--
-- Indexes for table `vendor_services`
--
ALTER TABLE `vendor_services`
  ADD PRIMARY KEY (`id`),
  ADD KEY `vendor_id` (`vendor_id`),
  ADD KEY `category_id` (`category_id`);

--
-- AUTO_INCREMENT for dumped tables
--

--
-- AUTO_INCREMENT for table `categories`
--
ALTER TABLE `categories`
  MODIFY `id` int(11) NOT NULL AUTO_INCREMENT, AUTO_INCREMENT=12;
--
-- AUTO_INCREMENT for table `vendors`
--
ALTER TABLE `vendors`
  MODIFY `id` int(11) NOT NULL AUTO_INCREMENT, AUTO_INCREMENT=4;
--
-- AUTO_INCREMENT for table `vendor_services`
--
ALTER TABLE `vendor_services`
  MODIFY `id` int(11) NOT NULL AUTO_INCREMENT, AUTO_INCREMENT=10;COMMIT;

分类表

$this->hasMany(‘Subcategories’, [ 'className'=> ‘Categories’,
            'foreignKey' => 'category_id',
            'conditions' => ['category_id!= 0']
        ]);

$this->belongsTo('MainCategories', [
                'className'=> ‘Categories’,
                    'foreignKey' => 'category_id',
        ]);

在此处输入图像描述

4

1 回答 1

1

下面是oracle中的查询，请看一下，按照mysql修改一下，

添加在插入下方，INSERT INTO VENDOR_SERVICES (ID, VENDOR_ID, CATEGORY_ID) VALUES(11, 3, 3);

SELECT FRM.CATEGORY_ID, FRM.VENDOR_ID, FRM.VENDER_NAME,COUNT(VENDOR_ID) counts FROM ( SELECT distinct CT.CATEGORY_ID, VS.VENDOR_ID, VS.CATEGORY_ID VS_CATEGORY_ID,vn.ID, vn.NAME VENDER_NAME FROM CATEGORIES CT INNER JOIN VENDOR_SERVICES VS ON VS.CATEGORY_ID=CT.CATEGORY_ID INNER JOIN VENDORS VN ON VS.VENDOR_ID=VN.ID) frm group by CATEGORY_ID, VENDOR_ID, VENDER_NAME

于 2018-12-04T12:41:01.573 回答