php - 从两个表中获取记录并转换为json

Question

嗨，我是 PHP 新手，尝试使用 php sql 获取以下响应，但我无法找到这样的期望输出

[{"Id":1, "name": "India", "Cities":[{"Id":1, "Name":"Mumbai", "country_id":1}, {"Id":2,"Name":"Delhi","country_id":1},
"id":3,"Name":Banglore","country_id":1}, {"Id":2, "Name":"USA", "Cities":[{"Id":6, "Name":"New York", "country_id":2},.....

我有两张表，一张是基于国家/地区的，另一张是基于城市的。

I tried
<?php

    include_once("config.inc.php");

    $sql = "SELECT * FROM country";
    $sqlCity = "SELECT * FROM city";

    $cityQuery = mysqli_query($conn, $sqlCity);
    $sqlQuery = mysqli_query($conn, $sql);

    $mainArray = array();

    if(mysqli_num_rows($cityQuery) > 0 ){
        $cityResponse = array();
        while($resCity = mysqli_fetch_assoc($cityQuery)){
            $cityResponse[] = $resCity;
        }

    if(mysqli_num_rows($sqlQuery) > 0 ){
        $response = array();
        while($res = mysqli_fetch_assoc($sqlQuery)){
            $response[] = $res;
        }
        foreach($cityResponse as $city){
            foreach($response as $country){
                if($city['country_id'] == $country['id']){

                    $mainArray = array("Cities" => $city);
                }
            }
        }
        echo '{"response": '.json_encode($response).', '.json_encode($mainArray).' "success": true}';
    }
    }else{
        echo '{"response": '.json_encode($response).' "success": false}';
    }


?>

目前我的回复显示

{"response": [{"id":"1","name":"India"},{"id":"2","name":"USA"},{"id":"3","name":"UK"}], {"Cities":{"id":"15","name":"Manchester","country_id":"3"}} "success": true}

Answer 1

有关详细代码说明，请查看内嵌注释

• 使用相关列名称修改 SQL 查询
• 你必须照顾的内存。默认情况下，5.3 中 php 内存限制为 128MB

• 检查代码并告诉我结果

  <?php
  
  $data = array();
  
  //include your database configuration files
  include_once("config.inc.php");
  
  //execute the join query to fetch the result
  $sql = "SELECT country.country_id, country.name AS country_name,".
  " city.city_id, city.name AS city_name FROM country ".
  " JOIN city ON city.country_id=country.country_id ".
  " ORDER BY country.country_id ";
  
  //execute query
  $sqlQuery = mysqli_query($conn, $sql) or die('error exists on select query');
  
  //check the number of rows count
  if(mysqli_num_rows($sqlQuery) > 0 ){
  
  //country id temprory array
  $country_id = array();
  
  //loop each result
  while($result = mysqli_fetch_assoc($sqlQuery)){
  
      //check the country id is already exist the only push the city entries
      if(!in_array($result['country_id'],$country_id)) {
  
          //if the city is for new country then add it to the main container
          if(isset($entry) && !empty($entry)) {
              array_push($data, $entry);
          }
  
          //create entry array
          $entry = array();   
          $entry['Id'] = $result['country_id'];
          $entry['name'] = $result['country_name'];
          $entry['Cities'] = array();
  
          //create cities array
          $city = array();
          $city['Id'] = $result['city_id'];
          $city['name'] = $result['city_name'];
          $city['country_id'] = $result['country_id'];
  
          //append city entry
          array_push($entry['Cities'], $city);
          $country_id[] = $result['country_id'];
      }
      else {
  
          //create and append city entry only
          $city = array();
          $city['Id'] = $result['city_id'];
          $city['name'] = $result['city_name'];
          $city['country_id'] = $result['country_id'];
          array_push($entry['Cities'], $city);
      }
  
  }
  
  }
      //display and check the expected results
              echo json_encode($data);