запись Извлечение из двух таблицы и конвертировать в формат JSON

Question

Привет Я новичок в PHP и пытается получить ответ ниже с помощью PHP SQL, но я не быть в состоянии найти такой вывод Дезире

[{"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},..... 

I есть две таблицы, одна из которых основана на стране, а другая - городская.

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}'; 
    } 


?> 

В настоящее время мой ответ, показывая

{"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

Для детального кода объяснения проверить встроенные комментарии

• Изменить SQL запрос с соответствующими именами столбцов
• Память, которую вы должны позаботиться. По пределу памяти по умолчанию в РНР 128 в 5.3

• Проверьте код и дайте мне знать результат

  <?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); 
  

Answer 2

использовать как этот

<?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 json_encode(array('result'=>'true','Cities'=>$mainArray)); 
    } 
    }else{ 

     echo json_encode(array('result'=>'false','Cities'=>$response)); 
    } 


?>