Question

嗨，我是PHP新手并试图使用php sql获得以下响应，但我无法找到这样的欲望输出

[{"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},.....

我有两个表，一个是基于国家的，另一个是基于城市的。

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

像这样使用

<?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));
    }


?>

Answer 2

有关详细代码说明，请检查内联注释

使用相关列名修改SQL查询
你需要注意的记忆。默认情况下，php中的内存限制为5.3 MB

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);

Answer 3

您可以尝试使用此方法。它实际上对我有用，很酷。您可以通过编辑代码将表扩展到3个或更多。

try {

    $results = array();

    $query = "SELECT * FROM country";
    $values = array();

    $stmt = $datab->prepare($query);
    $stmt->execute($values);

    while($country = $stmt->fetch(PDO::FETCH_ASSOC)){

        $cities = null;

        $query2 = "SELECT * FROM city WHERE country_id = ?" ;
        $values2 = array($country['id']);

        $stmt2 = $datab->prepare($query2);
        $stmt2->execute($values2);
        $cities = $stmt2->fetchAll();

        if($cities){
            $country['cities'] = $cities;
        } else {
            $country['cities'] = '';
        }

        array_push($results, $country);  
    }

    echo json_encode(array("Countries" => $results));

} catch (PDOException $e) {
    throw new Exception($e->getMessage());
}

从两个表中获取记录并转换为json

3 个答案: