Question

我想要什么

我希望数组看起来像这个

  array
  (
  [0]=>country
  (
    [id] => 
    [name] => 
    [url] => 
      )
 [city] => Array
    (
        [0] => Array
            (
                [id] => 68870
                [name] => Abu Dhabi
                [url] => abudhabi-ae
            )

        [1] => Array
            (
                [id] => 68872
                [name] => Das Island
                [url] => dasisland-ae
            )

        [2] => Array
            (
                [id] => 68873
                [name] => Dubai
                [url] => dubai-ae
            )
       [1]=>country
         (
        [id] => 
        [name] => 
       [url] => 
      )

      [city] => Array
    (
        [0] => Array
            (
                [id] => 68870
                [name] => Abu Dhabi
                [url] => abudhabi-ae
            )

        [1] => Array
            (
                [id] => 68872
                [name] => Das Island
                [url] => dasisland-ae
            )

        [2] => Array
            (
                [id] => 68873
                [name] => Dubai
                [url] => dubai-ae
            )
         )

我写了这样的代码

   <?php
      require('/home/indiamart/public_html/serve-biztradeshows-com/db.php');
     $query="select distinct ct.name name , listing_combo.country country from country ct join listing_combo on ct.id=listing_combo.country ";
  $result=mysql_query($query);
  while ($row=mysql_fetch_assoc($result)) 
   {
        $country_id= $row['country'];

       $arr = array();
       $arr[]=$row;
       $arr1['country']=json_encode($arr);
       $select_city="select id,name,url from city where country like'$country_id' ";
        $city=mysql_query($select_city);

       while($row1=mysql_fetch_assoc($city))
       {
        $arr[]=$row1;
        }
         $arr1['city']=json_encode($arr);
       }
       print_r($arr1);
       echo "</br>";
       ?>

但它只打印查询的最后结果和ites格式是这样的

数组也可以创建，所以我可以最后编码它，但它的格式应该没问题。

阵列（

[国家] =＆GT;

[城市] =＆GT;

）

请帮助

Answer 1

是。这是因为您总是在循环中重新初始化$arr = array();。将此行放在while循环之外。

另外请不要使用mysql函数，因为它们已被弃用。请改用mysqli或PDO。

避免sql注入，转义查询中的变量。

我无法测试，但你需要这样的东西：

require('/home/indiamart/public_html/serve-biztradeshows-com/db.php');
//In your db.php create a $conn variable, what is a connection (resource) to 
//your database with mysqli_connect function.
//See mysqli in php manual.

$countryQuery = "SELECT DISTINCT ct.name name , listing_combo.country country "
        . "FROM country ct "
        . "JOIN listing_combo ON ct.id=listing_combo.country";
//Use mysqli instead of mysql functions since they are deprecated.
$result = mysqli_query($conn, $countryQuery);
$countries = array();
$i = 0;
while ($row = mysqli_fetch_assoc($result)) {
    //Do not need to create a new variable, what you do not use anywhere else.
    //$country_id= $row['country']; You can use $row['country'] 
    //in your query cityQuery later.

    //$arr1['country']=json_encode($arr);
    //Json do not need here, since, you want an array, not a string.
    $countries[$i]['country'] = $row;

    //Avoid sql injection, so use mysqli_real_escape_string
    $cityQuery = "SELECT id,name,url "
            . "FROM city "
            . "WHERE country like'" . mysqli_real_escape_string($conn, $row['country']) . "'";
    $cityResult = mysqli_query($conn, $cityQuery);
    $j = 0;
    while ($cityRow = mysql_fetch_assoc($cityResult)) {
        //Add all cities to the current country
        $countries[$i]['country']['city'][$j] = $cityRow;
        //Incrase the key of the array cities
        $j++;
    }
    //Incrase the key of countries
    $i++;
}
print_r($countries);

我在你的代码中做了一些修改。重命名所有变量，告诉它们包括什么。

我正在使用mysqli函数而不是mysql函数。

删除未使用的变量。

添加了评论，以便您可以看到我在做什么。

Answer 2

我无法连接到您的数据库，所以我无法测试。但感觉应该是：

     <?php
    require('/home/indiamart/public_html/serve-biztradeshows-com/db.php');
    $query="select distinct ct.name name , listing_combo.country country from country ct join listing_combo on ct.id=listing_combo.country ";
    $result=mysql_query($query);

    while ($row=mysql_fetch_assoc($result)) 
    {
        $countries[$row['country']]=$row;
    }

    foreach ($countries as $country_id => $country_row) {
        $select_city="select id,name,url from city where country like'$country_id' ";
        $city=mysql_query($select_city);
        while($row1=mysql_fetch_assoc($city))
        {
            $cities[]=$row1;
        }
        foreach ($cities as $city_id => $city_row) {
            $countries[$country_id][$city_id]=$city_row;
        }
    }

    print_r($countries);
    echo "</br>";
?>

您最好先查询国家/地区以避免重置数据。感谢这位优秀的评论者。

使用不同的键合并/附加json

2 个答案: