PHP和MySQL到XML

Question

我有2张桌子，第一张桌子存放餐厅信息，第二张桌子存放每张桌子。 它们使用res_id链接。

1）info_main [id，res_id，res_name，res_pc] 2）菜肴[id，dishName，price，res_id（外键）]

我的SQL查询是

$query = "SELECT *  FROM info_main LEFT JOIN dishes ON info_main.res_id = dishes.res_id"; 

我将查询的结果插入到工作正常的xml文件中。 下面是代码：

    $query = "SELECT *  FROM info_main LEFT JOIN dishes ON info_main.res_id = dishes.res_id";
$result = mysql_query($query);

if (!$result) {
  die('Invalid query: ' . mysql_error());
}
header("Content-type: text/xml");

echo '<markers>';

// Iterate through the rows, printing XML nodes for each
while ($row = @mysql_fetch_assoc($result)){
  // ADD TO XML DOCUMENT NODE
  echo '<marker> ';

      echo '<detail1>';
        echo '<resdetails ';
            echo 'name="' . parseToXML($row['res_name']) . '" ';
            echo 'id="' . parseToXML($row['res_ID']) . '" ';
            echo 'pc="' . parseToXML($row['res_pc'] ). '" ';
        echo '/>';

        echo '<dishdetails ';
            echo 'name="' . parseToXML($row['dishName']) . '" ';
            echo 'price="' . parseToXML($row['price']) . '" ';
        echo '/>';

      echo '</detail1>';

      echo '</marker>';
    }

但是，如果一家餐馆的数据库中有3道菜，那么此方法就可以正常工作，然后xml创建3个节点：

我想要像这样的xml结构

<detail1>
<resdetails name="spoted dog" id="xyz" pc="xyz"/>
<dishdetails name="bean burger" price="1" />
<dishdetails name="cheese and tomato panini" price="3" />
<dishdetails name="veg salad" price="2" />
</details1>

我想不出一种方法来实现上述xml结构。 非常感谢您的帮助。 谢谢

Answer 1

 $query = "SELECT *  FROM info_main LEFT JOIN dishes ON info_main.res_id = dishes.res_id";
$result = mysql_query($query);

if (!$result) {
  die('Invalid query: ' . mysql_error());
}
header("Content-type: text/xml");

echo '<markers>';

// Iterate through the rows, printing XML nodes for each
$resname = "";
while ($row = @mysql_fetch_assoc($result)){
  // ADD TO XML DOCUMENT NODE
   if ($resname!=$row['res_name']) {
   //restname isn't populated or doesn't match current so output new headers
     echo '<marker> ';
      echo '<detail1>';
        echo '<resdetails ';
            echo 'name="' . parseToXML($row['res_name']) . '" ';
            echo 'id="' . parseToXML($row['res_ID']) . '" ';
            echo 'pc="' . parseToXML($row['res_pc'] ). '" ';
        echo '/>';
   }
        //this bit needs to always happen
        echo '<dishdetails ';
            echo 'name="' . parseToXML($row['dishName']) . '" ';
            echo 'price="' . parseToXML($row['price']) . '" ';
        echo '/>';


   if ($resname!=$row['res_name']) {
   //restname isn't populated or doesn't match current so output new headers
      echo '</detail1>';

      echo '</marker>';
   }
   $resname = $row['res_name'];  //set resname to this res_name as this is our check to see if we've already put out required headers for this item that way every change it'll put this back in
    }

这样的事情（注意可能需要整理一下）

Answer 2

对于该结构，只需执行以下操作：

  echo '<marker> ';

  echo '<detail1>';

 while ($row = @mysql_fetch_assoc($result)){

 echo '<dishdetails ';
        echo 'name="' . parseToXML($row['dishName']) . '" ';
        echo 'price="' . parseToXML($row['price']) . '" ';
    echo '/>';

 }

echo '</detail1>';

  echo '</marker>';

真正不同的是，我将代码的一部分移出了while循环。

Answer 3

终于我得到了这个工作，我使用了两个查询，一个得到了不同的要求，另一个得到了菜名。 我在主循环中使用了循环。 下面是代码。

$query = "SELECT DISTINCT  *  FROM info_main";

$result = mysql_query($query);


if (!$result) {
  die('Invalid query: ' . mysql_error());
}
header("Content-type: text/xml");

echo '<markers>';

// Iterate through the rows, printing XML nodes for each
while ($row = @mysql_fetch_assoc($result)){
  $id=$row['res_id'];
  // ADD TO XML DOCUMENT NODE
  echo '<marker> ';

      echo '<detail1>';
        echo '<resdetails ';
            echo 'name="' . parseToXML($row['res_name']) . '" ';
            echo 'id="' . parseToXML($id) . '" ';
            echo 'pc="' . parseToXML($row['res_pc'] ). '" ';
        echo '/>';

        $dishes = "SELECT * FROM dishes WHERE res_id = '" . $id . "'";

        $dish = mysql_query($dishes);

        while ($row2 = @mysql_fetch_assoc($dish)){
          $id2 = $row2['res_id'];

            echo '<dishdetails ';
                echo 'name="' . parseToXML($row2['dishName']) . '" ';
                echo 'price="' . parseToXML($row2['price']) . '" ';
            echo '/>';

        }

      echo '</detail1>';

      echo '</marker>';
    }
echo '</markers>';