嘗試獲取XML getElementsByTagName + getAttribute調用循環工作

Question

我有這個過分複雜的Eatec XML轉儲，很遺憾，每個菜單都有它自己的特殊屬性，用於在早餐/午餐/晚餐時間提供膳食。我試圖從他們的描述中提取物品，並做一個循環來顯示早餐的食譜，以及其他菜單時間等。

這裏是瘦了一個示例XML

<data> 
    <menu name="location breakfast" servedate="20160626" location="food court 1" mealperiodname="Breakfast" > 
    <recipes> 
     <recipe id="5626935" category="HOT MORNING GRAINS" description="Oatmeal RD"> 
     </recipe> 
     <recipe id="5371796" category="HOT MORNING GRAINS" description="Rice Brown RD"> 
     </recipe> 
    </recipes> 
    </menu> 
    <menu name="location lunch" servedate="20160626" location="food court 2" mealperiodname="Lunch"> 
    <recipes> 
     <recipe id="4430587" category="SOUPS" description="Soup Tomato w/Garden Vegetables"> 
     </recipe> 
     <recipe id="4210899" category="SOUPS" description="Soup Beef Barley w/Vegetables"> 
     </recipe> 
    </recipes> 
    </menu> 
</data> 

而且我是比較新的PHP/XML，還是想在這裏學習我的繩子，這裏就是我想出了，但沒能保持在自己的用餐時間內環狀物品。

<?php 

    $menu = new DOMDocument(); 
    $breakfast = 'Breakfast'; 
    $lunch = 'Lunch'; 
    $menu->load("http://amphl.org/test.xml"); 

    // Get location name 
    $menu_get = $menu->getElementsByTagName("menu"); 
    $location_name = $menu_get->item(0)->getAttribute("location"); 
    $menu_period = $menu_get->item(0)->getAttribute("name"); 

    // Get menus 
    $recipes = $menu->getElementsByTagName("menu"); 
    $recipe_items = $menu->getElementsByTagName("recipe"); 

    // echo tests 
    echo '<h3 class="location_date">'.$menu_period.'</h3>'; 
    echo '<h3 class="location_date">'.$location_name.'</h3>'; 

    echo '<div class="meal_items">'; 
    // echo '<h3 class="food_name"> Breakfast </h3>'; 
     foreach($recipes as $recipe) 
     { 
     // Get recipe name 
      $recipe_type = $recipe->getAttribute("mealperiodname"); 

      echo '<h3 class="location_date">'.$recipe_type.'</h3>'; 
      if ($recipe_type == $breakfast) { 
       foreach($recipe_items as $recipe_item) 
       { 
        $recipe_name = $recipe_item->getAttribute("description"); 
         echo '<p class="item_list"><a alt="" href="#">'.$recipe_name.'</a></p>'; 
       } 

       } 
      else if ($recipe_type == $lunch) { 
       foreach($recipe_items as $recipe_item) 
       { 
        $recipe_name = $recipe_item->getAttribute("description"); 
         echo '<p class="item_list"><a alt="" href="#">'.$recipe_name.'</a></p>'; 
       } 

       } 

    } 
echo '</div>'; 

而不是顯示在自己的循環早餐和午餐的飯菜，它的加載關於什麼是就餐時間每天的食譜。我是否過於複雜？我被自己的代碼迷失了？

Answer 1

我覺得你糾結自己

$menu = new DOMDocument(); 
$breakfast = 'Breakfast'; 
$lunch = 'Lunch'; 
libxml_use_internal_errors(true); 
$menu->load("http://amphl.org/test.xml"); 

foreach($menu->getElementsByTagName("menu") as $recipes) { 
    echo '<h3 class="location_date">'. $recipes->getAttribute('servedate') .'</h3>' ."\n"; 
    echo '<h3 class="location_date">'.$recipes->getAttribute('location').'</h3>' ."\n"; 
    $recipe_type = $recipes->getAttribute("mealperiodname"); 
    echo '<h3 class="location_date">'.$recipe_type.'</h3>' ."\n"; 

    echo ' <div class="meal_items">' . "\n"; 
    foreach($recipes->getElementsByTagName("recipe") as $recipe_item) { 
     echo ' <p class="item_list"><a alt="" href="#">' . 
      $recipe_item->getAttribute("description") . 
      '</a></p>' ."\n"; 
    }   
    echo ' </div>' ."\n"; 
} 

結果

<h3 class="location_date">20160626</h3> 
<h3 class="location_date">food court 1</h3> 
<h3 class="location_date">Breakfast</h3> 
    <div class="meal_items"> 
    <p class="item_list"><a alt="" href="#">Oatmeal RD</a></p> 
    <p class="item_list"><a alt="" href="#">Rice Brown RD</a></p> 
    </div> 
<h3 class="location_date">20160626</h3> 
<h3 class="location_date">food court 2</h3> 
<h3 class="location_date">Lunch</h3> 
    <div class="meal_items"> 
    <p class="item_list"><a alt="" href="#">Soup Tomato w/Garden Vegetables</a></p> 
    <p class="item_list"><a alt="" href="#">Soup Beef Barley w/Vegetables</a></p> 
    </div> 

demo

Answer 2

如果您命名一個變量$recipes = $menu->getElementsByTagName("menu");但我想你想使用$mrecipes = $recipe->getElementsByTagName('recipe')到包含在menu元素內recipe元素選擇foreach($recipes as $recipe)內這是相當混亂。

Answer 3

此外，對於DOM API方法，您可以使用XPath來獲取DOM樹的部分。 Xpath允許條件，因此您可以獲取特定用餐期間的所有菜單節點。

以下是將信息作爲文本輸出的示例。

$mealPeriods = [ 
    'Breakfast', 
    'Lunch' 
]; 

$document = new DOMDocument(); 
$document->loadXml($xml); 
$xpath = new DOMXpath($document); 

// iterate the meal periods 
foreach ($mealPeriods as $mealPeriod) { 
    echo $mealPeriod, "\n=====================\n"; 

    // fetch all menu element nodes for a specific meal period 
    $expression = "/data/menu[@mealperiodname = '$mealPeriod']"; 
    foreach ($xpath->evaluate($expression) as $menu) { 
    echo "\n", $menu->getAttribute('location'), "\n---------------------\n"; 

    // iterate the recipe element nodes for a specific menu 
    foreach ($xpath->evaluate('recipes/recipe', $menu) as $recipe) { 
     echo '#', $recipe->getAttribute('id'), ' '; 
     echo $recipe->getAttribute('description'), "\n"; 
    } 
    } 
    echo "\n"; 
} 

輸出：

Breakfast 
===================== 

food court 1 
--------------------- 
#5626935 Oatmeal RD 
#5371796 Rice Brown RD 

Lunch 
===================== 

food court 2 
--------------------- 
#4430587 Soup Tomato w/Garden Vegetables 
#4210899 Soup Beef Barley w/Vegetables