未定義的索引數組不能引用我的其他PHP頁面？

Question

<?php 
define("TITLE", "Menu Item | Risque"); 

include('includes/header.php'); 

$menuItem = ''; 
$dish = ''; 

if (isset($_GET['item'])) { 

    $menuItem = $_GET['item']; 

    $dish = $menuItems[$menuItem]; 
} 

function suggestedTip ($price, $tip){ 

    $totalTip = $price * $tip; 
} 
?> 

<center><img src = "hr.png"> 
<br> 

<div id = "dish"> 


    <h1><?php $dish = array(); echo $dish["title"]; ?> <span class="price"><sup>$</sup><?php echo $dish["price"]; ?></span></h1> 
    <p><?php echo $dish["blurb"]; ?></p> 
    <br> 
    <p><strong>Suggested beverage: <?php echo $dish["drink"]; ?></strong></p> 
    <p><em>Suggested tip: <sup>$</sup><?php suggestedTip($dish["price"], 0.20); ?></em></p> 

</div> 

arrays.php代碼如下所示 莫名其妙的數組中的值不會在我的網頁顯示

<?php 

$navItems = array (

       array(
        'slug' => "index.php", 
        'title' => "Home" 
        ), 

       array(
        'slug' => "menu.php", 
        'title' => "Menu" 
        ), 

       array(
        'slug' => "contact.php", 
        'title' => "Contact" 
        ), 

       ); 



$menuItems = array(

       "club-sandwich" => array(

        "name" => "Club Sandwich", 
        "price" => 11, 
        "blurb" => "Bacon ipsum dolor sit amet fatback landjaeger ullamco pariatur. Nostrud laboris et, duis drumstick eiusmod kevin ut aliquip. Filet mignon short ribs tenderloin short loin kielbasa non pork chop aliqua. Jerky shank tri-tip capicola, non andouille tenderloin cupidatat turducken meatball pork beef eu shoulder jowl.", 
        "drink" => "Club Soda" 

       ), 

       "dill-salmon" => array(

        "name" => "Lemon &amp; Dill Salmon", 
        "price" => 18, 
        "blurb" => "Pork belly tempor ground round qui exercitation, jowl leberkas sed voluptate excepteur jerky. Reprehenderit veniam cow, quis in ribeye andouille eu pastrami eiusmod exercitation dolor.", 
        "drink" => "Fancy Wine" 

       ), 

       "super-salad" => array(

        "name" => "The Super Salad<sup>&reg;</sup>", 
        "price" => 34, 
        "blurb" => "Gumbo beet greens corn soko endive gumbo gourd. Parsley shallot courgette tatsoi pea sprouts fava bean collard greens dandelion okra wakame tomato. Dandelion cucumber earthnut pea peanut soko zucchini.", 
        "drink" => "Jug o' Water" 

       ), 

       "mexican-barbacoa" => array(

        "name" => "Mexican Barbacoa", 
        "price" => 23, 
        "blurb" => "I love fruitcake danish caramels. Tart danish pastry liquorice chocolate cake fruitcake. Bear claw gingerbread muffin I love apple pie apple pie tiramisu brownie chocolate. Sweet roll cotton candy cupcake gingerbread gummies jelly-o. Muffin I love croissant I love jelly-o brownie jelly beans. Toffee I love pastry.", 
        "drink" => "Beer with a lime" 

       ), 

      ); 

？>

我是新來的PHP，但我有一個項目，是因爲很快就會發生，但是我一直收到title，price和blurb的未定義錯誤。

如果有人能幫到，我將不勝感激。

Answer 1

您已經爲變量$ dish聲明瞭空數組。所以沒有辦法從空陣列獲得像標題，價格，blurb索引。如果您試圖從數據庫獲取值，請將數據庫結果存儲在變量$ dish中並繼續。在使用isset（）函數進行回顯之前，還要始終檢查數組索引是否存在。否則它會拋出錯誤最大模式和錯誤日誌文件中的通知錯誤。請參閱下面的代碼。

<h1><?php if(isset($dish["title"])) echo $dish["title"]; ?> <span class="price"><sup>$</sup><?php if(isset($dish["price"])) echo $dish["price"]; ?></span></h1> 
<p><?php echo if(isset($dish["blurb"])) $dish["blurb"]; ?></p> 
<br> 
<p><strong>Suggested beverage: <?php if(isset($dish["drink"])) echo $dish["drink"]; ?></strong></p> 
<p><em>Suggested tip: <sup>$</sup><?php if(isset($dish["price"])) suggestedTip($dish["price"], 0.20); ?></em></p> 

所以這段代碼不會顯示錯誤。