顯示從數據庫中選擇的下拉值（egmysql）

Question

我已經從選擇下拉列表中將蘋果的值保存到mysql中。 現在在編輯頁面中，我想顯示相應的值。 但我不似乎得到它的權利

<?php $food = $mysqli->real_escape_string($_POST['food']); ?> 
 

 
<select id="food" required name="food" > 
 

 
    <option value="" disabled selected>Select a Food</option> 
 
    <option value="apple" <?php if($food == apple) 
 
    echo "selected='selected'"; ?>Apple</option> 
 
    
 
    <option value="kiwi" <?php if($food == kiwi) 
 
    echo "selected='selected'"; ?>Kiwi</option> 
 
</select></div>

Answer 1

<select id="food" required name="food" > 
 

 
    <option value="" >Select a Food</option> 
 
    <option value="apple" <?php if($food == "apple") 
 
    echo "selected='selected'"; ?>Apple</option> 
 
    
 
    <option value="kiwi" <?php if($food == "kiwi") 
 
    echo "selected='selected'"; ?>Kiwi</option> 
 
</select></div>

Answer 2

String應該由single報價或報價doubleif($food == 'apple')

封閉10

<select id="food" required name="food" > 

    <option value="" disabled selected>Select a Food</option> 
    <option value="apple" <?php if($food == 'apple') 
    echo "selected='selected'"; ?>Apple</option> 

    <option value="kiwi" <?php if($food == 'kiwi') 
    echo "selected='selected'"; ?>Kiwi</option> 
</select></div>