如果isset語句沒有調出輸入字段

Question

我有以下代碼，我試圖讓'Finalize Draft Order'提交按鈕僅在'Create Draft Order'按鈕被按下/設置時出現。現在，按下Create Create Order按鈕後，按鈕不會顯示出來。它只有當我把它拿出來，如果（isset功能的顯示

我在做什麼錯

<form method="POST" name="form"> 
<input type="submit" value="Create Draft Order" name="shuffle"> 
</form> 

    Shuffled results: <br> 
    <div class="main-bag">  
    <div class="shuffle_results" id="results"></div> 
    <form method="post"> 
<?php 
$count = 0; 
    foreach ($array as $result) : 
    $count++; 
    $shuffle_count = $count; 
     $shuffle_firstname = htmlentities($result['firstname']); 
     $shuffle_lastname = htmlentities($result['lastname']); 
     $shuffle_id = htmlentities($result['id']); 
     $shuffle_username = htmlentities($result['username']); 
     $shuffle_email = htmlentities($result['email']); 
?> 
    <input type="hidden" name="count[]" value="<?php echo $shuffle_count; ?>"> 
     <input type="hidden" name="firstname[]" value="<?php echo $shuffle_firstname; ?>"> 
     <input type="hidden" name="lastname[]" value="<?php echo $shuffle_lastname; ?>"> 
     <input type="hidden" name="id[]" value="<?php echo $shuffle_id; ?>"> 
     <input type="hidden" name="username[]" value="<?php echo $shuffle_username; ?>"> 
     <input type="hidden" name="email[]" value="<?php echo $shuffle_email; ?>"> 

<?php 

    endforeach; 
    // only show this button if we have done a shuffle 
    if (isset($_POST['shuffle'])) : 
     echo '<input type="submit" value="Finalize Draft Order" name="insert">'; 
    endif; 

?> 

更新：？

$array = array(); 

while ($row = mysqli_fetch_assoc($query)) { 
$array[] = array( 
'id' => $row['id'], 
'firstname' => $row['firstname'], 
'lastname' => $row['lastname'], 
'username' => $row['username'], 
'email' => $row['email'] 
); 

if (isset($_POST['shuffle'])) { 

    } 
} 
shuffle($array); 
    echo json_encode($array); 

Answer 1

我不認爲$ _ POST [」洗牌']設置，而不是使用提交按鈕，它改變你的代碼：

<form method="POST" name="form"> 
<input type="hidden" name="shuffle" value="true"> 
<input type="submit" value="Create Draft Order"> 
</form> 

我敢肯定，提交的名稱不COUN t表示POST值。

Answer 2

這是你的問題。您的輸入按鈕被命名爲shuffle，在提交時不會傳遞。如果您添加一個名爲「shuffle」的新隱藏輸入元素，它將傳遞它。

編輯：

你已經在你的問題其創建草案按下按鈕後說。因此，按您的例子：

<form method="POST" name="form"> 
<input type="submit" value="Create Draft Order" name="form"> 
<input type="hidden" name="shuffle" value="1"> 
</form> 

    Shuffled results: <br> 
    <div class="main-bag">  
    <div class="shuffle_results" id="results"></div> 
    <form method="post"> 
<?php 
$count = 0; 
    foreach ($array as $result) : 
    $count++; 
    $shuffle_count = $count; 
     $shuffle_firstname = htmlentities($result['firstname']); 
     $shuffle_lastname = htmlentities($result['lastname']); 
     $shuffle_id = htmlentities($result['id']); 
     $shuffle_username = htmlentities($result['username']); 
     $shuffle_email = htmlentities($result['email']); 
?> 
    <input type="hidden" name="count[]" value="<?php echo $shuffle_count; ?>"> 
<input type="hidden" name="firstname[]" value="<?php echo $shuffle_firstname; ?>"> 
     <input type="hidden" name="lastname[]" value="<?php echo $shuffle_lastname; ?>"> 
     <input type="hidden" name="id[]" value="<?php echo $shuffle_id; ?>"> 
     <input type="hidden" name="username[]" value="<?php echo $shuffle_username; ?>"> 
     <input type="hidden" name="email[]" value="<?php echo $shuffle_email; ?>"> 

<?php 

    endforeach; 
    // only show this button if we have done a shuffle 
    if (isset($_POST['shuffle'])) : 
     echo '<input type="submit" value="Finalize Draft Order" name="insert">'; 
endif; 
?> 

Answer 3

我不知道用的是什麼版本的PHP meame69和ScottyMcGready的，但你很可以檢查，看看是否按鈕被點擊通過爲isset檢查（$ _ POST [ 「submit」]）是真實的。

http://www.learningaboutelectronics.com/Articles/How-to-check-if-the-submit-button-is-clicked-in-PHP.php

我不能發表評論，但我會編輯此一次，我找出什麼是錯貝基的代碼。