Question

我有以下代码，我试图让'Finalize Draft Order'提交按钮仅在按下/设置'Create Draft Order'按钮时出现。现在，按下“创建草稿订单”按钮后，按钮不会显示。它只显示我是否从if（isset函数。

）中取出

我做错了什么？

<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

我认为没有设置$ _POST ['shuffle']。而不是使用提交按钮，它将您的代码更改为：

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

我很确定提交的名字不计入POST值。

Answer 2

这是你的问题。您的输入按钮名为shuffle，在提交时不会传递。如果添加一个名为“shuffle”的新隐藏输入元素，它将通过它。

修改

在按下创建草稿按钮后，您已在问题中说过了。因此，根据您的例子：

<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PHP meame69和ScottyMcGready正在使用什么版本的PHP，但您可以通过检查isset（$ _ POST [“submit”]）来检查是否单击该按钮是真的。

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

我还不能评论，但是一旦我弄清楚Becky的代码有什么问题，我会编辑这个。

如果isset语句没有显示输入字段

3 个答案: