Php POST不传递变量
这是我的密码。单击按钮转到playlist.php文件后,变量不会像post命令中那样被带出。帮忙 编辑:这是发送变量的文档中的代码Php POST不传递变量,php,post,curl,get,Php,Post,Curl,Get,这是我的密码。单击按钮转到playlist.php文件后,变量不会像post命令中那样被带出。帮忙 编辑:这是发送变量的文档中的代码 if(empty($name) || empty($email)){ echo '<h1>Please fill out all fields.</h1>'; }else{ $dup = mysql_query("SELECT name FROM sets WHERE name='".$_POST
if(empty($name) || empty($email)){
echo '<h1>Please fill out all fields.</h1>';
}else{
$dup = mysql_query("SELECT name FROM sets WHERE name='".$_POST['name']."'");
if(mysql_num_rows($dup) >0){
echo '<h1>Username Already Used.</h1>';
}
else{
$sql = mysql_query("INSERT INTO sets VALUES ('','$name','$email')");
if($sql){
$to = $email;
$email_subject = "You've created a playlist!";
$email_body = "Thanks for signing up! ".
" Here are the details you used to sign up with:\n Name: $name \n Email: $email";
$headers = "From: email@email.com";
$headers .= "Reply-To: email@email.com";
mail($to,$email_subject,$email_body,$headers);
echo '<form action="playlist.php" method="post">
<input type="hidden" name="name" value="<?php echo $name; ?>">
<p class="text-center"><button type="submit" class="btn btn-primary btn-large">Visit my Playlist!</button></p>
</form>';
}
else{
echo '<h1>Error in Registration.</h1>';
}
}
}
我在您的代码中看到的至少一个错误如下:
echo '<form action="playlist.php" method="post">
<input type="hidden" name="name" value="<?php echo $name; ?>">
<p class="text-center"><button type="submit" class="btn btn-primary btn-large">Visit my Playlist!</button></p>
</form>';
您没有在任何地方检查$\u POST['name'],这是发送变量的文档,而不是接收变量的文档。那么,您所说的发送是什么意思呢?在此代码中只发送邮件您的html表单代码是什么样子的?
echo '<form action="playlist.php" method="post">
<input type="hidden" name="name" value="' . $name . '">
<p class="text-center"><button type="submit" class="btn btn-primary btn-large">Visit my Playlist!</button></p>
</form>';