Question

在表单提交之前，必须单击两次我的表单提交按钮。我的教授给了我一些Java脚本代码，我已经在下面实现了，它将自动点击两次提交按钮，但它不起作用。

我的表单操作如下所示：

<form name="com" id="com" action="<?php if (($_POST['email'] != NULL) && ($_POST['comment'] != NULL) && ($_POST['fname'] != NULL) && ($_POST['lname'] != NULL)) {
echo "leavehandle.php";
$submit == "true";
} else {
echo "leave.php";
} ?>" method="post">

我的结尾表单标签之外还有这个PHP脚本：

<?php
if ($submit == "true") {
echo"<script>document.getElementById('com').submit();</script> ";
} ?>

在表单提交之前，仍然需要单击两次我的表单提交按钮。任何帮助都会非常感激 - 我没有多少Java Script经验！

谢谢！

编辑：

我将表单操作中的代码调整为以下内容：

<form name="com" id="com" action="<?php echo "leavehandle.php";
if ($submit == "true"){
} else {
echo "leave.php";
} ?>" method="post">

我是否需要在表单操作的第一部分中使用PHP开头标记？此外，我不确定在我的leavehandle.php中放置其他代码来验证表单的位置。

Leavehandle.php：

<!DOCTYPE html>
<html> <!--Opening HTML tag-->
<div align="center"> <!--Aligns content center-->
<head>
<?php //pulls in my header from another PHP page using Inlcude()
include("header.php");
?>
</head>
<style>
html {
font-family:arial;
background-color:#ffcc99;
}
</style>
<body>
<?php
include('mysqli_connect.php');

$fname = $_POST['fname'];
$lname = $_POST['lname'];
$email = $_POST['email'];
$comment = $_POST['comment'];

$query = "INSERT INTO guestbook (id, email, fname, lname, date, comment)     VALUES (NULL, '$email', '$fname', '$lname', NOW(), '$comment');";

$result = mysqli_query($dbc, $query);

if ($result) {
echo "Thank you for submitting a comment $fname!";
}
else {
echo "ERROR";
}

?>
</body>
<?php //pulls in my footer from another PHP page using Inlcude()
include("footer.php");
?>
</div>
</html>

Answer 1

所以这里很热，它将是你的表格部分

<form name="com" id="com" action="leavehandle.php" method="post">

你的leavehandle.php已修复

＆＃13;

<!DOCTYPE html>
<html> <!--Opening HTML tag-->
<div align="center"> <!--Aligns content center-->
<head>
<?php //pulls in my header from another PHP page using Inlcude()
include("header.php");
?>
</head>
<style>
html {
font-family:arial;
background-color:#ffcc99;
}
</style>
<body>
<?php
include('mysqli_connect.php');

//set vars to null 
$fname = $lname = $email = $comment = "";

//validade post 



if(!empty($_POST['fname'])&&!empty($_POST['lname'])&&!empty($_POST['email'])&&!empty($_POST['comment'])) {
   

$fname = $_POST['fname'];
$lname = $_POST['lname'];
$email = $_POST['email'];
$comment = $_POST['comment'];

$query = "INSERT INTO guestbook (id, email, fname, lname, date, comment)     VALUES (NULL, '$email', '$fname', '$lname', NOW(), '$comment');";
 
$result = mysqli_query($dbc, $query);
 
} else {
	
	//redirect to error page is is null 
	header("Location: leave.php");
}

if ($result) {
echo "Thank you for submitting a comment $fname!";
}
else {
echo "ERROR";
}

?>
</body>
<?php //pulls in my footer from another PHP page using Inlcude()
include("footer.php");
?>
</div>
</html>

＆＃13;

必须单击“提交”按钮两次才能提交表单 - Java Script fix无法正常工作

1 个答案: