如何在简单的 PHP 中在同一页面中发布？
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How to Post in the Same Page in Simple PHP?
提问by Nirjhor
I am trying to create a Registration System using only PHP. This is an example of that. But I think I have done something wrong. I tried to find similar solution in StackOverFlow posts but didn't get any exact solution. It would be really get if someone would help me to find the error in my code below.
我正在尝试仅使用 PHP 创建一个注册系统。这是一个例子。但我觉得我做错了什么。我试图在 StackOverFlow 帖子中找到类似的解决方案，但没有得到任何确切的解决方案。如果有人能帮我在下面的代码中找到错误，那将是真的。
<?php
// POST HANDLER -->
if(isset($_POST['registerForm']))
{
conFunc(); // Connection Function
$userid = $_POST['userid'];
$name = $_POST['name'];
$getUserId = mysql_query("SELECT * FROM `user` WHERE `userid` = '".$userid."'");
$id = mysql_fetch_array($getUserId);
if($id)
{
echo "This User ID is Already Available on the System, Please Choose Something Else!";
}
else
{
$query = mysql_query("INSERT INTO `user` (`userid`, `name`");
if($query)
{
echo "A New User is Registered Successfully:<br /><br />";
echo "<b>User ID:</b> " . $userid . "<br />";
echo "<b>User Name:</b> " . $name . "<br />";
}
else
{
echo "There is an Error while Saving: " . mysql_error();
echo "<br />Please click on Create User from menu, and try again<br /><br />.";
}
}
exit;
}
// POST HANDLER -->
?>
<!-- FORM GOES BELOW -->
<form action="<?php echo $_SERVER['PHP_SELF']?>" method="post" name="registerForm">
<table style="width: 100%">
<tr>
<td>User ID</td>
<td><input name="userid" type="text" style="width: 300px" /><br /></td>
</tr>
<tr>
<td>Name</td>
<td><input name="name" type="text" style="width: 300px" /><br /></td>
</tr>
<tr>
<td></td>
<td><input style="width: 130px; height: 30px" type="submit" name="submit" value="Register Now" /><br /></td>
</tr>
</table>
</form>
回答by Yogesh Suthar
You have to check the submit button is set or not.
您必须检查提交按钮是否设置。
if(isset($_POST['registerForm']))
should be
应该
if(isset($_POST['submit'])) {
// your php code
} else {
// your html code
}
回答by Nishu Tayal
registerform element is not treated as post element so check with submit button.
registerform 元素不被视为 post 元素，因此请使用提交按钮进行检查。
Try following code :
尝试以下代码：
<?php
// POST HANDLER -->
if(isset($_POST['submit'])){
conFunc(); // Connection Function
$userid = $_POST['userid'];
$name = $_POST['name'];
$getUserId = mysql_query("SELECT * FROM `user` WHERE `userid` = '".$userid."'");
$id = mysql_fetch_array($getUserId);
if($id)
{
echo "This User ID is Already Available on the System, Please Choose Something Else!";
}
else
{
$query = mysql_query("INSERT INTO `user` (`userid`, `name`");
if($query)
{
echo "A New User is Registered Successfully:<br /><br />";
echo "<b>User ID:</b> " . $userid . "<br />";
echo "<b>User Name:</b> " . $name . "<br />";
}
else
{
echo "There is an Error while Saving: " . mysql_error();
echo "<br />Please click on Create User from menu, and try again<br /><br />.";
}
}
exit;
}else{
// POST HANDLER -->
?>
<!-- FORM GOES BELOW -->
<form action="<?php echo $_SERVER['PHP_SELF']?>" method="post" name="registerForm">
<table style="width: 100%">
<tr>
<td>User ID</td>
<td><input name="userid" type="text" style="width: 300px" /><br /></td>
</tr>
<tr>
<td>Name</td>
<td><input name="name" type="text" style="width: 300px" /><br /></td>
</tr>
<tr>
<td></td>
<td><input style="width: 130px; height: 30px" type="submit" name="submit" value="Register Now" /><br /></td>
</tr>
</table>
</form>
<?php } ?>
回答by Rakesh Sharma
you need to read about mysql queries http://php.net/manual/en/book.mysql.php
你需要阅读关于 mysql 查询http://php.net/manual/en/book.mysql.php
also check your insert query not any data values inserted.
还要检查您的插入查询，而不是插入的任何数据值。
<?php if(isset($_POST['submit']))
{
conFunc(); // Connection Function
$userid = $_POST['userid'];
$name = $_POST['name'];
$getUserId = mysql_query("SELECT * FROM user WHERE userid = '".$userid."'");
$id = mysql_fetch_array($getUserId);
if($id)
{
echo "This User ID is Already Available on the System, Please Choose Something Else!";
}
else
{
$query = mysql_query("INSERT INTO user(userid, name) values('" .$userid . "','" . $name . "')";
if($query)
{
echo "A New User is Registered Successfully:<br /><br />";
echo "<b>User ID:</b> " . $userid . "<br />";
echo "<b>User Name:</b> " . $name . "<br />";
}
else
{
echo "There is an Error while Saving: " . mysql_error();
echo "<br />Please click on Create User from menu, and try again<br /><br />.";
}
}
exit;
}
// POST HANDLER -->
?>
<!-- FORM GOES BELOW -->
<form action="" method="post" name="registerForm">
<table style="width: 100%">
<tr>
<td>User ID</td>
<td><input name="userid" type="text" style="width: 300px" /><br /></td>
</tr>
<tr>
<td>Name</td>
<td><input name="name" type="text" style="width: 300px" /><br /></td>
</tr>
<tr>
<td></td>
<td><input style="width: 130px; height: 30px" type="submit" name="submit" value="Register Now" /><br /></td>
</tr>
</table>
</form>
回答by Gautam3164
Your script should be like
你的脚本应该像
$getUserId = mysql_query("SELECT id FROM `user` WHERE `userid` = '".$userid."'");
because you are getting all the results and you need to retrive the idonly and your form action should be your same page itself
因为你得到了所有的结果，你需要检索id唯一的结果，你的表单操作应该是你的同一个页面本身
