Php 我填写登录表单，点击提交按钮，页面被重新加载，电子邮件id为空。如何填写电子邮件id文本？

我填写登录表单并单击提交按钮，页面被重新加载，电子邮件id为空。如何填写电子邮件id文本？
这是php代码

<?php 
if(isset($_POST["btnSubmit"]))
{   
if($_POST["txtEmail"]=="")
{
    $errormsg = "Please enter Email.";

}
else if($_POST["txtPassword"]=="")
{
    $errormsg = "Please enter Password.";
}

else
{

    $Email =  $_POST["txtEmail"];

    $Password =  md5($_POST["txtPassword"]);

    $info = mysql_query("select * from tbluser where Email ='$Email' and Password='$Password'");

    $count_info=mysql_num_rows($info);

    if($count_info>0)
    {
            while($a = mysql_fetch_array($info))
            {
                $b = strcmp($a["Email"],$Email);

                if($b==0 && $a["Password"]=="$Password" && $a["IsVisible"]==1)
                {   
                    $_SESSION["UserId"] = $a["user_id"];
                    $_SESSION["UserName"] = $a["user_name"];
                    header("Location:index.php");
                }
                else
                {
                    $errormsg="Register Your Account First.";
                }
            }
    }
    else
    {
        $errormsg="Invalid email or password.";
    }

}

}
?>

---

我想您正在寻找类似于：

<input type="text" id="txtEmail" name="txtEmail" placeholder="Email" value="<?php if(isset($Email)){echo $Email;} ?>" />
<!-- you might/want to escape $Email with something like htmlspecialchars -->

PHP部分

<?php
$Email=''; //added here so the email is not confined to the else block (better practice to initilize).<------
if(isset($_POST["txtEmail"])){$Email =  $_POST["txtEmail"];} // this will ensure that even if you have not typed in the password the email still shows up.
if(isset($_POST["btnSubmit"]))
{   
if($_POST["txtEmail"]=="")
//rest of your code

那你为什么不能填写呢？这个页面可能会帮你解决这个问题。如果我知道，那么我不会问它…现在我们可以解决它：）只有电子邮件是空白的？对不起，它无法工作。重新加载页面后仍然显示空白。你能详细阐述一下吗？你的英语不太好理解你想说的话。我读了你的英语，但我认为它有几个方面。抱歉，如果你同时填写电子邮件和密码（已测试），它也不起作用。现在检查，即使你没有键入任何密码，它也会显示警告：无法修改标题信息-标题已由发送（输出开始于C:\xampp\htdocs\meetitors\connection.php:5）在第42行（标题（“Location:index.php”）；）的C:\xampp\htdocs\meetintors\login.php中，您是否收到任何警告？您正在从
connection.php的第5行发送一些输出，请检查该文件（即使
标记外的额外空间也算作输出）。
<?php
$Email=''; //added here so the email is not confined to the else block (better practice to initilize).<------
if(isset($_POST["txtEmail"])){$Email =  $_POST["txtEmail"];} // this will ensure that even if you have not typed in the password the email still shows up.
if(isset($_POST["btnSubmit"]))
{   
if($_POST["txtEmail"]=="")
//rest of your code

<input type="text" id="txtEmail" name="txtEmail" placeholder="Email" value="<?php echo htmlspecialchars($Email);?>" />
                                                                          // ^^ added here