Question

我在PHP工作。我创建了一个名为Donor.php的表单并将其连接到数据库。现在我正在尝试在PHP中应用检查。但他们是一个问题。因为我已在表单上应用PHP中的空字段检查，但这些检查不起作用。请查看我的代码。因为我的工作因为这个问题而陷入困境。我的代码文件在这里：

Donor.php

<?php
//error_reporting(0);
if(isset($_POST['submit'])){

 $first_name=$_POST['firstname'];
 $last_name=$_POST['lastname'];
 $Country=$_POST['country'];
 $City=$_POST['city'];
 $Gender=$_POST['gender'];
 $Email=$_POST['email'];
 $Password=$_POST['pwd'];

include_once "connectionn.php";
$emailChecker=mysql_real_escape_string($Email);
$sql_email_check=mysql_query("Select Email FROM user WHERE Email='$emailChecker'");
$email_check=mysql_num_rows($sql_email_check);


if((empty($first_name)) ||(empty($last_name)) ||(empty($City)) ||(empty($Gender)) ||(empty($Email)) ||(empty($Password))) {
    $errorMsg='We are sorry, but there appears to be a problem with the form you submitted.'; 

    if (empty($first_name)) {
   $errorMsg.='$var is either 0, empty, or not set at all';
   header('Location: Donor.php');
}
    if(empty($last_name)){
        $errorMsg.='lastname';
        header('Location: Donor.php');
        }
        if(empty($City)){
        $errorMsg.='City';
        header('Location: Donor.php');
        }
        if(empty($Gender)){
        $errorMsg.='Gender';
        header('Location: Donor.php');
        }
        if(empty($Email)){
        $errorMsg.='email';
        header('Location: Donor.php');
        }
        if(empty($Password)){
        $errorMsg.='Password';
        echo "$errorMsg.";
        header('Location: Donor.php');
        }
    }else if($email_check>0){
        $errorMsg="invalid";
        }else{
            $sql="INSERT INTO user (User_ID,First_Name, Last_Name, gender, city, Email, Password) VALUES (NULL,'$first_name', '$last_name','$Gender','$City','$Email','$Password')";
$result=mysql_query($sql);


$UserID="SELECT max(User_ID) as usr from user";
$userIDResult=mysql_query($UserID);
if($userIDResult === false)
{
    die(mysql_error());
    }
while($R=mysql_fetch_array($userIDResult)){
    $usrID= $R['usr'];

    }
    $donor="INSERT INTO donor(User_ID, Country)Values('".$usrID."','$Country')";
    $resultdonor=mysql_query($donor);




mysql_close();
header('Location: DonorPro.php');

            }
}

    ?>
<?php 
include "Header.php";
//include "registration.php";
 ?>
<div class="DonorDiv">
<h1>Lets Join:</h1>
<form name="input" action="" method="post" <?php print"$errorMsg"; ?>>

First Name: <input type="text" name="firstname" placeholder="First Name"  id="r">
<?php print "$first_name"; 
// if (!isset($_POST['firstname'])) { 
    //echo '$var is either 0, empty, or not set at all';
//}
    ?>


Last Name: <input type="text" name="lastname" placeholder="Last Name"  id="u" <?php print "$last_name";?>> <br>
Institution: <input type="text" name="country" placeholder="Institution"  id="" <?php print "$Institution";?>>
City: <input type="text" name="city" placeholder="City"  id="" <?php print "$City";?>><br>
Country: <input type="text" name="country" placeholder="Country" id="" <?php print "$Country";?>><br>
Gender: <input type="text" name="gender" placeholder="Gender"  id="" <?php print "$Gender";?>><br>
Email Address: <input type="Email" name="email" placeholder="Email" id="g" <?php print "$Email";?>><br>
Password:<input type="Password" name="pwd" placeholder="Password" id="v" <?php print"$Password";?>><br>

<input type="submit" src="images/button(9).png" alt="Submit" id="q">
</form>
</div>

<?php include "Footer.php"; ?>

Answer 1

不推荐使用PHP mysql lib，你应该考虑使用myslqi或php PDO。

Here is a tutorial

您还应该小心：$first_name以及显示表单时未定义的其他变量，因此您将收到警告。

无论如何，你的问题是这个检查总是错误的：

if(isset($_POST['submit'])){

最简单（但不是最好）的方法是在表单中添加隐藏的输入：

<input type="hidden" name="hidden">

Answer 2

在告诉浏览器重定向到另一个页面后，您必须退出PHP脚本：

header('Location: Donor.php');
exit;

（除了SQL注入和其他一些问题。）

Checks In PHP无效

2 个答案: