Question

这是注册表，我希望其数据存储在MySQL数据库中

它总是打印其他部分，即提交失败

我的Php文件如下：我做了一个简单的注册表。我希望将其数据存储在$ db变量中提到的数据库中。我已经在文件中提到的phpmyadmin中创建了数据库。但是我不知道为什么这段代码不起作用。请告诉我哪一行代码是错误的，我应该如何纠正。

<?php
 $host="localhost";
 $user="root";
 $pwd="";
 $db="mohita";

$conn=mysqli_connect($host,$user,$pwd,$db) or die("unable to connect");
?>

<html>
   <head>
      <title>Try here</title>
   </head>

   <body>
      <form method="post">
       <input type="text" name="yourname"/>
       <input type="email" name="mailid"/>       
       <input type="password" name="mypass"/>
       <input type="number" name="phn"/>
       <textarea rows="5" cols="20" name="address"/></textarea>
       <input type="submit" name="reg"/>
      </form>
   </body>
</html>

<?php
  if(isset($_POST['reg'])){
       $yourname=$_POST['yourname'];
       $mailid=$_POST['mailid'];
       $mypass=$_POST['mypass'];
       $phn=$_POST['phn'];
       $address=$_POST['address'];

     $query="insert into login(Name,Email,Pass,Mobile,Address) values('$yourname','$mailid','$mypass','$phn','$address')";

    $run=mysqli_query($conn,$query);

    if($run){
       echo "<h1>Data submitted successfully</h1>";
     }

    else{
       echo "<h1>Failed to submit data</h1>";
     }
}
?>

Answer 1

尝试一下，让我知道是否可行！

  <html>
  <head>
  <title>Try here</title>
 </head>

 <body>
  <form method="post" action="">
   <input type="text" name="yourname"/>
   <input type="email" name="mailid"/>       
   <input type="password" name="mypass"/>
   <input type="number" name="phn"/>
   <textarea rows="5" cols="20" name="address"/></textarea>
   <input type="submit" name="reg"/>
  </form>
<?php
if(isset($_POST['reg'])){
   $host="localhost";
    $user="root";
    $pwd="";
   $db="mohita";
   $yourname=$_POST['yourname'];
   $mailid=$_POST['mailid'];
   $mypass=$_POST['mypass'];
   $phn=$_POST['phn'];
   $address=$_POST['address'];

  $conn=mysqli_connect($host,$user,$pwd,$db) or die("unable to connect");
 $query="insert into login(Name,Email,Pass,Mobile,Address) values('$yourname','$mailid','$mypass','$phn','$address')";

 $run=mysqli_query($conn,$query);
   if($run){
   echo "<h1>Data submitted successfully</h1>";
 }

else{
   echo "<h1>Failed to submit data</h1>";
 }
}
?>

  </body>
 </html>

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1 个答案: