Question

我在php中尝试一个联系表单，其中的详细信息将存储在数据库中。如果我没有输入任何值，它会显示错误消息，但它会存储在数据库中。如何在错误消息显示时验证表单，不应在数据库中输入数据。这是代码

<?php
 $username = "root";
$password = "";
$hostname = "localhost"; 
$db = "abc";

//connection to the database
$name="";
$email="";
$batch="";
$mobile="";

    if (isset($_POST['submit'])) {
    $error = "";

    if (!empty($_POST['name'])) {
    $name = $_POST['name'];
    } else {
    $error .= "You didn't type in your name. <br />";
    }

    if (!empty($_POST['email'])) {
    $email = $_POST['email'];
      if (!preg_match("/^[_a-z0-9]+(\.[_a-z0-9-]+)*@[a-z0-9-]+(\.[a-z0-9-]+)*(\.[a-z]{2,3})$/i", $email)){ 
      $error .= "The e-mail address you entered is not valid. <br/>";
      }
    } else {
    $error .= "You didn't type in an e-mail address. <br />";
    }
if (!empty($_POST['batch'])) {
    $batch = $_POST['batch'];
    } else {
    $error .= "You didn't type batch. <br />";
    }
     if(($_POST['code']) == $_SESSION['code']) { 
    $code = $_POST['code'];
    } else { 
    $error .= "The captcha code you entered does not match. Please try again. <br />";    
    }



    if (!empty($_POST['mobile'])) {
    $mobile = $_POST['mobile'];
    } else {
    $error .= "You didn't type your Mobile Number. <br />";
    }







    if (empty($error)) {





 $success = "<b>Thank you! Your message has been sent!</b>";


    }
    }
    ?>

              <div id="contactForm">



                <?php
      if (!empty($error)) {
      $dbhandle = mysql_connect($hostname, $username, $password) or die("Unable to connect to MySQL");
mysql_select_db($db,$dbhandle) or die('cannot select db');

mysql_query("INSERT INTO contact (name,batch,email,mobile) 
                VALUES('$name','$batch','$email','$mobile') ") or die(mysql_error());
      echo '<p class="error"><strong>Your message was NOT sent<br/> The following error(s) returned:</strong><br/>' . $error . '</p>';
      } elseif (!empty($success)) {
      echo $success;
      }
    ?>

Answer 1

这与应该是什么相反

if (!empty($error)) {
    ^
     // your database stuff here
}

您应该在错误为空时运行该查询，而不是在其不为空时运行

if (empty($error)) {
      // now save to database    
 }

同时浏览How can I prevent SQL injection in PHP?

Answer 2

检查数据库中插入数据的条件。您正在检查(!empty($error))是否应该表示存在错误。此外，由于$error是一个字符串，我建议您将值检查为if(trim($error) != "")，而不是使用empty()

Answer 3

// also correct !empty ()
mysql_query(" INSERT INTO contact (`name`,`batch`,`email`,`mobile`) 
            VALUES('".$name."','".$batch."','".$email."','".$mobile."');

您需要连接字符串。如果您将$email放在引号中，它将被视为字符串而不是变量。

Answer 4

您应该使用else if检查每个条件..

if(isset($POST['submit'])){
if(empty($_POST['email'])){
$error[] = "email is required";
}
elseif(empty($_POST['name'])){
$error[]= "name is required;";
}
...
else{
 $email = $_POST['email'];
 $name = $_POST['name'];
// do all the stuff here
}
}

值无需输入数据即可插入数据库

4 个答案: