Question

    <div class="contact-forum">

    <h1>Contact Forum</h1>
    <div class="contact-enters">
    <form action="Home.php" method="post">
        <div class="firstname">
            <p>First Name:</p> </div>  <input type="text" name="firstname" placeholder="FirstName" /> <br />

        <div class="lastname">
            <p> LastName: </p></div> <input type="text" name="lastname" placeholder="LastName" /> <br />

        <div class="Email">
            <p> Email: </p></div>  <input type="text" name="email" placeholder="E-mail" /> <br />

        <div class="topic">
            <p>Topic: </p> </div><input type="text" name="topic" placeholder="Topic" /> <br />


     <div class="message">
        <p> Message: </p> </div> <div class="message-box"><input type="text" name="message" /></div> <br /> 

        <input type="submit" name="Send" />
    </form>
    </div>
</div>

<div class="php">
<?php

$con = mysqli_connect("localhost", "KyleLongrich", "Cash7144") or die('can not connect to server');

mysqli_select_db($con,"kylelongrich") or die("could not connect to database");


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

$Firstname = strip_tags($_POST['firstname']);
$Lastname = strip_tags($_POST['lastname']);
$Email = strip_tags($_POST['email']);
$Topic = strip_tags($_POST['topic']);
$Message = strip_tags($_POST['message']);

$email_query = mysqli_query($con,"SELECT Email FROM emailquestions WHERE Email ='$Email'") or die("Could not check email");
$count_email = mysqli_num_rows($email_query);

if ($count_email > 0)
{
    echo 'Your email is alread in use';
}


$sql = "INSERT INTO emailquestions (Firstname, Lastname, Email, Subject, Message) VALUES ('$Firstname','$Lastname','$Email','$Topic','$Message')";

mysqli_query($con, $sql);

$message = 'Your message has been sent';

}


$message='';



?>

显示时，提示按钮在发送时不会发送任何内容或工作。没有PHP代码错误回来但它将无法正常工作。可能是因为造型？任何帮助都会很棒

Answer 1

在if语句后尝试使用else，并在查询中使用SET。

if ($count_email > 0)
{
echo 'Your email is alread in use';
}
else
{
$sql = "INSERT INTO emailquestions SET         
 (Firstname, Lastname, Email, Subject,   
 Message) VALUES   
     ('$Firstname','$Lastname','$Email','$Topic','$Message')";

mysqli_query($con, $sql);

$success_message = 'Your message has been        
sent';
echo $success_message;
}

希望有所帮助

Answer 2

这有点无关，但我认为这会有所帮助。您正在使用mysqli，但我建议您使用PDO连接到您的数据库。它将来更安全，更容易扩展。以下是如何使用PDO连接数据库的示例。

<?php

  $myUser = "XXXXXX";
  $myPass = "XXXXXX";

  try{

    $dbPDO = new PDO('mysql:host=localhost;dbname=xxxxxxxx', $myUser, $myPass);
    $dbPDO->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);

    echo "Connection was successful";

  } catch(PDOException $e){

    print "Error!: " . $e->getMessage() . "<br />";
    die();

  }

?>

只需将X更改为服务器的设置即可。我会将其存储在自己的文件中，并使用include_once('dbConn.php');之类的内容将其包含在以后的所有文件中。

当您想要启动查询时，您可以这样做：

$query = $dbPDO->prepare("SELECT * FROM Table_Name");
$query->execute();

当然，您希望将变量传递给您的查询，以便您可以这样做：

$query = $dbPDO->prepare("SELECT * FROM Table_Name WHERE ID = :id");
$query->bindParam(':id', $id);
$query->execute();

这可以避免SQL注入。只需确保在将变量绑定到查询之前清理变量。

将变量直接传递到查询中会使您容易受到SQL注入的攻击，PDO通过将变量绑定到查询来帮助防止这种情况。

Answer 3

您必须关闭contact-enters div。尝试流动代码。

<h1>Contact Forum</h1>
    <div class="contact-enters">
    <form action="Home.php" method="post">
        <div class="firstname">
            <p>First Name:</p> </div>  <input type="text" name="firstname" placeholder="FirstName"> <br />

        <div class="lastname">
            <p> LastName: </p></div> <input type="text" name="lastname" placeholder="LastName"> <br />

        <div class="Email">
            <p> Email: </p></div>  <input type="text" name="email" placeholder="E-mail"> <br />

        <div class="topic">
            <p>Topic: </p> </div><input type="text" name="topic" placeholder="Topic"> <br />


     <div class="message">
        <p> Message: </p> </div> <div class="message-box"><input type="text" name="message"></div> <br /> 

        <input type="submit" name="Send">
    </form>
    </div>
</div>

PHP提交按钮不能正常工作

3 个答案: