2017-09-25 76 views
0

我想在提交后的表单下的同一页面上显示消息。提交后显示消息(php,html)

我的PHP

header('Location: services.php'); 
$formSucc = "Ďakujeme vaša správa bola úspešne odoslaná."; 

我的HTML

<form class="quote" action="kontakt.php" method="POST"> 
    <div> 
     <label>Vaše Meno</label><br> 
     <input type="text" placeholder="Meno..." name="name" required="" oninvalid="this.setCustomValidity('Prosím zadajte meno')" oninput="setCustomValidity('')"> 
    </div> 
    <div> 
     <label>Email</label><br> 
     <input type="email" placeholder="E-mail..." name="email" required="" oninvalid="this.setCustomValidity('Prosím zadajte platný e-mail')" oninput="setCustomValidity('')"> 
    </div> 
    <div> 
     <label>Správa</label><br> 
     <textarea placeholder="Napíšte nám" name="text" required="" oninvalid="this.setCustomValidity('Prosím napíšte čo si želáte')" oninput="setCustomValidity('')"></textarea><br> 
     <span class="error"> <?php echo $formSucc;?></span> 
    </div> 
    <input type="submit" value="Odoslať" name="submit" id="submit"> 
    </div> 
</form> 
+1

向我们展示您的整个表单页面。 – GrumpyCrouton

+0

http://textuploader.com/djppv这是表单 – Georgo

+0

您可能想要阅读以下基本示例:http://w3schools.sinsixx.com/php/php_mail.asp.htm – user3366016

回答

0

你应该改变你的头重定向到包含GET变量。

header('Location: services.php?msg=This is a message'); 

然后,改变你的HTML检查这个消息,如果它存在然后显示它。

<span class="error"><?php if(!empty($_GET['msg'])) echo $_GET['msg']; ?></span> 
+0

感谢它的作品:) – Georgo