PHPエコーメッセージをフェードアウト

Question

PHPエコーメッセージをフェードアウトしたい。ここに私のPHPコードは次のとおりです。

フォームデータが送信されます

<?php 
    $name = $_POST['name']; 
    $email = $_POST['email']; 
    $message = $_POST['message']; 
    $subject = $_POST['subject']; 

    $to = 'ajay.k@enexl.com'; 
    if (filter_var($email, FILTER_VALIDATE_EMAIL)) { // this line checks that we have a valid email address 
     $mailSubject = "Contact request from " .$name; 
     $txt = "name : ".$name.".\n\nSubject : ".$subject.".\n\nMail id : ".$email."\n\nMessage : ".$message; 
     $headers = "From: ".$email ; 
     mail($to,$mailSubject,$txt,$headers); 
     $data = array(); 
     $data['status'] = 'success'; 
     //echo json_encode($data); 
     echo "<script src='http://ajax.googleapis.com/ajax/libs/jquery/1.9.0/jquery.min.js'></script>"; 
     echo "<p id='#text'>Your email was sent! One of our team members would contact you shortly!</p>"; // success message 
     echo "<script type='text/javascript'>"; 
     echo "$(function(){"; 
     echo "$('#text').fadeOut(5000);"; 
     echo "});"; 
     echo "</script>"; 
    } 
    else{ 
     echo "Mail was not sent, make sure that all fields are filled in"; 
    } 

?> 

、首尾よく、私はYour email was sent! One of our team members would contact you shortly!としての応答を取得します。しかし、期待どおりにフェードアウトしません。どのように私はそれをフェードアウトさせることができますか？

Answer 1

あなたのID修飾子に#を使用しないでください。 #textにid textない#text

echo "<p id='#text'>Your email was sent! One of our team members would contact you shortly!</p>"; // success message 

を持つ要素を見つけるために言う

echo "<p id='text'>Your email was sent! One of our team members would contact you shortly!</p>"; // success message