這裏是我的javascript
讀碼變量值:
function submitForm() {
var name = document.getElementsByName('name').value
,email = document.getElementsByName('email').value
,subject = document.getElementsByName('subject').value
,body = document.getElementsByName('body').value;
$.post('php/sendForm.php', {name: name,email: email,subject: subject,body: body});
}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<form>
<div class="form-group inline-block" method="post" action="php/sendForm.php" >
<input type="text" class="form-control" name="name" placeholder="Name"></div>
<div class="form-group inline-block">
<input type="email" class="form-control" name="email" aria-describedby="emailHelp" placeholder="Enter email">
<small id="emailHelp" class="form-text text-muted">Send me an email!</small>
<input type="text" class="form-control" name="subject" aria-describedby="emailHelp" placeholder="Subject">
</div>
<div class="form-group">
<label for="exampleTextarea">Message</label>
<textarea class="form-control" name="body" rows="3">
</textarea>
</div>
<button onclick="submitForm()" type="submit" class="btn btn-primary">Submit </button>
</form>
所有這一切都意味着由php/sendForm.php
讀取。 但是,我似乎無法讀取我的php
文件中的名稱,電子郵件,主題和身體變量。 這裏是php
代碼:
$name = $_POST['name'];
$email = $_POST['email'];
$subject = $_POST['subject'];
$body = $_POST['body'];
// The message
$message .= $email. $name. $body ;
你一定形式不提交,並重新加載頁面。 – adeneo
@adeneo雅,它重新加載和所有內容 –
你怎麼知道你不能讀取值?你輸出什麼東西? –