Yani JQuery ajax işlevini kullanarak form verilerini yolluyorum. Bu tüm i veriler mesajları yakalamak mümkün değil ne kadar hiç iyi iş gibi görünüyor.
Ben $ dize kullanmaya çalışıyorum = $ _POST ['isim'] hiçbir şans ile gönderir sayfadaki sonuç yakalamak için.
JQUERY-
$(function()
{
$('.error').hide();
$(".button").click(function() {
// validate and process form here
$('.error').hide();
var name = $("input#name").val();
if (name == "") {
$("label#name_error").show();
$("input#name").focus();
return false;
}
var email = $("input#email").val();
if (email == "") {
$("label#email_error").show();
$("input#email").focus();
return false;
}
var phone = $("input#phone").val();
if (phone == "") {
$("label#phone_error").show();
$("input#phone").focus();
return false;
}
var dataString = 'name='+ name + '&email=' + email + '&phone=' + phone;
//alert (dataString);return false;
$.ajax({
type: "POST",
url: "from_text_script.php",
data: dataString,
success: function() {
window.location.href="from_text_script.php"
}
});
return false;
});
});
DAN-
<form name="contact" action="">
<fieldset>
<label for="name" id="name_label">Name</label>
<input type="text" name="name" id="name" size="30" value="" class="text-input" />
<label class="error" for="name" id="name_error">This field is required</label><br/>
<label for="email" id="email_label">Return Email</label>
<input type="text" name="email" id="email" size="30" value="" class="text-input" />
<label class="error" for="email" id="email_error">This field is required.</label> <br/>
<label for="phone" id="phone_label">Return Phone</label>
<input type="text" name="phone" id="phone" size="30" value="" class="text-input" />
<label class="error" for="phone" id="phone_error">This field is required.</label> <br/>
<br />
<input type="submit" name="submit" class="button" id="submit_btn" value="Send" />
</fieldset>
</form>
/div> -->
Sorry about the typed code not being a screen shot. The site won't let me post images yet. Kind of silly really. Anyway... Cheers guys.