Manually sending a post in PHP - php

Question

I have a form that will be validated client side before being submitted via an ajax request to the server for server-side validation. Should the validation fail server side then a postback will need to be made containing all the error messages. Is there some way I can do this?
For example:
if ((!empty($nameError) && (!empty($emailError)) {
$_POST['nameError'] = $nameError;
$_POST['emailError'] = $emailError;
// send postback with values
}
else {
echo 'No errors';
}
UPDATE ------------------------------------------------
Here is the javascript that handles the submission of the form:
$(".button").click(function() {
$(".error").hide();
var name = $(":input.name").val();
if ((name == "") || (name.length < 4)){
$("label#nameErr").show();
$(":input.name").focus();
return false;
}
var email = $(":input.email").val();
if (email == "") {
$("label#emailErr").show();
$(":input.email").focus();
return false;
}
var phone = $(":input.phone").val();
if (phone == "") {
$("label#phoneErr").show();
$(":input.phone").focus();
return false;
}
var comment = $.trim($("#comments").val());
if ((!comment) || (comment.length > 100)) {
$("label#commentErr").show();
$("#comments").focus();
alert("hello");
return false;
}
var info = 'name:' + name + '&email:' + email + '&phone:' + phone + '&comment:' + comment;
var ajaxurl = '<?php echo admin_url("admin-ajax.php"); ?>';
alert(info);
jQuery.ajax({
type:"post",
dataType:"json",
url: myAjax.ajaxurl,
data: {action: 'submit_data', info: info},
success: function(response) {
if (response.type == "success") {
alert("success");
}
else {
alert("fail");
}
}
});
$(":input").val('');
return false;
});
And here is the php function that the ajax posts to:
function submit_data() {
$nameErr = $emailErr = $phoneErr = $commentErr = "";
$full = explode("&", $_POST["info"]);
$fname = explode(":", $full[0]);
$name = $fname[1];
$femail = explode(":", $full[1]);
$email = $femail[1];
$fphone = explode(":", $full[2]);
$phone = $fphone[1];
$fcomment = explode(":", $full[3]);
$comment = $fcomment[1];
if ((empty($name)) || (strlen($name) < 4)){
$nameErr = "Please enter a name";
}
else if (!preg_match("/^[a-zA-Z ]*$/", $name)) {
$nameErr = "Please ensure you have entered your name and surname";
}
if (empty($email)) {
$emailErr = "Please enter an email address";
}
else if (!preg_match("/([\w\-]+\#[\w\-]+\.[\w\-]+)/", $email)) {
$emailErr = "Please ensure you have entered a valid email address";
}
if (empty($phone)) {
$phoneErr = "Please enter a phone number";
}
else if (!preg_match("/(?:\(?\+\d{2}\)?\s*)?\d+(?:[ -]*\d+)*$/",$phone)) {
$phoneErr = "Please ensure you have entered a valid phone number";
}
if ((empty($nameErr)) && (empty($emailErr)) && (empty($phoneErr)) && (empty($commentErr))) {
$conn = mysqli_connect("localhost", "John", "Change9", "plugindatadb");
mysqli_query($conn, "INSERT INTO data (Name, Email, Phone, Comment) VALUES ('$name', '$email', '$phone', '$comment')");
}
else {
// display error messages
}
die();
}

Answer 1

Your answer will be in two parts:
Pseudo code:
Part1: PHP
if ($error) {
$reply["status"]=false;
$reply["message"]="Fail message"; //Here you have to put your own message, maybe use a variable from the validation you just did before this line: $reply["message"] = $fail_message.
}
else {
$reply["status"]=true;
$reply["message"]="Success message"//$reply["message"] = $success_message;
}
echo json_encode($reply);//something like {"status":true, "message":"Success message"}
Part2 AJAX: modify you ajax response to this.
success: function(response) {
if (response.status == true) {
alert("success: "+response.message);
}
else {
alert("fail: " + response.message);
}
}

Answer 2

Use json ajax request. In case error exists show the error message. I generally put a flag for success or fail .
$message='';
if ((!empty($nameError) && (!empty($emailError)) {
$errorArray=array();
$errorArray['nameError'] = $nameError;
$errorArray['emailError'] = $emailError;
// send postback with values
}
else {
$message='No errors';
}
echo json_encode(array(
"message"=>$message,
"errors"=>$errorArray
));