[英]How do I call the PHP mail() function in my JS?
我有一個困境。 我為我正在構建的網站創建了此訂單請求頁面,其工作原理如下:
order.js
的布爾值。 同樣,它會為其他項目和您的信息獲取所需的單位計數,並將它們應用於變量。console.log()
它們。這就是問題所在……這是 order.js 文件的一個片段。
function compileInfo() {
console.log("compileInfo active");
var name = "Name: " + $("#name").val() + "\n";
var email = "Email: " + $("#email").val() + "\n";
var phone = "Phone: " + $("#phone").val() + "\n";
var weddingDate = "Wedding Date: " + $("#date").val() + "\n";
var comments = "Comments: " + $("#comments").val() + "\n";
var base = "Base Experience: " + $("#base").hasClass("active") + "\n";
var special = "Special Edition: " + $("#special").hasClass("active") + "\n";
var teaser = "Teaser Trailer: " + $("#teaser").hasClass("active") + "\n";
var raw = "Raw Footage: " + $("#raw").hasClass("active") + "\n";
var standard = "Standard Shipping: " + $("#standard").hasClass("active") + "\n";
var expedited = "Expedited Shipping: " + $("#expedited").hasClass("active") + "\n";
var dvd = "Standard DVD: " + a + "\n";
var br = "Standard Blu-Ray: " + b + "\n";
var dvdSe = "Special DVD: " + x + "\n";
var brSe = "Special Blu-Ray: " + y + "\n";
var info = new Array();
info[0] = name;
info[1] = email;
info[2] = phone;
info[3] = weddingDate;
info[4] = comments;
var services = new Array();
services[0] = base;
services[1] = special;
services[2] = teaser;
services[3] = raw;
var delivery = new Array();
delivery[0] = standard;
delivery[1] = expedited;
var extras = new Array();
extras[0] = dvd;
extras[1] = br;
extras[2] = dvdSe;
extras[3] = brSe;
var dataVar = info + "\n" + services + "\n" + delivery + "\n" + extras;
var dataVarJSON = JSON.stringify(dataVar);
console.log(dataVar);
$.ajax({
type: "POST",
url: "order.php",
data: {data : dataVarJSON},
success: function() {
console.log("SUCCESS");
}
});
}
function validate() {
var name = $("#name").val();
var email = $("#email").val();
var weddingDate = $("#date").val();
if (name === "" || email == "" || weddingDate == "") {
alert("You must complete all required fields to send this request.");
} else {
console.log("working");
compileInfo();
return true;
}
}
這是我收到的 PHP:
<?php
header('content-type: application/json; charset=utf-8');
header("access-control-allow-origin: *");
$body = json_decode(stripslashes($_POST['data']));
$to = "thekevinhaube@gmail.com";
$subject = "Order Request";
function sendInfo() {
mail($to, $subject, $body);
}
?>
現在,我遠非 PHP 專家。 事實上,這是我第一次遇到它。 我怎樣才能讓它發送。 這似乎POST
就好,但不能發送到上市的電子郵件地址。 任何和所有的幫助表示贊賞! 同樣,這是我第一次使用 PHP,所以......
你需要給 sendInfo 一些參數:
function sendInfo($to, $subject, $body) {
...
}
並且不要在本地主機上嘗試 mail() ,請嘗試 - 如果可能 - 在線。
聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.