Question

我有一个困境。我为我正在建立的网站创建了这个订单请求页面，以下是它的工作原理：

您检查/取消选中核心项目，并根据其状态（活动或非活动），它会将true / false值应用于相应order.js中的布尔值。同样，它会为其他项目和您的信息获取所需的单位数，并将其应用于变量。
它将这些变量分组到数组中，现在，它们console.log()。

这就是麻烦来的地方......这是order.js档案的一个片段。

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;
    }

}

这是我收到的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);
    }
 ?>

现在，我远非PHP专家。事实上，这是我第一次遇到它。我怎样才能发送它。似乎POST似乎没问题，但是没有发送到列出的电子邮件地址。任何和所有的帮助表示赞赏！再次，这是我第一次使用PHP，所以......

Answer 1

您需要为sendInfo提供一些参数：

function sendInfo($to, $subject, $body) {
  ...
}

并且不要在您的localhost上尝试mail（），如果可能的话，尝试在线。

如何在JS中调用PHP mail（）函数？

1 个答案: