Question

我在这里遇到了一些javascript。我需要将动态变量/变量传递给外部函数。这是生成PHP网页URL的函数，我最终尝试将一些数据发布到（不是完整函数，只有重要部分）：

var id = "";
var name = "";
var url = "";


function initialize () {
  downloadUrl("phpsqlajax_genxml1.php", function (data) {
    var xml = data.responseXML;
    var markers = xml.documentElement.getElementsByTagName("marker");
    for (var i = 0; i < markers.length; i++) {
        var id = markers[i].getAttribute("id");
        var name = markers[i].getAttribute("name");
        var description1 = markers[i].getAttribute("description");
        var type = markers[i].getAttribute("type");
        var point = new google.maps.LatLng(
                parseFloat(markers[i].getAttribute("lat")),
                parseFloat(markers[i].getAttribute("lng")));
        var url = "markerpages.php?name=" + name + "&id=" + id;
        var html = "<b>" + name + "</b> <br/>" + description1;
        var contentString = '<div id="content">' +
                '<div id="siteNotice">' +
                '</div>' +
                '<h1 id="firstHeading" class="firstHeading">' + name + '</h1>' +
                '<div id="bodyContent">' +
                '<p>' + description1 + '</p>' + 
                '<p><a href="'+ url +'" onclick="postData(url)">Link to PHP web page</a>' +
                '</p>' +
                '</div>' +
                '</div>';`

我有＆＃39; url＆＃39;变量全局定义，然后我的postData函数看起来像这样，并且在initialize函数之外：

function postData(arg) {
  alert("Url: "+url+"Name: "+name+"ID: "+id);
    $(document).ready(function () { 

            var href = $('.postData').attr('href');
            var id = markers[i].getAttribute("id");
            var name = markers[i].getAttribute("name");
            var url = "markerpages.php?name=" + name + "&id=" + id;

            window.location.href = href;
            console.log("outside ajax is working");
            console.log(url);
            $.ajax({
                type: "POST",
                url: ' + url + ',
                data: {
                    source1: "some text",
                    source2: "some text 2"},
                success: function (data) {
                    console.log(data);
                    console.log(url);
                    console.log("inside ajax is working");
                }
            });

        });
}

jquery正在触发onclick，但url以未定义的形式返回，所以我只需要引入根据存储在数据库中的标记名生成的url变量。我是初学者，所以这可能很简单。我环顾了论坛和文档，但几个小时都未能弄明白，哈哈。

真诚地感谢您的帮助！

Answer 1

将变量声明在所有函数范围之外，如下所示：

 //Declare the three variables globally
 var url = "";
 var name = "";
 var id = "";
 ...

 function initialize () {

   ...
   //id and name are local to this function, if you want them to be global as well
   //declare them outside the scope of all functions along with url variable

   //Assign them suitable values
   id = markers[i].getAttribute("id");
   name = markers[i].getAttribute("name");
   url = "markerpages.php?name=" + name + "&id=" + id;

  ...
 }

function postData(arg) {

   //Use all the variables here directly for example you could just try alerting them
   alert("Url: "+url+"Name: "+name+"ID: "+id);

 }

您的案例中的问题是“url”变量不全局。您分别在postData(arg)和initialize()函数中声明了url变量，因此该变量在这些函数的范围内是本地的。

此外，在postData(arg)功能中，您可以执行此操作：

  var id = markers[i].getAttribute("id");
  var name = markers[i].getAttribute("name");
  var url = "markerpages.php?name=" + name + "&id=" + id;

但是，markers数组遍历initialize()函数，而不是postData函数。如果您想在此上下文中访问“id”和“name”变量。只需在所有函数范围之外声明它们一次，并在稍后的任何上下文中初始化或更新它们，就像我用'url'变量显示一样。

将动态变量传递给外部函数

1 个答案: