Question

我想将一个对象传递给函数flipIt(obj)。我将图片ID存储在名为globe的变量中。

当我将globe传递给flipIt()时，它无效，因为globe是包含图片ID且flipIt()需要对象的变量。

我已尝试('#'+globe)将其设为obj，但它也无效：

flipIt('#'+globe);

功能定义是：

function flipIt(obj) {
    console.log("value before Function status   " + status);
    alert('FlipIT Called' + obj);

    $(obj).css("-webkit-transform-style","preserve-3d");
    $(obj).css("-webkit-transition","all 1.0s linear");
    $(obj).css("transform-style","preserve-3d");
    $(obj).css("transition","all 1.0s linear");
}

我尝试打印obj值...对于变量它是打印id值而不是它应该打印HTMLimage元素。

的Javascript

$(document).ready(function () {
    var globe;

    /*   Reading the data from XML file*/
    $.ajax({
        type: "GET",
        url: "photos.xml",
        dataType: "xml",
        success: function(xml) {
            $(xml).find('item').each(function() {
                var path = $(this).attr('path');
                var width = $(this).attr('width');
                var height = $(this).attr('height');
                var id = $(this).attr('id');
                var alt = $(this).attr('alt');
                var longdesc = $(this).find('longdesc').text();
                var description = $(this).find('desc').text();
                $('#myImageFlow').prepend('<img src="'+path+'" id='+id+'  height="'+height+'"  width="'+ width+'" longdesc="'+longdesc+'" alt="'+alt+'"/>');
                imgArr[i] = description;
                i = i+1;
            });
        });
    });
});

XML文件：

<items id = "items">
   <item path="img/1_bankofamerica.jpg" width="300" height="360" id="id1" alt="img1" type="bitmapfile">
      <back>swf/backcard_0.swf </back>
      <longdesc>img/img1.png</longdesc>
      <desc>Decription about Image # 1 </desc>
   </item>
   <item path="img/2_harbourfront.jpg" width="300" height="360" id="id2" alt="img2" type="bitmapfile">
      <back>swf/backcard_1.swf </back>
      <longdesc>img/img2.png</longdesc>
      <desc>Decription about Image # 2 </desc>
   </item>
   <item path="img/2_harbourfront3.jpg" width="300" height="360" id="id3" alt="img3" type="bitmapfile">
      <back>swf/backcard_2.swf </back>
      <longdesc>img/img3.png</longdesc>
      <desc>Decription about Image # 3 </desc>
   </item>
   <item path="img/3_harbourfront.jpg" width="300" height="360" id="id4" alt="img4" type="bitmapfile">
      <back>swf/backcard_3.swf </back>
      <longdesc>img/img4.png</longdesc>
      <desc>Decription about Image # 4 </desc>
   </item>
   <item path="img/5_lighthouse.jpg" width="300" height="360" id="id5" alt="img5" type="bitmapfile">
      <back>swf/backcard_4.swf </back>
      <longdesc>img/img5.png</longdesc>
      <desc>Decription about Image # 5 </desc>
   </item>
</items>

Answer 1

假设globe是一个包含您要选择的元素的id的字符串（不带#前缀），您需要在传递之前将其转换为jQuery对象它对你的功能：

flipIt($('#'+globe));

然后，您不需要在$()函数内包围另一个flipIt，因为它已经是一个对象。所以在函数内部，只需执行：

obj.css("-webkit-transform-style","preserve-3d");
...

接下来，我想你想在全局范围内声明globe。现在，您将在文档就绪函数的范围内声明它。所以把它放在外面：

var globe;

$(document).ready(function () {
    // document ready
});

准备好的文件还有一个较短的符号：

$(function() {
    // document ready
});

将对象传递给函数

1 个答案: