Question

我有以下代码发出我的所有javascript：

@using SquishIt.Framework.Minifiers.JavaScript
@{
    string[] scriptLinks = new[]
    {
        Url.Content("~/content/script/json2.js"),
        Url.Content("~/content/script/b.extensions.js"),
        Url.Content("~/content/script/jquery.extensions.js"),
        Url.Content("~/content/script/jquery.validate.min.js"),
        Url.Content("~/content/script/jquery.validate.unobtrusive.min.js"),
        Url.Content("~/content/script/jquery.validate.tooltip.js"),
        Url.Content("~/content/script/jquery.form.js"),
        Url.Content("~/content/script/jquery.placeholder.js"),
        Url.Content("~/content/script/jquery.tooltip.js"),
        Url.Content("~/content/script/jquery.signalr-0.5.2.js"),
        // TODO: Url.Content("~/signalr/hubs"), // SignalR auto-generated hubs proxy ? Get Programatically??
        Url.Content("~/content/script/b.core.js"),
        Url.Content("~/content/script/b.facebook.js"),
        Url.Content("~/content/script/b.twitter.js"),
        Url.Content("~/content/script/b.realtime.js"),
        Url.Content("~/content/script/b.init.js"),
        Url.Content("~/content/script/views/user/login.js"),
        Url.Content("~/content/script/views/post/create.js"),
        Url.Content("~/content/script/views/post/list.js")
    };
}
@{
@* we don't bundle external javascript links nor localization messages. *@
    const string jQueryVersion = "1.7.2";
    string jQueryExternal = "//ajax.googleapis.com/ajax/libs/jquery/{0}/jquery.min.js".FormatWith(jQueryVersion);
    string jQueryLocal = @Url.Content("~/content/script/jquery-{0}.min.js".FormatWith(jQueryVersion));
}
<script src="@Url.Action("Localization", "Resource")"></script>
<script src="@jQueryExternal"></script>
<script>!window.jQuery && document.write(unescape("%3Cscript src='@jQueryLocal'%3C/script%3E"))</script>
@(Bundle.JavaScript()
    .Add(scriptLinks)
    .WithoutTypeAttribute()
    .WithMinifier<YuiMinifier>()
    .WithConfiguration()
    .MvcRender("~/all.js?_=1")
)
@JavaScript.Emit()
@* TODO: analytics service (pixel goes to noscript partial) *@

除了我删除google缓存的外部jQuery引用的情况之外，这总是很好用：<script src="@jQueryExternal"></script>并且其余的脚本都在一个脚本标记中进行了修改。

我还想通过，如果我在document.write标记之后添加了任何脚本标记，例如：

<script>!window.jQuery && document.write(unescape("%3Cscript src='@jQueryLocal'%3C/script%3E"))</script>
<script></script> @* work around an issue where javascript minified into a single script tag won't find references to jQuery if loaded in the previous script tag. *@

确实有效。

否则我得到“任何未定义的东西”

在chrome 20和firefox 11中都进行了测试，那么这个额外的空脚本标签到底做了什么？为什么它会缓解这个角落错误？

Answer 1

结果证明问题是在document.write元素中未正确关闭。

unescape("%3Cscript src='@jQueryLocal'%3C/script%3E")

到

unescape("%3Cscript src='@jQueryLocal'%3E%3C/script%3E")

缩小脚本的奇怪含义

1 个答案: