如何从源码中抽取一个link?
How to draw a link from the source?
来源:
<div class="filmPosterBox">
<div id="filmPosterLink" class="hide"/>
<div class="posterLightbox">
<a class="film_mini" href="http://1.fwcsd.pl/df/2eee.jpg" rel="v:image">
<img title="tytulek(2011)" alt="tytul film" src="http://1.cos.pl/coes/cod2scdos.jpg"/>
</a>
</div>
</div>
想画一样的link:(坏了link)
我创建了一些东西,但没有正常工作:
sPosterFile = oInternetExplorer.Document.getElementsByClassName("posterLightbox")(0) _
.getElementsByTagName("a")(0).innerHTML
link 的目标存储在标签的 href
属性中,而不是内部 HTML.
sPosterFile = oInternetExplorer.Document _
.getElementsByClassName("posterLightbox")(0) _
.getElementsByTagName("a")(0).<b>href</b>
来源:
<div class="filmPosterBox">
<div id="filmPosterLink" class="hide"/>
<div class="posterLightbox">
<a class="film_mini" href="http://1.fwcsd.pl/df/2eee.jpg" rel="v:image">
<img title="tytulek(2011)" alt="tytul film" src="http://1.cos.pl/coes/cod2scdos.jpg"/>
</a>
</div>
</div>
想画一样的link:(坏了link)
我创建了一些东西,但没有正常工作:
sPosterFile = oInternetExplorer.Document.getElementsByClassName("posterLightbox")(0) _
.getElementsByTagName("a")(0).innerHTML
link 的目标存储在标签的 href
属性中,而不是内部 HTML.
sPosterFile = oInternetExplorer.Document _
.getElementsByClassName("posterLightbox")(0) _
.getElementsByTagName("a")(0).<b>href</b>