考虑以下HTML页面摘录:
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="UTF-8">
<title>Document</title>
</head>
<body>
<div class="BoxBody">
<span class="txt">20 Records found. </span>
<p style="text-align: right;"><span class="txt">[First/Previous] 1 , <a class="page" href="javascript:paginacao('paginar','2');" title="Go to page 2">2</a> [<a class="page" title="Next page" href="javascript:paginacao('paginar','next');">Next</a>/<a class="page" title="Last page" href="javascript:paginacao('paginar','last');">Last</a>]</span></p>
<br>
<span class="txt">25 Records found. </span>
<p style="text-align: right;"><span class="txt">[First/Previous] 1 , <a class="page" href="javascript:paginacao('paginar2','2');" title="Go to page 2">2</a> [<a class="page" title="Next page" href="javascript:paginacao('paginar2','next');">Next</a>/<a class="page" title="Last page" href="javascript:paginacao('paginar2','last');">Last</a>]</span></p>
</div>
</body>
</html>
我正在尝试获取带有“下一页”页面anchor
(如果有的话)的href
标签。
我在使用Firefox的控制台中对此进行了尝试,并且可以正常工作:
document.querySelector(".BoxBody > p:nth-child(2) > span:nth-child(1)").querySelector("a[title='Next page']")
我也使用querySelector
编写了一个示例VBA代码,但是失败了,出现了Invalid argument
。
Sub test()
Dim oFSO As Object, paginator As Object
Dim oFS As Object, sText As String
Set oFSO = CreateObject("Scripting.FileSystemObject")
Set oFS = oFSO.OpenTextFile(ThisWorkbook.Path & "\example.html")
Do Until oFS.AtEndOfStream
sText = oFS.ReadAll()
Loop
Dim html As HTMLDocument, html2 As Object
Set html = New HTMLDocument
Set html2 = html
html2.Write sText
Set paginator = html.querySelector(".BoxBody > p:nth-child(2) > span:nth-child(1)").querySelector("a[title='Next page']")
End Sub
是什么原因造成的? p:nth-child(2)
标识符?
我应该如何使用VBA提取该元素?
答案 0 :(得分:3)
nth-child(2)
在VBA中不受支持,并确实导致了错误消息。您不能使用:nth-child()
或:nth-of-type()
。在处理伪类的库中几乎没有实现。您可以有趣地使用first-child
。您还会发现在哪些对象上可以链接querySelector受到限制。
Dim ele As Object, iText As String
Set ele = html.querySelector(".BoxBody > p > span:first-child > a[title='Next page']")
On Error Resume Next
iText = ele.href
On Error GoTo 0
If iText = vbNullString Then '<== This assumes that the href has a value otherwise use an On Error GoTo which will then handle the error and print "no href"
Debug.Print "No href"
Else
Debug.Print "href"
End If