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How to get Links count in an web application,without
descriptive programming?
- 8 Answers
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Answers were Sorted based on User's Feedback
Answer / saidev
You can just palce page check point on that page,it will
give you the details like number of images ,number of links
etc in the page.
| Is This Answer Correct ? | 4 Yes | 0 No |
Answer / chaitu
This Script is Helpfull to count the No.of links in a web
page
Step1: Set Objects = Browser("BrowserName").Page
("PageName").ChildObjects
Step2: Objectcount = Objects.count
Linkcount = 0
Step3: For i = 1 to Objectcount-1
Link = Objects(i).GetROProperty("micclass")
If Link = "Link" Then
Linkcount = Linkcount+1
msgbox Linkcount
End If
Next
msgbox Linkcount ' It show total number links
| Is This Answer Correct ? | 7 Yes | 5 No |
Answer / raghava
Hi guys
Here we have to use the description object.Its working with
me.Just try it out.
Set oDesc = Description.Create()
oDesc("micclass").Value = "Link"
Set Lists = Browser("Google").Page("Google").ChildObjects
(oDesc)
NumberOfLists = Lists.Count()
linkcnt = 0
For i = 0 To NumberOfLists - 1
linkcnt = linkcnt+1
Next
msgbox linkcnt
Regards
raghava
| Is This Answer Correct ? | 2 Yes | 0 No |
Answer / priya
here is an example to get the link count in google page
Just try out this
SystemUtil.Run ("www.google.com")
Set Links = Browser("title:=.*").Page("title:=.*").Object.Links
Msgbox "Total links: " & Links.Length
| Is This Answer Correct ? | 1 Yes | 0 No |
Answer / a
POSTED BY -CHAITU
This Script is Helpfull to count the No.of links in a web
page
Step1: Set Objects = Browser("BrowserName").Page
("PageName").ChildObjects
Step2: Objectcount = Objects.count
Linkcount = 0
Step3: For i = 1 to Objectcount-1
Link = Objects(i).GetROProperty("micclass")
If Link = "Link" Then
Linkcount = Linkcount+1
msgbox Linkcount
End If
Next
msgbox Linkcount ' It show total number links
--------------
This code is not working.
Check it out.
Link = Objec(i).GetROProperty("micclass")
Error is here.
| Is This Answer Correct ? | 1 Yes | 1 No |
'Hi, we can use "innertext" as a unique property of link.
'Name may be changes in the runtime. And we can use the
index number to identify the link in the execution.
(NOTE:-We dont use index to retrive the child objects.why
because each child object have different index number.)
TRY THIS CODE:--------THIS WIL WORK FINE.
set olink=description.create()
oDesc("micclass").Value = "Link"
set chobj=browser().page().childobjects(olink)
n=chobj.count
msgbox n
for i=0 to n-1
val1=chobj(i).getroproperty("innertext")
msgbox val1
Next
Any doubts----mail to------ mskalyan.smile@gmail.com
| Is This Answer Correct ? | 1 Yes | 1 No |
Answer / venkatesh.m
Hi friends,
set odesc=description.create()
set odesce("htmltag").value="a"
set lists=browser("b.name").page("p.name").childobjects
(odesc)
set counter=lists.count
msgbox "total no.of links"& counter
for i=0 to counter-1 step 1
set link=lists(i).getroproperty("name")
msgbox link,0,"curren link is"
Next
The above script works correctly.
| Is This Answer Correct ? | 0 Yes | 2 No |
Answer / sudhir
Hi Friends ,
Above Ans #5 gives syntax mismatch error.
Find below correct code.
set odesc=description.create()
oDesc("micclass").Value = "Link"
set lists=browser("Welcome: Mercury Tours").page("Find a
Flight: Mercury").childobjects(odesc)
counter=lists.count
msgbox counter
for i=0 to counter-1 step 1
name=lists(i).getroproperty("name")
msgbox name,0,"curren link is"
Next
| Is This Answer Correct ? | 0 Yes | 2 No |
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