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Mar 1st, 2024, 02:13 PM
#1
Thread Starter
Junior Member
VB6 'XML' Receive file name (unknown) from an url address
Hello, I hope you explain it well, I have a problem receiving the name of a file by entering a URL address.
I don't understand the XML language much, although I found a code that received with the 'status' command if the file existed, now I want to know by putting an address of any server which name of the file is behind that address.
Example: the address I send is:
https://server/tgha3ynvylrx1g434
File name behind this address is:
FileDown.rar
My problem question is, if I sent this URL address, the response would be the name of the file and what would be the command to receive said response.
-------Find code in VB6-------------------------
Dim sUrl as string
url="https://server/tgha3ynvylrx1g434"
xmlhttp.open "GET", sUrl, False ' "GET", sUrl, False 'GET
xmlhttp.setRequestHeader "Content-Type", "application/xml"
xmlhttp.send ""
If xmlhttp.Status = 200 Then
Msgbox "YES"
Else
Msgbox "NO"
End If
Set xmlhttp = Nothing
---------------------------------------------------
This is the code, although it returns all the code in xml with the file name.
But it is impossible for me since the texts sent on different servers are varied with different characters and I cannot capture the name of the file, I hope I have explained myself. Thank you.
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