Hi,
How to open a single specific file from a specific folder which contains many files?
can you please elaborate?
@ADITYA_MUKHERJEE
in other hands you can use the click activity.
indicate on screen option=> indicate that specific file (which one is you want to open)…
Use Start Process Activity Pass the path of the file.
Use Assign Activity
filelist = Directory.GetFiles(“Path”)
filelist data type is a string of array
For each activity
item filelist
Use start process activity pass item.
hi @ADITYA_MUKHERJEE you can also use Directory.GetFiles(“PathOfFile”) this expression in assign activity…
in to field create one variable and use the above expression in VB expression field… as @kommijeevan told
this will open all the files in that path right?
but i want to open only a specific file in that path.
Use Start Process Activity Pass the path of the file.
specific file means what will be the extension of that file ?
@ADITYA_MUKHERJEE
.xlsx, excel files.
No worries.
- assign Variable “str_getfiles=new FileInfo(item.tostring)”
2.assign another varibale “str_type=Path.GetExtension(item.tostring)”
3.IF
str_type=“.xlsx”
then your workflow.
chrees
mark solution if you got your answer so that it will help others.
@ADITYA_MUKHERJEE
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