VB.NET Extract Filename and Path
•
Ditulis oleh Tim Pasti Oke
Get filename
pIni("data", "file") = System.IO.Path.GetFileName(y.OpenFileDialog1.FileName)
Get File Path
pIni("data", "path") = System.IO.Path.GetDirectoryName(y.OpenFileDialog1.FileName)
pIni("data", "file") = System.IO.Path.GetFileName(y.OpenFileDialog1.FileName)
Get File Path
pIni("data", "path") = System.IO.Path.GetDirectoryName(y.OpenFileDialog1.FileName)
Butuh solusi instan siap pakai?
Dapatkan modul & script server production-ready langsung di web store kami.
Komentar
Belum ada komentar. Jadilah yang pertama memberikan komentar!