Gonen Shoham

This is where I came up to: D:\temp>"C:\Program Files\7-Zip\7z" a "d:\ri_core\0801m01\FRM_CRP2\ARIA0801m01.zip" "d:\ARIA0801m01\bin\packages\0801m01\skel\config.txt" "\ARIA0801m01\bin\packages\0801m01\skel" output on screen 7-Zip 19.00 (x64) : Copyright (c) 1999-2018 Igor Pavlov : 2019-02-21 Open archive: d:\ri_core\0801m01\FRM_CRP2\ARIA0801m01.zip Path = d:\ri_core\0801m01\FRM_CRP2\ARIA0801m01.zip Type = zip Physical Size = 1986022630 Scanning the drive: 1 folder, 2 files, 518 bytes (1 KiB) Updating...

thank for a quick reply.... I tried many options with no luck..... it aways creates the file in the root folder of the zip file..... I ask for more details.... the zip file is located in : D:\ricore\0801m01\FRMCRP2\ARIA0801m01.zip The target config.txt file is located in the zip file in: *\ARIA0801m01\bin\packages\0801m01\skel* where should I place the new config.txt file that I would like to use as input for the replace? How would the command look like? thanks a lot for help !!!!

I need help on replacing single file in specific folder within an archive file. step 1 I extract config.txt file form folder ARIA0801m01\bin\packages\0801m01\skel\config.txt in archive D:\ricore\0801m01\FRMCRP2\ARIA0801m01.zip into D:\temp using the following command: 7z x D:\ri_core\0801m01\FRM_CRP2\ARIA0801m01.zip ARIA0801m01\bin\packages\0801m01\skel\config.txt -oD:\temp\unzip -y Step 2 then I edit and update the file manually I would like to replace it back in the original location within the...