Iexpress extract the files and then run the script to reference the extracted file

Question

I tried to deploy my project using IExpress. I have the following scenario. I want to extract some files into place, preferably c: \ program files \. Then after it copies all the files, I want it to run the .cmd file (which is a script). The script is also added in the project itself, and it refers to the file that IExpress is copying. Now how to access the path where the file was extracted. So that I can access it in my script.

Answer 1

If the script is in the project itself, so it is extracted in the same directory when sending the files, it should start in the same directory.

Test it easily, create cmd as follows:

cmdsetup.cmd:

@echo Starting path:% ~ dp0 →% temp% \% ~ n0.log

Put this in your package and when it is done, go to the% temp% directory, find the cmdsetup.log file and see it. This should be the path where your files are.
If so, from there. If I am wrong, come back and comment, also correct your question to make it clearer.

Hope this helps.

Ps: I voted for the question, because I do not understand why a negative vote was given.

Answer 2

The answer is to use this format:

start /wait .\hello.cmd 

I did this with the following two files. One of the key points is that the file I called should have been in 8.3 format. In other words, he could not find hello.cmd for the first time, because I called him hello.world.cmd.

First file (start.cmd)

 @echo off cls echo this is start.cmd pause dir pause echo going to hello world start /wait .\hello.cmd echo back in start.cmd pause 

Second file (hello.cmd)

 @echo off echo HELLO WORLD! pause exit 

Directions

Use IEXPRESS to create a package containing both files above. Run the START.CMD program.