# call batch file and run using its own path/directory

I have a batch files that calls other batch files like this:

e:\foo\master.bat has the content:

call e:\bar\run1.bat


and e:\bar\run1.bat has the content

app1.exe


the problem is that when I run the master.bat app1.exe will not be executed, because it will expect it to be in the e:\foo directory instead of it being in e:\bar directory

## 2 Answers

You are a bit unclear where app1.exe is located.

If it shares the folder with run1.bat change run1.bat

to either

@Echo off
Pushd "%~dp0"
app1.exe
popd


or

@Echo off
"%~dp0app1.exe"


%0 refers to the currently running batch and the modifier ~dp returns drive and path (with a trailing backslash.)

• the Pushd method works best for me because, app1.exe internally also uses current dir, and using Pushd it will use the dir where app1.exe is located, without Pushd (but with the prefix) it will use the master.bat location
– Omu
May 31, 2018 at 18:18
• With this type of implications in mind, that method was the first in my answer intentionally. Thanks for feedback. May 31, 2018 at 18:22

The answer to your question can be drawn from a similar question on Stack Overflow.

What is the current directory in a batch file?

Using the variables mentioned here, you can update run1.bat to call app1.exe with the following line: %~dp0app1.exe. (The %~dp0 variable includes a trailing slash.) This will tell the batch file to run the executable from the current batch file's location.