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I'm on Windows 7. I'm trying to run a .bat file where I stop and start MySQL and run some other commands.

However, running "C:\Program Files (x86)\EasyPHP\binaries\mysql\bin\mysqld.exe" results in starting mysqld, which then occupies the prompt, and the other .bat commands are not run.

  • I tried using Start → mysqld does not start
  • Start /B → mysqld does not start
  • CMD /C → mysqld just starts in the same window
  • Start /B startMysql.bat (a seperate bat that starts mysql) → mysqld does not start

I use this to test the commands:

@echo  on
CMD /C "C:\Program Files (x86)\EasyPHP\binaries\mysql\bin\mysqld.exe"
timeout 2:
share|improve this question
Did you try: start cmd /c "C:\Program Files (x86)\EasyPHP\binaries\mysql\bin\mysqld.exe" The combination is sometimes needed to get things to work right. – krowe Aug 12 '13 at 16:26
As you can see from my question I tried that, its in the list of things I tried. It just starts mysqld in the same window, losing the option to run other commands – saratis Aug 13 '13 at 8:15

I solved it by installing MYSQL as a service and running net start mysql and net stop mysql. That doesn't solve the issue, so I will leave this open for someone to answer. Otherwise I will accept this as the answer.

share|improve this answer

use "start":

c:>start c:\xampp\mysql\bin\mysqld --defaults-file=c:\xampp\mysql\bin\my.ini

Tested alone and inside a .bat; I am on Windows 8.1 but I expected it works on Windows 7 too

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