Run command for each file in a directory in MS-DOS

Question

I’m making a small bat to break a branch over an application I’m mounting, where it should run sql scripts within the Mysql database. The problem is that I have to edit this bat every time I insert a new script and it ends up getting boring...

Is there any way to make through the prompt I can pick up the name of all the files inside a folder and from this run a command at the prompt for each file?

Answer 1

One possibility is to use FOR:

FOR %a IN (*) DO echo %a

The above syntax is to execute directly in CMD. To use in . bat use two %:

FOR %%a IN (*) DO echo %%a

Replace the echo by the desired command, and add the parameters you want. Instead of the * you can specify the wildcards system normal, for file name and extension.

If you want to do some operation with listing in text files, you can get the directory names like this:

DIR *.* /b > listagem.txt

The /b flag is to take the file path. Use the /s flag if you want to include subdirectories.

Answer 2

If you need to make a script plus increment in the future you can use the powershell, the line below returns the names of all the files in the specified folder.

(Get-ChildItem "C:\Program Files\MySQL\MySQL Server 5.6"  -File).Name

For a script with menu and option from which file to select follows an example:

$arquivos = (Get-ChildItem "C:\Program Files\MySQL\MySQL Server 5.6\data"  -File).Name
Write-Host "Arquivos encontrados:"
Write-Host "99 - para executa todos."
$i = 0
foreach($item in $arquivos){
    Write-Host $i "-" $item
    $i++
}

$opcao = Read-Host "selecione uma opção: "

if($opcao -eq 99){
    Write-Host "executar todos os arquivos"
}else{
    Write-Host "Arquivo selecionado " $arquivos[$opcao]
}

The result in the terminal is