I am making a small bat to break a branch on an application that I am mounting, where it should execute sql scripts inside the MySQL database. The problem is that I have to edit this bat every time I insert a new script and this ends getting annoying ...
Is there any way I can get the name of all files inside a folder through the prompt and from there execute a command at the prompt for each file?
If you need to make a script more increment in the future you can use 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 of 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
One possibility is to use FOR:
FOR %a IN (*) DO echo %a
The above syntax is to run directly in the CMD. To use in .bat use two % :
FOR %%a IN (*) DO echo %%a
Replace echo with the desired command, and add the parameters you want. Instead of * you can specify the normal