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>From address@hidden Mon Aug 30 07:04:53 2004
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From: "fpasa" <address@hidden>
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Subject: Re: Arquivo zip
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Olha s=F3, fiz o seguinte, mas queria outra coisa q n=E3o tah rolando=20
id=E9ia. Queria verificar da lista "anunc.lst" quais os arquivos q=20
est=E3o na lista, mas n=E3o est=E3o no zip. Alguma id=E9ia ?

arquivo=3D`ls *ANUNC* | tail -1`

for arq in `cat anunc.lst`
do
        nome=3D`zipinfo -1 $arquivo $arq`
        tam=3D`zipinfo -Z $arquivo $arq | awk '{ print $4 }'`
        linhas=3D`unzip -p $arquivo $arq | wc -l`

echo "
"
echo "NOME DO ARQUIVO:                          " $nome
echo "TAMANHO DO ARQUIVO(BYTES):                " $tam
echo "QUANTIDADE DE LINHAS:                     " $linhas
echo "----------------------------------------------------------------
------
"
done


--- Em address@hidden, "lferreir1" <lferreir@e...>=20
escreveu
> Fabiano,=20
> Utilize unzip -v arquivo.zip
> O resultado s=F3 n=E3o ir=E1 mostrar a qtde de linhas que cada arquivo=20
> cont=E9m.
> Se algu=E9m da lista souber...
> At=E9!
> Luiz
> --- Em address@hidden, "fpasa" <shell.script@t...>=20
> escreveu
> > Pessoal,
> >=20
> > tenho um arquivo zip que dentro tem v=E1rios arquivos.
> > Como poderia fazer para extrair o nome, tamanho e n=FAmero de=20
linhas=20
> de=20
> > cada arquivo ? Alguma id=E9ia ?
> >=20
> > abra=E7os,
> > Fabiano



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