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RE: [shell-script] Alterando conteudo de arquivos em massa
From: |
Gemayel Alves |
Subject: |
RE: [shell-script] Alterando conteudo de arquivos em massa |
Date: |
Tue, 15 Jan 2008 14:13:08 +0000 |
fala Ráyner,
veja se isso resolve:
ls -1|while read arquivos;do
sed 's/explorer/firefox/g' $arquivos> $arquivos.bk
mv $arquivos.bk $arquivos;
done
________________________________
> To: address@hidden
> From: address@hidden
> Date: Mon, 14 Jan 2008 07:13:58 -0300
> Subject: [shell-script] Alterando conteudo de arquivos em massa
>
>
> Há alguma maneira de se alterar o conteúdo dos
> arquivos em grande escala? O que eu quero na verdade e
> entrar em vários arquivos e substituindo uma palavra
> por outra.
>
> Tipo:
>
> ls | cat |ed s/internetExplorer/Fiferox/ g
> #isso não da certo
>
> Já tentei alguns comandos mas nenhum da certo.
>
> Obrigado
>
> Ráyner Carvalho
> (61) 9137-6842
>
> Abra sua conta no Yahoo! Mail, o único sem limite de espaço para
> armazenamento!
> http://br.mail.yahoo.com/
>
>
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From: Gemayel Alves <address@hidden>
Subject: Fatiador de Wordlist.
X-Yahoo-Group-Post: member; u5456689
Pessoal segue o source q fiz essa manha, espero q me ajudem a melhorar certo.
a minha idéia era colocar uma barra de progresso no sed enquanto ele trabalhava
para mostrar o progresso do mesmo mas não sei se dá, mas como ja dizia o
filozofo tudo é possível em shellscript.
++++++++++++++cut+++++++++++++++++
#!/bin/bash
if [ "$1" = "" ];
then
echo " Fatiador de wordlist by Gemayellira v.1
ex:
$0 wordlist.txt
"
exit
fi
echo "Fatiador de wordlist by Gemayellira v.1"
qtd=`cat $1|wc -l`
echo fatiando total $qtd
divisao1=$(( $qtd / 10 ))
sed '1,'$divisao1'!d' $1> $1.1
echo 1 - $divisao1 `du -hs $1.1` gerado! LIVRE `df -h .|sed '/^S/d'|awk
'{print $4}'`
divisao2=$(( $qtd / 10 * 2 ))
divisao1=`expr $divisao1 + 1`
sed ''$divisao1','$divisao2'!d' $1> $1.2
echo 2 - $divisao2 `du -hs $1.2` gerado! LIVRE `df -h .|sed '/^S/d'|awk
'{print $4}'`
divisao3=$(( $qtd / 10 * 3 ))
divisao2=`expr $divisao2 + 1`
sed ''$divisao2','$divisao3'!d' $1> $1.3
echo 3 - $divisao3 `du -hs $1.3` gerado! LIVRE `df -h .|sed '/^S/d'|awk
'{print $4}'`
divisao4=$(( $qtd / 10 * 4 ))
divisao3=`expr $divisao3 + 1`
sed ''$divisao3','$divisao4'!d' $1> $1.4
echo 4 - $divisao4 `du -hs $1.4` gerado! LIVRE `df -h .|sed '/^S/d'|awk
'{print $4}'`
divisao5=$(( $qtd / 10 * 5 ))
divisao4=`expr $divisao4 + 1`
sed ''$divisao4','$divisao5'!d' $1> $1.5
echo 5 - $divisao5 `du -hs $1.5` gerado! LIVRE `df -h .|sed '/^S/d'|awk
'{print $4}'`
divisao6=$(( $qtd / 10 * 6 ))
divisao5=`expr $divisao5 + 1`
sed ''$divisao5','$divisao6'!d' $1> $1.6
echo 6 - $divisao6 `du -hs $1.6` gerado! LIVRE `df -h .|sed '/^S/d'|awk
'{print $4}'`
divisao7=$(( $qtd / 10 * 7 ))
divisao6=`expr $divisao6 + 1`
sed ''$divisao6','$divisao7'!d' $1> $1.7
echo 7 - $divisao7 `du -hs $1.7` gerado! LIVRE `df -h .|sed '/^S/d'|awk
'{print $4}'`
divisao8=$(( $qtd / 10 * 8 ))
divisao7=`expr $divisao7 + 1`
sed ''$divisao7','$divisao8'!d' $1> $1.8
echo 8 - $divisao8 `du -hs $1.8` gerado! LIVRE `df -h .|sed '/^S/d'|awk
'{print $4}'`
divisao9=$(( $qtd / 10 * 9 ))
divisao8=`expr $divisao8 + 1`
sed ''$divisao8','$divisao9'!d' $1> $1.9
echo 9 - $divisao9 `du -hs $1.9` gerado! LIVRE `df -h .|sed '/^S/d'|awk
'{print $4}'`
divisao10=$qtd
divisao9=`expr $divisao9 + 1`
sed ''$divisao9','$divisao10'!d' $1> $1.10
echo 10- $divisao10 `du -hs $1.10` gerado! LIVRE `df -h .|sed '/^S/d'|awk
'{print $4}'`
+++++++++++++++++eof++++++++++++++++++++++
Gemayellira
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Subject: Formar CSV
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-0000
Pessoal, tenho um arquivo com o seguinte formato
uid: CPF
cn: NOME
mail: E-MAIL
ou: SETOR
Preciso fazer um csv com estas informações. No VI fiz com as seguintes
linhas
:%s/^\(uid: .*\)\n\(cn:.*\)/\1;\2/g #substitui a linha que começa com
uid, possui um \n e segue com cn pelo primeiro-resultado;segundo-resultado
:%s/^\(uid: .*\)\n\(mail:.*\)/\1;\2/g
:%s/^\(uid: .*\)\n\(ou:.*\)/\1;\2/g
Ok. Ao final, tenho o CSV.
Mas, quando fiz no bash, não funcionou.
ldapsearch -x -h IP -b BASE -LLL '(&(ou=*string*)(&(atributo=x)))' uid cn
mail ou|grep -v dn:|sed 's/^(uid: .*)\n(cn: .*)/"\1;\2"/g'
sed: -e expressão #1, caractere 32: invalid reference \2 on `s' command's
RHS
Alguma dica?
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