I have a file with strings I need at first to filter for example only rows that have 5 characters And secondly filter only those that have at least 2 characters repeated together Example: Display aazyx or zyaax but does not display azyxa And thirdly filter q lines have at least one vowel.
The latter I know is possible, the other two I do not know if it is possible with thirst. Thanks in advance for your help and attention.
I do not know if I understand very well, but supposing you have a text file in the following format:
file.txt
abcde
aabbc
kkkkk
ggggggggggggg
ccdde
We can filter as follows:
cat arquivo.txt | grep -x '.\{5,5\}' | grep -E '(.){1}' | grep -E '[aeiouAEIOU]{1}'
Note 1: I think it would be better to use egrep instead of grep for regular expressions.
Note: I have no time for the details of a man in grep to search for parameters.
Hello, good afternoon.
I hope this is what you need:)
#!/bin/bash
vCharsVogal=(aa ee ii oo uu)
vCharsConso=(bb cc dd ff gg hh ii jj kk ll mm nn pp qq rr ss tt vv xx zz ww yy)
vCharVogal=(a e i o u)
for line in $(cat $1); #//$1 eh o parametro para o arquivo de entrada
do
#1 -Primeiro testa se possui 5 chars
if (("${#line}" == "5"));
then
#2 - Depois verifica se possui repeticao de chars vogais
continua=1
for char in ${vCharsVogal[*]}
do
countChar=$(echo "${line}" | awk -F"${char}" '{print NF-1}')
if (("$countChar" >= "1"));
then
#3 - Se entrou aqui, quer dizer que jah possui vogal :)
echo "$line"
continua=0
break
fi
done
#Se nao teve nas vogais, ai faz as consoantes
if(("${continua}" == "1"));
then
for char in ${vCharsConso[*]}
do
countChar=$(echo "${line}" | awk -F"${char}" '{print NF-1}')
if (("$countChar" >= "1"));
then
#3 - Vai verificar se existe vogal
for vogal in ${vCharVogal[*]}
do
countVog=$(echo "${line}" | awk -F"${vogal}" '{print NF-1}')
if (("$countVog" >= "1"));
then
echo "$line"
continua=0
break
fi
done
if(("${continua}" == "0"));
then
break
fi
fi
done
fi
fi
done