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I have a some strings like below in Linux.
abc_def_fgh_2018_08_11.zip
I want to extract just 2018_08_11 as a variable. All the strings have the same pattern 3 underscores and some dates or characters and .zip
How can I do that?
Take a look at this parameter expansion:
$ foo='abc_def_fgh_2018_08_11.zip'
$ foo=${foo#*_*_*_} # remove the prefix
$ echo "${foo%.*}" # remove the sufix and print
2018_08_11
Or using regular expression:
$ foo='abc_def_fgh_2018_08_11.zip'
$ regex='^([^_]*_){3}(.*)\.'
$ [[ $foo =~ $regex ]] && echo "${BASH_REMATCH[2]}"
2018_08_11
BASH_REMATCH is a special array where the matches from [[ ... =~ ... ]] are assigned to.
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I have a string with in a file:
"1.0.0.0.5";
I would like to replace 5 with 6 so the output should look like
"1.0.0.0.6";
could you please help me in this to achieve the output as above in bash
It seems to me that what you really want to do is increment the 5th dot-delimited field in that line.
line='"1.0.0.0.5";'
if [[ $line =~ \"([^\"]+) ]]; then
IFS="." read -ra fields <<< "${BASH_REMATCH[1]}"
((++fields[-1]))
(
IFS="."
printf '"%s";\n' "${fields[*]}"
)
fi
"1.0.0.0.6";
bash doesn't have an operator for assigning to a string index. So you'll need to concatenate the portions before and after the index you want to replace.
s=1.0.0.0.5
s=${s:0:8}6${s:9:}
${s:0:8} means the first 8 characters, and ${s:9:} means all the characters starting from index 9. So this replaces the character at index 8.
You can use a Bash regex:
s="1.0.0.0.5"
replacement="6"
[[ $s =~ ^(.*\.)[^.]*$ ]]
echo "${BASH_REMATCH[1]}$replacement"
Prints:
1.0.0.0.6
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What is a command that produces the same output as formatted printf in bash
If the echo on your System supporting -ne then you can define a function like...
$ puts(){(echo -ne "sometext ${1}\t ${2}\n ${3}\n")}
$ puts hello world bye
sometext hello world
bye
( Typed in a bash )
For looping through the List of Arguments this is possible...
puts(){
for string in ${#}; do
echo -ne ${string}
done
}
Than you can do...
puts 0 1 "\n" 3 "\t" 4 5 6 "\n" 7 8 9 "\n"
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experts i have many files in a directory and these files contain some numerical numbers.
Probably not POSIX-compatible; tested on Bash 5.0. Assuming that the filenames don't contain whitespaces and have fixed length.
for file_with_a in *A.??
do
target_left=${file_with_a:0:15} # e.g., 2019__01_NDV.NT
target_right=${file_with_a:16:3} # e.g., .AS
cat $target_left[A-C]$target_right > ${target_left}_ABC$target_right
done
With mapfile aka readarray which is a bash4+ feature.
#!/usr/bin/env bash
files=([0-9][0-9][0-9][0-9]__*[ABC]*.??)
while mapfile -d '' -n3 array && ((${#array[*]} == 3)); do
if [[ ${array[0]%%_*} == ${array[1]%%_*} && ${array[0]%%_*} == ${array[2]%%_*} ]]; then
paste "${array[0]}" "${array[1]}" "${array[2]}" > "${array[0]%${array[0]:(-4)}}_ABC.${array[0]:(-2)}"
fi
done < <(printf '%s\0' "${files[#]}")
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I know how to print all letters
{a..z} and {A..Z} and {0..9}
But is there a way to print all possible ASCII Characters via a bash loop?
You don't need a loop
echo -e \\x{0..7}{{0..9},{A..F}}
It prints all chars from 0 to 127.
If it is okay to use awk:
awk 'BEGIN{for (i=32;i<127;i++) printf("%c", i)}'
Or using printf:
for((i=32;i<127;i++)) do printf "\x$(printf %x $i)"; done
use this:
for ((i=32;i<127;i++)) do printf "\\$(printf %03o "$i")"; done;printf "\n"
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I got a list containing filenames that match the following two patterns:
one is like XXX_01.fastq
another is XXX_01_001.fastq
I am going to write a for loop (in bash) to loop over all the filenames with different patterns and I need to determine which ones match the patterns above. Any help about it?
Contents of list.txt:
$ cat list.txt
AAA_01.fastq
AA_01_001.fastq
BBB_01_002.fastq
BBB_02.fastq
Example using bash pattern matching:
for file in `cat list.txt`; do
if [[ $file =~ [A-Z]{3}_[0-9]{2}\.fastq || $file =~ [A-Z]{3}_[0-9]{2}_[0-9]{3}\.fastq ]]; then
echo "MATCH $file";
fi;
done
Output:
MATCH: AAA_01.fastq
MATCH: BBB_01_002.fastq
MATCH: BBB_02.fastq