This question already has answers here:
Bash : extracting part of a string
(4 answers)
Closed 6 years ago.
Say I have a string 0.0.25, how do I delete the last part after dot (including it) to make it like 0.0? Note that last part can have variable number of digits.
In bash you can do the following:
echo "${var%.*}"
See the Shell Parameter Expansion section of the manual.
Using awk you could:
echo "0.0.25" | awk -F. '{print $1"."$2}'
Related
This question already has answers here:
How do I set a variable to the output of a command in Bash?
(15 answers)
How do I remove the file suffix and path portion from a path string in Bash?
(15 answers)
Closed 1 year ago.
I have a file 20210823_Club_Member_Name_by_ID.txt. I want to get only the first element of the file name which is 20210823 and store it into a variable using shell script.
Currently, my code can print out the first element in the terminal but I also want to store it into a variable for further usage.
file='20210823_Club_Member_Name_by_ID.txt'
echo "$file" | awk -F'[_.]' '{print $1}'
// I try to store it like below, but it does not work
fileDate= echo "$file" | awk -F'[_.]' '{print $1}'
echo $fileDate
As Jetchisel commented, you can use shell parameter expansion to safely extract the value. The %% operator removes as much of the matching text as possible, starting from the end of the string; if we use _* then this will essentially remove everything after and including the first underscore.
file='20210823_Club_Member_Name_by_ID.txt'
fileDate="${file%%_*}"
The fileDate variable will now contain 20210823.
This question already has answers here:
Extract substring in Bash
(26 answers)
Closed 6 years ago.
How to retrieve the first 10 characters of a variable with Bash?
FOO="qwertzuiopasdfghjklyxcvbnm"
I need to get qwertzuiop.
If the variable is: FOO="qwertzuiopasdfghjklyxcvbnm"
then
echo ${FOO:0:10}
will give the first 10 characters.
Use the head command.
echo $FOO | head -c 10
=> qwertzuiop
This question already has answers here:
Replacing some characters in a string with another character
(6 answers)
Difference between sh and Bash
(11 answers)
Closed 4 years ago.
Trying to write a bash script and in one part of it I need to take whatever parameter was passed to it and replace the hyphens with underscores if they exist.
Tried to do the following
#!/usr/bin/env bash
string=$1
string=${string//-/_}
echo $string;
It's telling me that this line string=${string//-/_} fails due to "Bad substitution" but it looks like it should do it? Am I missing something?
There is nothing wrong with your script, and it should work in modern versions of Bash.
But just in case you can simplify that to :
#!/bin/bash
echo "$1" | tr '-' '_'
This is in case that parameter substitution does not work ( which seems to be your case ).
Regards!
This question already has answers here:
How to find/replace and increment a matched number with sed/awk?
(5 answers)
Closed 4 years ago.
Suppose I have the following Bash script:
#!/bin/bash
INCREMENT_BY=5
sed 's/20000/&+$INCREMENT_BY/g' old > new
I expect all occurrences of 20000 to be replaced by 20005, but instead they are replaced with 20000+$INCREMENT_BY. How can I make this work?
You should use double quote for eval variable, like:
sed "s/20000/$(( $INCREMENT_BY + 2000))/g" old
This question already has answers here:
How to insert strings containing slashes with sed? [duplicate]
(11 answers)
Closed 7 years ago.
First let me apologize for asking a question that:
Has some examples, though I find them confusing
Has a man page, also find confusing
Problem:
I would like to replace text in a $STRING within bash for a script I am writing. I chose to combine date/time to allow for easier end user integration.
STARTTIME="2015-03-17/11:30:00"
sed "Unknown"
Attempted Solution:
sed '/s// /' "$STARTTIME"
Desired result is to remove the "/" and end up with 2015-03-17 11:30:00 to then be passed to a command.
Thank you for any assistance.
If you're using bash, I would suggest that you used built-in string manipulation:
$ s='2015-03-17/11:30:00'
$ echo "${s/\// }"
2015-03-17 11:30:00
The syntax inside the braces means "replace the first occurrence of a forward slash (which needs escaping) with a space".