I have a filepath like this: /bing/foo/bar/bin and I want to extract only the string between bing/ and the next slash.
So /bing/foo/bar/bin should just produce "foo".
I tried the following:
echo "/bing/foo/bar/bin" | sed -r 's/.*bing\/(.*)\/.*/\1/'
but this produces "foo/bar" instead of "foo".
Try this command
echo "/bing/foo/bar/bin" | sed -r 's|.*bing/([^/]*)/.*|\1|'
use | as delimiters instead of / is proper in your case, reference from "Delimiters in sed substitution",
sed can use any character as a delimiter, it will automatically use the character following the s as a delimiter.
or
echo "/bing/foo/bar/bin" | grep -oP "/bing/\K(\w+)"
Related
I have to get my db credentials from this configuration file:
# Database settings
Aisse.LocalHost=localhost
Aisse.LocalDataBase=mydb
Aisse.LocalPort=5432
Aisse.LocalUser=myuser
Aisse.LocalPasswd=mypwd
# My other app settings
Aisse.NumDir=../../data/Num
Aisse.NumMobil=3000
# Log settings
#Aisse.Trace_AppliTpv=blabla1.tra
#Aisse.Trace_AppliCmp=blabla2.tra
#Aisse.Trace_AppliClt=blabla3.tra
#Aisse.Trace_LocalDataBase=blabla4.tra
In particular, I want to get the value mydb from line
Aisse.LocalDataBase=mydb
So far, I have developed this
mydbname=$(echo "$my_conf_file.conf" | grep "LocalDataBase=" | sed "s/LocalDataBase=//g" )
that returns
mydb #Aisse.Trace_blabla4.tra
that would be ok if it did not return also the comment string.
Then I have also tryed
mydbname=$(echo "$my_conf_file.conf" | grep "Aisse.LocalDataBase=" | sed "s/LocalDataBase=//g" )
that retruns void string.
How can I get only the value that is preceded by the string "Aisse.LocalDataBase=" ?
Using sed
$ mydbname=$(sed -n 's/Aisse\.LocalDataBase=//p' input_file)
$ echo $mydbname
mydb
I'm afraid you're being incomplete:
You mention you want the line, containing "LocalDataBase", but you don't want the line in comment, let's start with that:
A line which contains "LocalDataBase":
grep "LocalDataBase" conf.conf.txt
A line which contains "LocalDataBase" but who does not start with a hash:
grep "LocalDataBase" conf.conf.txt | grep -v "^ *#"
??? grep -v "^ *#"
That means: don't show (-v) the lines, containing:
^ : the start of the line
* : a possible list of space characters
# : a hash character
Once you have your line, you need to work with it:
You only need the part behind the equality sign, so let's use that sign as a delimiter and show the second column:
cut -d '=' -f 2
All together:
grep "LocalDataBase" conf.conf.txt | grep -v "^ *#" | cut -d '=' -f 2
Are we there yet?
No, because it's possible that somebody has put some comment behind your entry, something like:
LocalDataBase=mydb #some information
In order to prevent that, you need to cut that comment too, which you can do in a similar way as before: this time you use the hash character as a delimiter and you show the first column:
grep "LocalDataBase" conf.conf.txt | grep -v "^ *#" | cut -d '=' -f 2 | cut -d '#' -f 1
Have fun.
You may use this sed:
mydbname=$(sed -n 's/^[^#][^=]*LocalDataBase=//p' file)
echo "$mydbname"
mydb
RegEx Details:
^: Start
[^#]: Matches any character other than #
[^=]*: Matches 0 or more of any character that is not =
LocalDataBase=: Matches text LocalDataBase=
You can use
mydbname=$(sed -n 's/^Aisse\.LocalDataBase=\(.*\)/\1/p' file)
If there can be leading whitespace you can add [[:blank:]]* after ^:
mydbname=$(sed -n 's/^[[:blank:]]*Aisse\.LocalDataBase=\(.*\)/\1/p' file)
See this online demo:
#!/bin/bash
s='# Database settings
Aisse.LocalHost=localhost
Aisse.LocalDataBase=mydb
Aisse.LocalPort=5432
Aisse.LocalUser=myuser
Aisse.LocalPasswd=mypwd
# My other app settings
Aisse.NumDir=../../data/Num
Aisse.NumMobil=3000
# Log settings
#Aisse.Trace_AppliTpv=blabla1.tra
#Aisse.Trace_AppliCmp=blabla2.tra
#Aisse.Trace_AppliClt=blabla3.tra
#Aisse.Trace_LocalDataBase=blabla4.tra'
sed -n 's/^Aisse\.LocalDataBase=\(.*\)/\1/p' <<< "$s"
Output:
mydb
Details:
-n - suppresses default line output in sed
^[[:blank:]]*Aisse\.LocalDataBase=\(.*\) - a regex that matches the start of a string (^), then zero or more whiespaces ([[:blank:]]*), then a Aisse.LocalDataBase= string, then captures the rest of the line into Group 1
\1 - replaces the whole match with the value of Group 1
p - prints the result of the successful substitution.
I have an text file, with line inside...
line: <version="AAA" email="...ANY..." file="BBB">
new desired line in text file to be: <version="AAA" email="NEW_TEXT" file="BBB">
I want to replace the ...ANY... expression with variable (replace entire line)
I have this script text-file script in #!/bin/bash, but I have problem when expanding double quotes in variables.
LINE_NUMBER="$(grep -nr 'email' *.txt | awk '{print $1}' | sed 's/[^0-9]*//g')"
VAR1="$(grep 'email' *.txt | cut -d '"' -f1-3)"
VAR2="$(grep 'email' *.txt | cut -d '"' -f5-)"
VAR3='NEW_TEXT'
NEW_LINE=$VAR1'"'$VAR3'"'$VAR2
new desired line in text file to be... <version="AAA" email="NEW_TEXT" file="BBB">
awk -i inplace 'NR=='"$LINE_NUMBER"'{sub(".*",'"'$NEW_LINE'"')},1' *.txt
but I get this new line:
<version="" email="NEW_TEXT" file="">
what do I do wrong?
How can I prevent expand duouble quotes inside variable?
please better write me an working example, I had tried other topics, forums, posts....but I have no luck.
You cas use sed :
VAR3='NEW_TEXT'
sed -i "s/email=\"[^\"]*\"/email=\"$VAR3\"/" myfile.xml
Suggesting:
var3="text space % special < chars"
Note var3 may not contain & which is special replacement meaning in sed
sed -E 's|email="[^"]*"|email="'"${var3}"'"|' input.1.txt
Explanation
[^"]* : Match longest string not having " till next ".
I have a variable which contains key/values separated by space:
echo $PROPERTY
server_geo=BOS db.jdbc_url=jdbc\:mysql\://mysql-test.com\:3306/db02 db.name=db02 db.hostname=/mysql-test.com datasource.class.xa=com.mysql.jdbc.jdbc2.optional.MysqlXADataSource server_uid=BOS_mysql57 hibernate33.dialect=org.hibernate.dialect.MySQL5InnoDBDialect hibernate.connection.username=db02 server_labels=mysql57,mysql5,mysql db.jdbc_class=com.mysql.jdbc.Driver db.schema=db02 hibernate.connection.driver_class=com.mysql.jdbc.Driver uuid=a19ua19 db.primary_label=mysql57 db.port=3306 server_label_primary=mysql57 hibernate.dialect=org.hibernate.dialect.MySQL5InnoDBDialect
I'd need to extract the values of the single keys, for example db.jdbc_url.
Using one code snippet I've found:
echo $PROPERTY | sed -e 's/ db.jdbc_url=\(\S*\).*/\1/g'
but that returns also other properties found before my key.
Any help how to fix it ?
Thanks
If db.name always follow db.jdbc_url, then use grep lookaround,
$ echo "${PROPERTY}" | grep -oP '(?<=db.jdbc_url=).*(?=db.name)'
jdbc\:mysql\://mysql-test.com\:3306/db02
or add the VAR to an array,
$ myarr=($(echo $PROPERTY))
$ echo "${myarr[1]}" | grep -oP '(?<=db.jdbc_url=).*(?=$)'
jdbc\:mysql\://mysql-test.com\:3306/db02
This is caused because you are using the substitute command (sed s/.../.../), so any text before your regex is kept as is. Using .* before db\.jdbc_url along with the begin (^) / end ($) of string marks makes you match the whole content of the variable.
In order to be totaly safe, your regex should be :
sed -e 's/^.*db\.jdbc_url=\(\S*\).*$/\1/g'
You can use grep for this, like so:
echo $PROPERTY | grep -oE "db.jdbc_url=\S+" | cut -d'=' -f2
The regex is very close to the one you used with sed.
The -o option is used to print the matched parts of the matching line.
Edit: if you want only the value, cut on the '='
Edit 2: egrep say it is deprecated, so use grep -oE instead, same result. Just to cover all bases :-)
I have a string in which I need to replace "," with "\," using shell script. I thought I can use sed to do this but no luck.
You can do that without sed:
string="${string/,/\\,}"
To replace all occurrences of "," use this:
string="${string//,/\\,}"
Example:
#!/bin/bash
string="Hello,World"
string="${string/,/\\,}"
echo "$string"
Output:
Hello\,World
You need to escape the back slash \/
I'm not sure what your input is but this will work:
echo "teste,test" |sed 's/,/\\/g'
output:
teste\test
Demo:
http://ideone.com/JUTp1X
If the string is on a file, you can use:
sed -i 's/,/\//g' myfile.txt
I want to get the string between <sometag param=' and '>
I tried to use the method from Get any string between 2 string and assign a variable in bash to get the "x":
echo "<sometag param='x'><irrelevant stuff='nonsense'>" | tr "'" _ | sed -n 's/.*<sometag param=_\(.*\)_>.*/\1/p'
The problem (apart from low efficiency because I just cannot manage to escape the apostrophe correctly for sed) is that sed matches the maximum, i.e. the output is:
x_><irrelevant stuff=_nonsense
but the correct output would be the minimum-match, in this example just "x"
Thanks for your help
You are probably looking for something like this:
sed -n "s/.*<sometag param='\([^']*\)'>.*/\1/p"
Test:
echo "<sometag param='x'><irrelevant stuff='nonsense'>" | sed -n "s/.*<sometag param='\([^']*\)'>.*/\1/p"
Results:
x
Explanation:
Instead of a greedy capture, use a non-greedy capture like: [^']* which means match anything except ' any number of times. To make the pattern stick, this is followed by: '>.
You can also use double quotes so that you don't need to escape the single quotes. If you wanted to escape the single quotes, you'd do this:
-
... | sed -n 's/.*<sometag param='\''\([^'\'']*\)'\''>.*/\1/p'
Notice how that the single quotes aren't really escaped. The sed expression is stopped, an escaped single quote is inserted and the sed expression is re-opened. Think of it like a four character escape sequence.
Personally, I'd use GNU grep. It would make for a slightly shorter solution. Run like:
... | grep -oP "(?<=<sometag param=').*?(?='>)"
Test:
echo "<sometag param='x'><irrelevant stuff='nonsense'>" | grep -oP "(?<=<sometag param=').*?(?='>)"
Results:
x
You don't have to assemble regexes in those cases, you can just use ' as the field separator
in="<sometag param='x'><irrelevant stuff='nonsense'>"
IFS="'" read x whatiwant y <<< "$in" # bash
echo "$whatiwant"
awk -F\' '{print $2}' <<< "$in" # awk