I'm trying to cut a string until it's first specified characters.
in this case, it's first Latin letter.
I tired this. It kind of works but sometimes shifts 1 or more characters.
f=$(echo $f | tail -c $((${#f}+-$(expr index "$f" [azertyuiopqsdfghjklmwxcvbnAZERTYUIOPQSDFGHJKLMWXCVBN])+2)))
take this string as example: ã%82¹ã%83%91ã%83¼ã%82¯ã%83«__original_ver.__-Your_name.mp3
I want to get: original_ver.__-Your_name.mp3
I tend to get this instead: ver.__-Your_name.mp3
Is there a better method? if so, some explanation is always welcome.
You can use extended globbing:
f=$(shopt -s extglob; LC_ALL=C; echo "${f##+([^[:alpha:]])}")
f=$(shopt -s extglob; LC_ALL=C; echo "${f/#+([^[:alpha:]])}")
or sed:
f=$(LC_ALL=C sed -r 's/^[^[:alpha:]]+//' <<< "$f")
Setting LC_ALL to C is mandatory, otherwise [[:alpha:]] might match wrong characters.
Related
I am using the below code for replacing a string
inside a shell script.
echo $LINE | sed -e 's/12345678/"$replace"/g'
but it's getting replaced with $replace instead of the value of that variable.
Could anybody tell what went wrong?
If you want to interpret $replace, you should not use single quotes since they prevent variable substitution.
Try:
echo $LINE | sed -e "s/12345678/${replace}/g"
Transcript:
pax> export replace=987654321
pax> echo X123456789X | sed "s/123456789/${replace}/"
X987654321X
pax> _
Just be careful to ensure that ${replace} doesn't have any characters of significance to sed (like / for instance) since it will cause confusion unless escaped. But if, as you say, you're replacing one number with another, that shouldn't be a problem.
you can use the shell (bash/ksh).
$ var="12345678abc"
$ replace="test"
$ echo ${var//12345678/$replace}
testabc
Not specific to the question, but for folks who need the same kind of functionality expanded for clarity from previous answers:
# create some variables
str="someFileName.foo"
find=".foo"
replace=".bar"
# notice the the str isn't prefixed with $
# this is just how this feature works :/
result=${str//$find/$replace}
echo $result
# result is: someFileName.bar
str="someFileName.sally"
find=".foo"
replace=".bar"
result=${str//$find/$replace}
echo $result
# result is: someFileName.sally because ".foo" was not found
Found a graceful solution.
echo ${LINE//12345678/$replace}
Single quotes are very strong. Once inside, there's nothing you can do to invoke variable substitution, until you leave. Use double quotes instead:
echo $LINE | sed -e "s/12345678/$replace/g"
Let me give you two examples.
Using sed:
#!/bin/bash
LINE="12345678HI"
replace="Hello"
echo $LINE | sed -e "s/12345678/$replace/g"
Without Using sed:
LINE="12345678HI"
str_to_replace="12345678"
replace_str="Hello"
result=${str//$str_to_replace/$replace_str}
echo $result
Hope you will find it helpful!
echo $LINE | sed -e 's/12345678/'$replace'/g'
you can still use single quotes, but you have to "open" them when you want the variable expanded at the right place. otherwise the string is taken "literally" (as #paxdiablo correctly stated, his answer is correct as well)
To let your shell expand the variable, you need to use double-quotes like
sed -i "s#12345678#$replace#g" file.txt
This will break if $replace contain special sed characters (#, \). But you can preprocess $replace to quote them:
replace_quoted=$(printf '%s' "$replace" | sed 's/[#\]/\\\0/g')
sed -i "s#12345678#$replace_quoted#g" file.txt
I had a similar requirement to this but my replace var contained an ampersand. Escaping the ampersand like this solved my problem:
replace="salt & pepper"
echo "pass the salt" | sed "s/salt/${replace/&/\&}/g"
use # if you want to replace things like /. $ etc.
result=$(echo $str | sed "s#$oldstr#$newstr#g")
the above code will replace all occurrences of the specified replacement term
if you want, remove the ending g which means that the only first occurrence will be replaced.
Use this instead
echo $LINE | sed -e 's/12345678/$replace/g'
this works for me just simply remove the quotes
I prefer to use double quotes , as single quptes are very powerful as we used them if dont able to change anything inside it or can invoke the variable substituion .
so use double quotes instaed.
echo $LINE | sed -e "s/12345678/$replace/g"
I want to extract version string (1_4_5) from my-app-1_4_5.img and then convert into dot version (1.4.5) without filename. Version string will have three (1_4_5) or four (1_4_5_7) segments.
Have this one liner working ls my-app-1_4_5.img | cut -d'-' -f 3 | cut -d'.' -f 1 | tr _ .
Would like to know if there is any better way rather than piping output from cut.
Here's an attempt with parameter expansion. I'm assuming you have a wildcard pattern you want to loop over.
for file in *-*.img; do
base=${file%.img}
ver=${base##*-}
echo "${ver//_/.}"
done
The construct ${var%pattern} returns the variable var with any suffix matching pattern trimmed off. Similarly, ${var#pattern} trims any prefix which matches pattern. In both cases, doubling the operator switches to trimming the longest possible match instead of the shortest. (These are POSIX-compatible pattenr expansion, i.e. not strictly Bash only.) The construct ${var/pattern/replacement} replaces the first match in var on pattern with replacement; doubling the first slash causes every match to be replaced. (This is Bash only.)
You can do it with sed:
sed -E "s/.*([0-9]+)_([0-9]+)_([0-9]+).*/\1.\2.\3/" <<< my-app-1_4_5.img
Assuming the version number will always be between the last dash and the file extension, you can use something like this in pure Bash:
name="file-name-x-1_2_3_4_5.ext"
version=${name##*-}
version=${version%%.ext}
version=${version//_/.}
echo $version
The code above will result in:
1.2.3.4.5
For a complete explanation about the brace expansions used above, please take a look at Bash Reference Manual: 3.5.1 Brace Expansion.
Remove everything but 0 to 9, _ and newline and then replace all _ with .:
echo "my-app-1_4_5.img" | tr -cd '0-9_\n' | tr '_' '.'
Output:
1.4.5
With bash and a regex:
echo "my-app-1_4_5.img" | while IFS= read -r line; do [[ "$line" =~ [^0-9]([0-9_]+)[^0-9] ]] && echo "${BASH_REMATCH[1]//_/.}"; done
Output:
1.4.5
A slightly shorter variant
name=my-app-1_4_5.img
vers=${name//[!0-9_]}
$ echo ${vers//_/.}
1.4.5
Goal is to take string from one file and replace the specific string in another file.
Almost done, but need to know the marked line, or maybe there is more relevant solution.
Thanks.
if grep -q "edem_pl" /sdcard/DiseqTree.ini ; then
grep edem_pl /sdcard/DiseqTree.ini > /sdcard/temp.txt
else
echo "no edemtv user"
fi
**********************
after making some actions
**********************
if [ -e "/sdcard/temp.txt" ]; then
**copy string from /sdcard/temp.txt and replace with it string that contains edemtv in /sdcard/DiseqTree.ini**
else
echo "no edemtv user"
fi
The command:
sed s/edem_pl/edemtv/g
will act as a filter that replaces (substitutes) all lines (globally) containing edem_pl that are passed through it with equivalent lines containing edemtv instead. You can insert this into your command like so:
grep edem_pl /sdcard/DiseqTree.ini | sed s/edem_pl/edemtv/g > /sdcard/temp.txt
and /sdcard/temp.txt will contain only the lines originally containing edem_pl, with edemtv substituted for them.
Keep in mind that edem_pl and edemtv are being treated as regular expressions here. In this case, no special escaping needs to be done, but if they contain regex special characters like [ and ^ those characters will need to be escaped with \.
the best solution
sed -i "s!.*edemtv.*!$(cat /sdcard/temp.txt)!g" /sdcard/DiseqTree.ini
also
replace=`grep edem_pl /sdcard/temp.txt`
sed -i "s/.*edemtv.*/$replace/g" /sdcard/DiseqTree.ini
one more
busybox sed -i '/.*edemtv.*/{
r /sdcard/temp.txt
d
}' /sdcard/DiseqTree.ini
thanks to 4pda community
I have a file, the content of file has a string like this:
'/ad/e','#'.base64_decode("ZXZhbA==").'($zad)', 'add'
I want to check the file has this string. But when I use grep to check, It always return false. I try some ways:
grep "'/ad/e','#'.base64_decode("ZXZhbA==").'($zad)', 'add'" foo.txt
grep "'/ad/e','#'\.base64_decode\("ZXZhbA\=\="\)\.'\(\$zad\)', 'add'" foo.txt
str="'/ad/e','#'\.base64_decode\("ZXZhbA\=\="\)\.'\(\$zad\)', 'add'"
grep "$str" foo.txt
Can you help me? Maybe, another command line.
This is my case:
while read str; do
if [ ! -z "$str" ]; then
if grep -Fxq "$str" "$file_path"; then
do somthing
fi
fi
done < <(cat /usr/local/caotoc/db.dat)
Thank you so much!
First, you need to make sure the string is quoted properly. This is a bit of an art form, since your string contains both single and double quotes.
One thought would be to use read and a here-document to avoid having to escape anything.
Second, you need to use -F to perform exact string matching instead of more general regular-expression matching.
IFS= read -r str <<'EOF'
'/ad/e','#'.base64_decode("ZXZhbA==").'($zad)', 'add'
EOF
grep -F "$str" foo.txt
Based on the update, you can use a simple loop to read them one at a time.
while IFS= read -r str; do
grep -F "$str" foo.txt
done < /usr/local/caotoc/db.dat
You may be able to simply use the -f option to grep, which will cause grep to output lines from foo.txt that match any line from db.dat.
grep -f /usr/local/caotoc/db.dat -F foo.txt
Instead of trying to workaround regexes, the simplest way is to turn off regular expressions using -F (or --fixed-strings) option, which makes grep act like a simple string search
-F, --fixed-strings PATTERN is a set of newline-separated strings
like this:
grep -F "'/ad/e','#'.base64_decode(\"ZXZhbA==\").'(\$zad)', 'add'" test
Note: because of the shell, you still need to escape:
double quotes
dollar sign or else $zad is evaluated as an environment variable
I Want to search variable and replace with its absolute path in file.
setenv ABC /home/xyz
cat file.txt
${ABC}/Test/Folder_1
${ABC}/Test/Folder_2
I want to replace all occurance of the ${ABC} by /home/xyz.
I tried by the below mentioned way, but does not work,
sed -i 's/\$ABC/echo $ABC/g' file.txt
I can do by below mentioned way, but I do not want to do this way.(I have to put so many back slash)
$ echo $ABC | sed -i 's/\$ABC/\/home\/xyz/g' file.txt
Please give me some suggestion for this question.
Thank You.
If you really want to use the value from a variable in your replacement string, you could use
sed "s#\${ABC}#$ABC#g" file.txt
Character after s in sed is the delimiter and it can be any one character of your choice and it works as long as it's not in the string-to-be-matched and string-to-be-replaced.
Example :
sed 's:string-to-be-matched:string-to-be-replaced:g' file-to-be-edited
: is the delimiter
g means global replacement.
In your case, as the string-to-be-replaced contains the / , the same you are using as sed delimiter.
Simple Solution will be :
sed -i 's:${ABC}:'"$ABC"':g' fill.txt
'" is at either end of $ABC in the replacement string. Purpose is to expand shell variable to use with sed
Another way to get the absolute path is readlink -f ${ABC}/Test/Folder_2
Or the perl alternative to your slash hungry command
$ echo $ABC | sed -i 's/\$ABC/\/home\/xyz/g' file.txt
would be
$ echo $ABC | perl -p -i -e 's!\$ABC!/home/xyz!g'
the first character after the 's above will be used as the delimiter in the replacement expression (i.e. 's#foo#bar#g')