Grep Line break in Indesign software - grep

does GREP in indesign can help to break the line
Example: (below is the formate i want grep help me to break line from complimentary & of equal) is that possible?
You and your guest are invited to enjoy one complimentary
MAIN COURSE when a MAIN COURSE
of equal or greater value is purchased.

If I understand your question, and you have one long line as follows:
You and your guest are invited to enjoy one complimentary MAIN COURSE when a MAIN COURSE of equal or greater value is purchased.
You want to break that long line into the following 3-lines:
You and your guest are invited to enjoy one complimentary
MAIN COURSE when a MAIN COURSE
of equal or greater value is purchased.
Then sed is what you want, not grep:
sed -e 's/COURSE\sof/COURSE\nof/' -e 's/complimentary\s/complimentary\n/'
Usage:
sed -e ... -e ... <<<"Your long line"
or
echo "Your long line" | sed -e ... -e ...
where ... are expression(s) 1 and 2 above (e.g. 's/COURSE\sof/COURSE\nof/')

Related

Grepping twice using result of first Grep in Large file

Am given a list if ID which I need to trace back a name in a file
file: ID contains
1
2
3
4
5
6
The ID are contained in a Large 2 GB file called result.txt
ABC=John,dhds,72828,73737,3939,92929
CDE=John,uubad,32424,ajdaio,343533
FG1=Peter,iasisaio,097282,iosoido
WER=Ann,97391279,89719379,7391739
result,**id=1**,iuhdihdio,ihwoihdoih,iuqhwiuh,ABC
result2,**id=2**,9729179,hdqihi,hidqi,82828,CDE
result3,**id=3**,biasi,8u9829,90u209w,jswjso,FG1
So I cat the ID file into a variable
I then use this variable in a loop to grep out the values to link back to the name using grep and cut -d from results.txt and output to a variable
so variable contains ABS CDE FG1
In the same loop I pass the output of the grep to perform another grep on results.txt, to get the name
ie regrets file for ABC CDE FG1
I do get the answer but takes a long time is their a more efficient way?
Thanks
Making some assumptions about your requirement... ID's that are not found in the big file will not be shown in the output; the desired output is in the format shown below.
Here are mock input files - f1 for the id's and f2 for the large file:
[mathguy#localhost test]$ cat f1
1
2
3
4
5
6
[mathguy#localhost test]$ cat f2
ABC=John,dhds,72828,73737,3939,92929
CDE=John,uubad,32424,ajdaio,343533
FG1=Peter,iasisaio,097282,iosoido
WER=Ann,97391279,89719379,7391739
result,**id=1**,iuhdihdio,ihwoihdoih,iuqhwiuh,ABC
result2,**id=2**,9729179,hdqihi,hidqi,82828,CDE
result3,**id=3**,biasi,8u9829,90u209w,jswjso,FG1
Proposed solution and output:
[mathguy#localhost test]$ sed 's/.*/\*\*id=&\*\*/' f1 | grep -Ff - f2 | \
> sed -E 's/^.*\*\*id=([[:digit:]]*)\*\*.*,([^,]*)$/\1 \2/'
1 ABC
2 CDE
3 FG1
The hard work here is done by grep -F which might be just fast enough for your needs. There is some prep work and some clean-up work done by sed, but those are both on small datasets.
First we take the id's from the input file and we output strings in the format **id=<number>**. The output is presented as the fixed-character patterns to grep -F via the option -f (take the patterns from file, in this case from stdin, invoked as -; that is, from the output of sed).
After we find the needed lines from the big file, the final sed just extracts the id and the name from each line.
Note: this assumes that each id is only found once in the big file. (Actually the command will work regardless; but if there are duplicate lines for an id, your business users will have to tell you how to handle. What if you get contradictory names for the same id? Etc.)

duplicate grep output when comparing two files

I have literally been at this for 5 hours, I have busybox on my device, and I unfortunately do not have -X in grep to make my life easier.
edit;
I have two list both of them have mac addresses, essentially I am just wanting to achieve offline mac address lookup so I don't have to keep looking it up online
list.txt has vendor mac prefix of course this isn't the complete list but just for an example
00:13:46
00:15:E9
00:17:9A
00:19:5B
00:1B:11
00:1C:F0
scan will have list of different mac addresses unknown to which vendor they go to. Which will be full length mac addresses. when ever there is a match I want the line in scan to be output.
Pretty much it does that, but it outputs everything from the scan file, and then it will output matching one at the end, and causing duplicate. I tried sort -u, but it has no effect its as if there is two different output from two different methods, the reason why I say that is because it will instantly output scan file that has everything in it, and couple seconds later it will output the matching one.
From searching I came across this
#!/bin/bash
while read line; do
grep -F 'list' 'scan'
done < list.txt
which displays the duplicate result when/if found, the output is pretty much echoing my scan file then displaying the matched pattern, this creating duplicate
This is frustrating me that I have not found a solution after click on all the links in google up to page 9.
Please someone help me.
I don't know if the Busybox sed supports this out of the box, but it should be easy to do in Awk or Perl instead then.
Create a sed script to print lines from file2 which are covered by a prefix in file1 by transforming each line in file1 into a sed command to print a match for that regular expression:
sed 's%.*%/&/p%' file1 | sed -n -f - file2
The same in Awk:
awk 'NR==FNR { a[++i]="^" $0; next }
{ for (j=1; j<=i; ++j) if ($0 ~ a[j]) print }' file1 file2
Ok guys I did a nested for loop (probably very in efficient) but I got it working printing the matching mac addresses using this
#!/usr/bin/bash
for scanlist in `cat scan | cut -d: -f1,2,3`
do
for listt in `cat list`
do
if [[ $scanlist == $listt ]]; then
grep $scanlist scan
fi
done
done
if anyone can make this more elegant but it works for me for now. I think the problem I had was one list contained just 00:11:22 while my other list contained 00:11:22:33:44:55 that is why I cut it on my scanlist to make same length as my other list. So this only output the matches instead of doing duplicate output.

grep within a line, show unmatched pattern

How do I grep within a line, and print the unmatched words?
For example, the line is something like " one two three ".
What I want is anything that is not one, and trimmed (leading and trailing spaces (could be space or tabs) removed ).
In this case, how do i get "two three"?
using sed instead:
sed -r 's/\bone\b//g;s/^\s*|\s*$//g'
E.g.
kent$ echo " one two three "|sed -r 's/\bone\b//g;s/^\s*|\s*$//g'
two three
Use sed instead:
sed -e 's/[ \t]*one[ \t]*//'

How to filter using grep on a selected word

grep (GNU grep) 2.14
Hello,
I have a log file that I want to filter on a selected word. However, it tends to filter on many for example.
tail -f gateway-* | grep "P_SIP:N_iptB1T1"
This will also find words like this:
"P_SIP:N_iptB1T10"
"P_SIP:N_iptB1T11"
"P_SIP:N_iptB1T12"
etc
However, I don't want to display anything after the 1. grep is picking up 11, 12, 13, etc.
Many thanks for any suggestions,
You can restrict the word to end at 1:
tail -f gateway-* | grep "P_SIP:N_iptB1T1\>"
This will work assuming that you have a matching case which is only "P_SIP:N_iptB1T1".
But if you want to extract from P_SIP:N_iptB1T1x, and display only once, then you need to restrict to show only first match.
grep -o "P_SIP:N_iptB1T1"
-o, --only-matching show only the part of a line matching PATTERN
More info
At least two approaches can be tried:
grep -w pattern matches for full words. Seems to work for this case too, even though the pattern has punctuation.
grep pattern -m 1 to restrict the output to first match. (Also doable with grep xxx | head -1)
If the lines contains the quotes as in your example, just use the -E option in grep and match the closing quote with \". For example:
grep -E "P_SIP:N_iptB1T1\"" file
If these quotes aren't in the text file, and there's blank spaces or endlines after the word, you can match these too:
# The word is followed by one or more blanks
grep -E "P_SIP:N_iptB1T1\s+" file
# Match lines ending with the interesting word
grep -E "P_SIP:N_iptB1T1$" file

extract a line from a file using csh

I am writing a csh script that will extract a line from a file xyz.
the xyz file contains a no. of lines of code and the line in which I am interested appears after 2-3 lines of the file.
I tried the following code
set product1 = `grep -e '<product_version_info.*/>' xyz`
I want it to be in a way so that as the script find out that line it should save that line in some variable as a string & terminate reading the file immediately ie. it should not read furthermore aftr extracting the line.
Please help !!
grep has an -m or --max-count flag that tells it to stop after a specified number of matches. Hopefully your version of grep supports it.
set product1 = `grep -m 1 -e '<product_version_info.*/>' xyz`
From the man page linked above:
-m NUM, --max-count=NUM
Stop reading a file after NUM matching lines. If the input is
standard input from a regular file, and NUM matching lines are
output, grep ensures that the standard input is positioned to
just after the last matching line before exiting, regardless of
the presence of trailing context lines. This enables a calling
process to resume a search. When grep stops after NUM matching
lines, it outputs any trailing context lines. When the -c or
--count option is also used, grep does not output a count
greater than NUM. When the -v or --invert-match option is also
used, grep stops after outputting NUM non-matching lines.
As an alternative, you can always the command below to just check the first few lines (since it always occurs in the first 2-3 lines):
set product1 = `head -3 xyz | grep -e '<product_version_info.*/>'`
I think you're asking to return the first matching line in the file. If so, one solution is to pipe the grep result to head
set product1 = `grep -e '<product_version_info.*/>' xyz | head -1`

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