I have a file "file.txt" that contains a number of if-statements as shown below:
if ( x == y)
if (x == (y + 1))
and so on.
How to display all the if-statements present in the file on the screen using grep.
I tried: grep -R "if ( * == * )" * but it's not showing the required result.
It looks like you're trying to glob. grep uses regular expressions.
Using GNU Grep:
grep -n '\<if\s*(' file.txt
This looks for if at the start of a word, followed by optional (*) whitespace (\s) and an opening parenthesis (().
If you only want to see tests for equivalence, you can do:
grep -n '\<if\s*(.*==' file.txt
...this adds a check for zero-to-many (*) (wildcard) characters (.) between ( and == but this won't catch multiline tests like:
if (status != OFF &&
volume == 11)
{
// [...]
}
If you only want to test for assignment:
grep -n '\<if\s*(.*[^!<>=+*/^-]=[^=]' test.txt
...this checks that the character before the = is not (^) and of !<>=+*/^- ([]) and that the character immediately afterward isn't an =.
This is not flawless, but I'm sure you can tweak the regular expression yourself to meet your needs.
Related
original string :
A/trunk/apple/B/trunk/apple/Z/trunk/orange/citrus/Q/trunk/melon/juice/venti/straw/
Depth of directories will vary, but /trunk part will always remain the same.
And a single character in front of /trunk is the indicator of that line.
desired output :
A /trunk/apple
B /trunk/apple
Z /trunk/orange
Q /trunk/melon/juice/venti/straw
*** edit
I'm sorry I made a mistake by adding a slash at the end of each path in the original string which made the output confusing. Original string didn't have the slash in front of the capital letter, but I'll leave it be.
my attempt :
echo $str1 | sed 's/\(.\/trunk\)/\n\1/g'
I feel like it should work but it doesn't.
With GNU awk for multi-char RS and RT:
$ awk -v RS='([^/]+/){2}[^/\n]+' 'RT{sub("/",OFS,RT); print RT}' file
A trunk/apple
B trunk/apple
Z trunk/orange
I'm setting RS to a regexp describing each string you want to match, i.e. 2 repetitions of non-/s followed by / and then a final string of non-/s (and non-newline for the last string on the input line). RT is automatically set to each of the matching strings, so then I just change the first / to a blank and print the result.
If each path isn't always 3 levels deep but does always start with something/trunk/, e.g.:
$ cat file
A/trunk/apple/banana/B/trunk/apple/Z/trunk/orange
then:
$ awk -v RS='[^/]+/trunk/' 'RT{if (NR>1) print pfx $0; pfx=gensub("/"," ",1,RT)} END{printf "%s%s", pfx, $0}' file
A trunk/apple/banana/
B trunk/apple/
Z trunk/orange
To deal with complex samples input, like where there could be N number of / and values after trunk in a single line please try following.
awk '
{
gsub(/[^/]*\/trunk/,OFS"&")
sub(/^ /,"")
sub(/\//,OFS"&")
gsub(/ +[^/]*\/trunk\/[^[:space:]]+/,"\n&")
sub(/\n/,OFS)
gsub(/\n /,ORS)
gsub(/\/trunk/,OFS"&")
sub(/[[:space:]]+/,OFS)
}
1
' Input_file
Explanation: Adding detailed explanation for above.
awk ' ##Starting awk program from here.
{
gsub(/[^/]*\/trunk/,OFS"&") ##Globally substituting everything from / to till next / followed by trunk/ with space and matched value.
sub(/^ /,"") ##Substituting starting space with NULL here.
sub(/\//,OFS"&") ##Substituting first / with space / here.
gsub(/ +[^/]*\/trunk\/[^[:space:]]+/,"\n&") ##Globally substituting spaces followed by everything till / trunk till space comes with new line and matched values.
sub(/\n/,OFS) ##Substituting new line with space.
gsub(/\n /,ORS) ##Globally substituting new line space with ORS.
gsub(/\/trunk/,OFS"&") ##Globally substituting /trunk with OFS and matched value.
sub(/[[:space:]]+/,OFS) ##Substituting spaces with OFS here.
}
1 ##Printing edited/non-edited line here.
' Input_file ##Mentioning Input_file name here.
With your shown samples, please try following awk code.
awk '{gsub(/\/trunk/,OFS "&");gsub(/trunk\/[^/]*\//,"&\n")} 1' Input_file
In awk you can try this solution. It deals with the special requirement of removing forward slashes when the next character is upper case. Will not win a design award but works.
$ echo "A/trunk/apple/B/trunk/apple/Z/trunk/orange" |
awk -F '' '{ x=""; for(i=1;i<=NF;i++){
if($(i+1)~/[A-Z]/&&$i=="/"){$i=""};
if($i~/[A-Z]/){ printf x""$i" "}
else{ x="\n"; printf $i } }; print "" }'
A /trunk/apple
B /trunk/apple
Z /trunk/orange
Also works for n words. Actually works with anything that follows the given pattern.
$ echo "A/fruits/apple/mango/B/anything/apple/pear/banana/Z/ball/orange/anything" |
awk -F '' '{ x=""; for(i=1;i<=NF;i++){
if($(i+1)~/[A-Z]/&&$i=="/"){$i=""};
if($i~/[A-Z]/){ printf x""$i" "}
else{ x="\n"; printf $i } }; print "" }'
A /fruits/apple/mango
B /anything/apple/pear/banana
Z /ball/orange/anything
This might work for you (GNU sed):
sed 's/[^/]*/& /;s/\//\n/3;P;D' file
Separate the first word from the first / by a space.
Replace the third / by a newline.
Print/delete the first line and repeat.
If the first word has the property that it is only one character long:
sed 's/./& /;s#/\(./\)#\n\1#;P;D' file
Or if the first word has the property that it begins with an upper case character:
sed 's/[[:upper:]][^/]*/& /;s#/\([[:upper:][^/]*/\)#\n\1#;P;D' file
Or if the first word has the property that it is followed by /trunk/:
sed -E 's#([^/]*)(/trunk/)#\n\1 \2#g;s/.//' file
With GNU sed:
$ str="A/trunk/apple/B/trunk/apple/Z/trunk/orange/citrus/Q/trunk/melon/juice/venti/straw/"
$ sed -E 's|/?(.)(/trunk/)|\n\1 \2|g;s|/$||' <<< "$str"
A /trunk/apple
B /trunk/apple
Z /trunk/orange/citrus
Q /trunk/melon/juice/venti/straw
Note the first empty output line. If it is undesirable we can separate the processing of the first output line:
$ sed -E 's|(.)|\1 |;s|/(.)(/trunk/)|\n\1 \2|g;s|/$||' <<< "$str"
A /trunk/apple
B /trunk/apple
Z /trunk/orange/citrus
Q /trunk/melon/juice/venti/straw
Using gnu awk you could use FPAT to set contents of each field using a pattern.
When looping the fields, replace the first / with /
str1="A/trunk/apple/B/trunk/apple/Z/trunk/orange"
echo $str1 | awk -v FPAT='[^/]+/trunk/[^/]+' '{
for(i=1;i<=NF;i++) {
sub("/", " /", $i)
print $i
}
}'
The pattern matches
[^/]+ Match any char except /
/trunk/[^/]+ Match /trunk/ and any char except /
Output
A /trunk/apple
B /trunk/apple
Z /trunk/orange
Other patterns that can be used by FPAT after the updated question:
Matching a word boundary \\< and an uppercase char A-Z and after /trunk repeat / and lowercase chars
FPAT='\\<[A-Z]/trunk(/[a-z]+)*'
If the length of the strings for the directories after /trunk are at least 2 characters:
FPAT='\\<[A-Z]/trunk(/[^/]{2,})*'
If there can be no separate folders that consist of a single uppercase char A-Z
FPAT='\\<[A-Z]/trunk(/([^/A-Z][^/]*|[^/]{2,}))*'
Output
A /trunk/apple
B /trunk/apple
Z /trunk/orange/citrus
Q /trunk/melon/juice/venti/straw
Assuming your data will always be in the format provided as a single string, you can try this sed.
$ sed 's/$/\//;s|\([A-Z]\)\([a-z/]*\)/\([a-z]*\?\)|\1 \2\3\n|g' input_file
$ echo "A/trunk/apple/pine/skunk/B/trunk/runk/bunk/apple/Z/trunk/orange/T/fruits/apple/mango/P/anything/apple/pear/banana/L/ball/orange/anything/S/fruits/apple/mango/B/rupert/cream/travel/scout/H/tall/mountains/pottery/barnes" | sed 's/$/\//;s|\([A-Z]\)\([a-z/]*\)/\([a-z]*\?\)|\1 \2\3\n|g'
A /trunk/apple/pine/skunk
B /trunk/runk/bunk/apple
Z /trunk/orange
T /fruits/apple/mango
P /anything/apple/pear/banana
L /ball/orange/anything
S /fruits/apple/mango
B /rupert/cream/travel/scout
H /tall/mountains/pottery/barnes
Some fun with perl, where you can using nonconsuming regex to autosplit into the #F array, then just print however you want.
perl -lanF'/(?=.{1,2}trunk)/' -e 'print "$F[2*$_] $F[2*$_+1]" for 0..$#F/2'
Step #1: Split
perl -lanF/(?=.{1,2}trunk)/'
This will take the input stream, and split each line whenever the pattern .{1,2}trunk is encountered
Because we want to retain trunk and the preceeding 1 or 2 chars, we wrap the split pattern in the (?=) for a non-consuming forward lookahead
This splits things up this way:
$ echo A/trunk/apple/B/trunk/apple/Z/trunk/orange/citrus/Q/trunk/melon/juice/venti/straw/ | perl -lanF'/(?=.{1,2}trunk)/' -e 'print join " ", #F'
A /trunk/apple/ B /trunk/apple/ Z /trunk/orange/citrus/ Q /trunk/melon/juice/venti/straw/
Step 2: Format output:
The #F array contains pairs that we want to print in order, so we'll iterate half of the array indices, and print 2 at a time:
print "$F[2*$_] $F[2*$_+1]" for 0..$#F/2 --> Double the iterator, and print pairs
using perl -l means each print has an implicit \n at the end
The results:
$ echo A/trunk/apple/B/trunk/apple/Z/trunk/orange/citrus/Q/trunk/melon/juice/venti/straw/ | perl -lanF'/(?=.{1,2}trunk)/' -e 'print "$F[2*$_] $F[2*$_+1]" for 0..$#F/2'
A /trunk/apple/
B /trunk/apple/
Z /trunk/orange/citrus/
Q /trunk/melon/juice/venti/straw/
Endnote: Perl obfuscation that didn't work.
Any array in perl can be cast as a hash, of the format (key,val,key,val....)
So %F=#F; print "$_ $F{$_}" for keys %F seems like it would be really slick
But you lose order:
$ echo A/trunk/apple/B/trunk/apple/Z/trunk/orange/citrus/Q/trunk/melon/juice/venti/straw/ | perl -lanF'/(?=.{1,2}trunk)/' -e '%F=#F; print "$_ $F{$_}" for keys %F'
Z /trunk/orange/citrus/
A /trunk/apple/
Q /trunk/melon/juice/venti/straw/
B /trunk/apple/
Update
With your new data file:
$ cat file
A/trunk/apple/B/trunk/apple/Z/trunk/orange/citrus/Q/trunk/melon/juice/venti/straw/
This GNU awk solution:
awk '
{
sub(/[/]$/,"")
gsub(/[[:upper:]]{1}/,"& ")
print gensub(/([/])([[:upper:]])/,"\n\\2","g")
}' file
A /trunk/apple
B /trunk/apple
Z /trunk/orange/citrus
Q /trunk/melon/juice/venti/straw
I have this text file:
# cat letter.txt
this
is
just
a
test
to
check
if
grep
works
The letter "e" appear in 3 words.
# grep e letter.txt
test
check
grep
Is there any way to return the letter printed on left of the selected character?
expected.txt
t
h
r
With shown samples in awk, could you please try following.
awk '/e/{print substr($0,index($0,"e")-1,1)}' Input_file
Explanation: Adding detailed explanation for above.
awk ' ##Starting awk program from here.
/e/{ ##Looking if current line has e in it then do following.
print substr($0,index($0,"e")-1,1)
##Printing sub string from starting value of index e-1 and print 1 character from there.
}
' Input_file ##Mentioning Input_file name here.
You can use positive lookahead to match a character that is followed by an e, without making the e part of the match.
cat letter.txt | grep -oP '.(?=e)'
With sed:
sed -nE 's/.*(.)e.*/\1/p' letter.txt
Assuming you have this input file:
cat file
this
is
just
a
test
to
check
if
grep
works
egg
element
You may use this grep + sed solution to find letter or empty string before e:
grep -oE '(^|.)e' file | sed 's/.$//'
t
h
r
l
m
Or alternatively this single awk command should also work:
awk -F 'e' 'NF > 1 {
for (i=1; i<NF; i++) print substr($i, length($i), 1)
}' file
This might work for you (GNU sed):
sed -nE '/(.)e/{s//\n\1\n/;s/^[^\n]*\n//;P;D}' file
Turn off implicit printing and enable extended regexp -nE.
Focus only on lines that meet the requirements i.e. contain a character before e.
Surround the required character by newlines.
Remove any characters before and including the first newline.
Print the first line (up to the second newline).
Delete the first line (including the newline).
Repeat.
N.B. The solution will print each such character on a separate line.
To print all such characters on their own line, use:
sed -nE '/(.e)/{s//\n\1/g;s/^/e/;s/e[^\n]*\n?//g;s/\B/ /g;p}' file
N.B. Remove the s/\B /g if space separation is not needed.
With GNU awk you can use empty string as FS to split the input as individual characters:
awk -v FS= '/[e]/ {for(i=2;i<=NF;i++) if ($i=="e") print $(i-1)}' file
t
h
r
Excluding "e" at the beginning in the for loop.
edited
empty string if e is the first character in the word.
For example, this input:
cat file2
grep
erroneously
egg
Wednesday
effectively
awk -v FS= '/^[e]/ {print ""} /[e]/ {for(i=2;i<=NF;i++) if ($i=="e") print $(i-1)}' file2
r
n
W
n
f
v
Please I have question: I have a file like this
#HWI-ST273:296:C0EFRACXX:2:2101:17125:145325/1
TTAATACACCCAACCAGAAGTTAGCTCCTTCACTTTCAGCTAAATAAAAG
+
8?8A;DDDD;#?++8A?;C;F92+2A#19:1*1?DDDECDE?B4:BDEEI
#BBBB-ST273:296:C0EFRACXX:2:1303:5281:183410/1
TAGCTCCTTCGCTTTCAGCTAAATAAAAGCCCAGTACTTCTTTTTTACCA
+
CCBFFFFFFHHHHJJJJJJJJJIIJJJJJJJJJJJJJJJJJJJIJJJJJI
#HWI-ST273:296:C0EFRACXX:2:1103:16617:140195/1
AAGTTAGCTCCTTCGCTTTCAGCTAAATAAAAGCCCAGTACTTCTTTTTT
+
#C#FF?EDGFDHH#HGHIIGEGIIIIIEDIIGIIIGHHHIIIIIIIIIII
#HWI-ST273:296:C0EFRACXX:2:1207:14316:145263/1
AATACACCCAACCAGAAGTTAGCTCCTTCGCTTTCAGCTAAATAAAAGCC
+
CCCFFFFFHHHHHJJJJJJJIJJJJJJJJJJJJJJJJJJJJJJJJJJJIJ
I
I'm interested just about the line that starts with '#HWI', but I want to count all the lines that are not starting with '#HWI'. In the example shown, the result will be 1 because there's one line that starts with '#BBB'.
To be more clear: I just want to know know the number of the first line of the patterns (that are 4 line that repeated) that are not '#HWI'; I hope I'm clear enough. Please tell me if you need more clarification
With GNU sed, you can use its extended address to print every fourth line, then use grep to count the ones that don't start with #HWI:
sed -n '1~4p' file.fastq | grep -cv '^#HWI'
Otherwise, you can use e.g. Perl
perl -ne 'print if 1 == $. % 4' -- file.fastq | grep -cv '^#HWI'
$. contains the current line number, % is the modulo operator.
But once we're running Perl, we don't need grep anymore:
perl -lne '++$c if 1 == $. % 4; END { print $c }' -- file.fastq
-l removes newlines from input and adds them to output.
I have two lists, one of which contains wildcards (in this case represented by *). I would like to compare the two lists and create an output of those that match, with each wildcard * representing a single character.
For example:
File 1
123456|Jane|Johnson|Pharmacist|janejohnson#gmail.com
09876579|Frank|Roberts|Butcher|frankie1#hotmail.com
092362936|Joe|Jordan|Joiner|joe#joesjoinery.com
928|Bob|Horton|Farmer|bhorton#farmernews.co.uk
File 2
1***6|Jane|Johnson|Pharmacist|janejohnson#gmail.com
09876579|Frank|Roberts|Butcher|f**1#hotmail.com
092362936|Joe|Jordan|J*****|joe#joesjoinery.com
928|Bob|Horton|Farmer|b*****n#f*********.co.uk
Output
092362936|Joe|Jordan|Joiner|joe#joesjoinery.com
928|Bob|Horton|Farmer|bhorton#farmernews.co.uk
Explanation
The first two lines are not considered matches because the number of *s is not equal to the number of characters shown in the first file. The latter two are, so they are added to output.
I have tried to reason out ways to do this in AWK and using Join, but I don't know any way to even start trying to achieve this. Any help would be greatly appreciated.
$ cat tst.awk
NR==FNR {
file1[$0]
next
}
{
# Make every non-* char literal (see https://stackoverflow.com/a/29613573/1745001):
gsub(/[^^*]/,"[&]") # Convert every char X to [X] except ^ and *
gsub(/\^/,"\\^") # Convert every ^ to \^
# Convert every * to .:
gsub(/\*/,".")
# Add line start/end anchors
$0 = "^" $0 "$"
# See if the current file2 line matches any line from file1
# and if so print that line from file1:
for ( line in file1 ) {
if ( line ~ $0 ) {
print line
}
}
}
$ awk -f tst.awk file1 file2
092362936|Joe|Jordan|Joiner|joe#joesjoinery.com
928|Bob|Horton|Farmer|bhorton#farmernews.co.uk
sed 's/\./\\./g; s/\*/./g' file2 | xargs -I{} grep {} file1
Explanation:
I'd take advantage of regular expression matching. To do that, we need to turn every asterisk * into a dot ., which represents any character in regular expressions. As a side effect of enabling regular expressions, we need to escape all special characters, particularly the ., in order for them to be taken literally. In a regular expression, we need to use \. to represent a dot (as opposed to any character).
The first step is perform these substitutions with sed, the second is passing every resulting line as a search pattern to grep, and search file1 for that pattern. The glue that allows to do this is xargs, where a {} is a placeholder representing a single line from the results of the sed command.
Note:
This is not a general, safe solution you can simply copy and paste: you should watch out for any characters, in your file containing the asterisks, that are considered special in grep regular expressions.
Update:
jhnc extends the escaping to any of the following characters: .\^$[], thus accounting for almost all sorts of email addresses. He/she then avoids the use of xargs by employing -f - to pass the results of sed as search expressions to grep:
sed 's/[.\\^$[]/\\&/g; s/[*]/./g' file2 | grep -f - file1
This solution is both more general and more efficient, see comment below.
I want a command that can match all the below criteria in Red Hat:
·number range between 0100xxxx to 0110xxxxx
·And have money over 300
·Status either X or Z
·id contains letter ‘a’
·Error_code starting with 2
number,money,status,error-code,id
010018739,13213,X,300,abcde
010523456,343,Z,500,xcvfe
010743576,563,X,201,fgsa
012095654,300,X,400,gcaz
019432343,300,X,402,dewa
011023324,200,X,206,dea
020023433,100,X,303,a
010832134,300,X,200,a
012244242,433,Z,204,ghfsa
Something like this:
awk -F, '($1>=1000000 && $1<11099999) && $2>300 && ($3 ~ "X" || $3 ~ "Z") && index($5,"a") && index($4,"2")==1' file
It doesn't cater for the status being lower-case (but you didn't ask for that), nor does it cater for there being spaces in front of the status or error code (but you didn't ask for that either).
grep only matches text, awk is much more flexible and should fit your case better. For instance:
awk 'BEGIN {FS=","} $2 > 300 {print;}' < yourfile
Basically this is saying that ',' is the field separator, and then for every line where the second field ($2) is > 300, the action (in this case just print the whole line, which could even be omitted IIRC) is executed.
You can have conditions as complex as you like, with a syntax that is similar to C. I would suggest reading man awk and googling for more complex examples, but you should get the idea.