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I have a file having the following structure
destination list
move from station d-435-435 to point place1
move from station d-435-435 to point place2
move from mainpoint
I want to extract the word "d-435-435"(Only the first match, this need not be same value always) in between the words "from station" and "to point"
How can I achieve this?
What I have tried so far?
id=$(sed 's/.*from station \(.*\) to.*/\1/' input.txt)
But this returns the following value: destination list d-435-435 move from mainpoint
1st solution: With your shown samples, please try following GNU awk code. Using match function of awk program here to match regex rom station\s+\S+\s+to point to get requested value by OP then removing from station\s+ and \s+to point from matched value and printing required value.
awk '
match($0,/from station\s+\S+\s+to point/){
val=substr($0,RSTART,RLENGTH)
gsub(/from station\s+|\s+to point/,"",val)
print val
exit
}
' Input_file
2nd solution: Using GNU grep please try following. Using -oP option to print matched portion and enabling PCRE regex respectively here. Then in main grep program matching string from station followed by space(s) then using \K option will make sure matched part before \K is forgotten(since e don't need this in output), Then matching \S+(non space values) followed by space(s) to point string(using positive look ahead here to make sure it only checks its present or not but doesn't print that).
grep -oP -m1 'from station\s+\K\S+(?=\s+to point)' Input_file
If GNU sed is available, how about:
id=$(sed -nE '0,/from station.*to/ s/.*from station (.*) to.*/\1/p' input.txt)
The -n option suppress the print unless the substitution succeeds.
The condition 0,/pattern/ is a flip-flop operator and it returns false
after the pattern match succeeds. The 0 address is a GNU sed extension which
makes the 1st line to match against the pattern.
With awk you can write the before and after conditions of
field $4, where d-435-435 is, and then print this field only the first match and exit with exit after print statement:
awk '$2=="from" && $3=="station" && $5=="to" && $6=="point" {print $4; exit}' file
d-435-435
or using GNU awk for the 3rd arg to match():
awk 'match($0,/from station\s+(.*)\s+to point/,a){print a[1];exit}' file
d-435-435
The regexp contains a parenthesis, so the integer-indexed element of array a[1] contain the portion of string between from station followed by space(s) \s+ and space(s) \s+ followed byto point.
This might work for you (GNU sed):
sed -nE '/.*station (\S+) to point.*/{s//\1/;H;x;/\n(\S+)\n.*\1/{s/\n\S+$//;x;d};x;p}' file
Turn off implicit printing and on extended regexps command line options -nE.
If a line matches the required criteria, extract the required string, append a copy to the hold space, check if the match has already been seen and if not print it. If the match has been seen, remove it from the hold space.
Otherwise, do not print anything.
This should work in any sed:
sed -e '/.*from station \([^ ]*\) to .*/!d' -e 's//\1/' -e q file
I have an input csv file that looks something like:
Name,Index,Location,ID,Message
Alexis,10,Punggol,4090b43,Production 4090b43
Scott,20,Bedok,bfb34d3,Prevent
Ronald,30,one-north,86defac,Difference 86defac
Cindy,40,Punggol,40d0ced,Central
Eric,50,one-north,aeff08d,Military aeff08d
David,60,Bedok,5d1152d,Study
And I want to write a bash shell script using awk and gsub to replace 6-7 alpha numeric character long strings under the ID column with "xxxxx", with the output in a separate .csv file.
Right now I've got:
#!/bin/bash
awk -F ',' -v OFS=',' '{gsub(/^([a-zA-Z0-9]){6,7}/g, "xxxxx", $4);}1' input.csv > output.csv
But the output from I'm getting from running bash myscript.sh input.csv doesn't make any sense. The output.csv file looks like:
Name,Index,Location,ID,Message
Alexis,10,Punggol,4xxxxx9xxxxxb43,Production 4090b43
Scott,20,Bedok,bfb34d3,Prevent
Ronald,30,one-north,86defac,Difference 86defac
Cindy,40,Punggol,4xxxxxdxxxxxced,Central
Eric,50,one-north,aeffxxxxx8d,Military aeff08d
David,60,Bedok,5d1152d,Study
but the expected output csv should look like:
Name,Index,Location,ID,Message
Alexis,10,Punggol,xxxxx,Production 4090b43
Scott,20,Bedok,xxxxx,Prevent
Ronald,30,one-north,xxxxx,Difference 86defac
Cindy,40,Punggol,xxxxx,Central
Eric,50,one-north,xxxxx,Military aeff08d
David,60,Bedok,xxxxx,Study
With your shown sample, please try the following code:
awk -F ',[[:space:]]+' -v OFS=',\t' '
{
sub(/^([a-zA-Z0-9]){6,7}$/, "xxxxx", $4)
$1=$1
}
1
' Input_file | column -t -s $'\t'
Explanation: Setting field separator as comma, space(s), then setting output field separator as comma tab here. Then substituting from starting to till end of value(6 to 7 occurrences) of alphanumeric(s) with xxxxx in 4th field. Finally printing current line. Then sending output of awk program to column command to make it as per shown sample of OP.
EDIT: In case your Input_file is separated by only , as per edited samples now, then try following.
awk -F ',' -v OFS=',' '
{
sub(/^([a-zA-Z0-9]){6,7}$/, "xxxxx", $4)
}
1
' Input_file
Note: OP has installed latest version of awk from older version and these codes helped.
The short version to your answer would be the following:
$ awk 'BEGIN{FS=OFS=","}(FNR>1){$4="xxxxxx"}1' file
This will replace all entries in column 4 by "xxxxxx".
If you only want to change the first 6 to 7 characters of column 4 (and not if there are only 5 of them, there are a couple of ways:
$ awk 'BEGIN{FS=OFS=","}(FNR>1)&&(length($4)>5){$4="xxxxxx" substr($4,8)}1' file
$ awk 'BEGIN{FS=OFS=","}(FNR>1)&&{sub(/.......?/,"xxxxxx",$4)}1' file
Here, we will replace 123456abcde into xxxxxxabcde
Why is your script failing:
Besides the fact that the approach is wrong, I'll try to explain what the following command does: gsub(/([a-zA-Z0-9]){6,7}/g,"xxxxx",$4)
The notation /abc/g is valid awk syntax, but it does not do what you expect it to do. The notation /abc/ is an ERE-token (an extended regular expression). The notation g is, at this point, nothing more than an undefined variable which defaults to an empty string or zero, depending on its usage. awk will now try to execute the operation /abc/g by first executing /abc/ which means: if my current record ($0) matches the regular expression "abc", return 1 otherwise return 0. So it converts /abc/g into 0g which means to concatenate the content of g to the number 0. For this, it will convert the number 0 to a string "0" and concatenate it with the empty string g. In the end, your gsub command is equivalent to gsub("0","xxxxx",$4) and means to replace all the ZERO's by "xxxxx".
Why are you getting always gsub("0","xxxxx",$4) and never gsub("1","xxxxx",$4). The reason is that your initial regular expression never matches anything in the full record/line ($0). Your reguar expression reads /^([a-zA-Z0-9]){6,7}/, and while there are lines that start with 6 or 7 characters, it is likely that your awk does not recognize the extended regular expression notation '{m,n}' which makes it fail. If you use gnu awk, the output would be different when using -re-interval which in old versions of GNU awk is not enabled by default.
I tried to find why your code behave like that, for simplicty sake I made example concering only gsub you have used:
awk 'BEGIN{id="4090b43"}END{gsub(/^([a-zA-Z0-9]){6,7}/g, "xxxxx", id);print id}' emptyfile.txt
output is
4xxxxx9xxxxxb43
after removing g in first argument
awk 'BEGIN{id="4090b43"}END{gsub(/^([a-zA-Z0-9]){6,7}/, "xxxxx", id);print id}' emptyfile.txt
output is
xxxxx
So regular expression followed by g caused malfunction. I was unable to find relevant passage in GNU AWK manual what g after / is supposed to do.
(tested in gawk 4.2.1)
I have a file that looks like this
ON,111111,TEN000812,Super,7483747483,767,Free
ON,262762,BOB747474,SuperMan,4347374,676,Free
ON,454644,FRED84848,Super Man,65757,555,Free
I need to match the values in the fourth column exactly as they are written. So if I am searching for "Super" I need it to return the line with "Super" only.
ON,111111,TEN000812,Super,7483747483,767,Free
Likewise, if I'm looking for "Super Man" I need that exact line returned.
ON,454644,FRED84848,Super Man,65757,555,Free
I have tried using grep, but grep will match all instances that contain Super. So if I do this:
grep -i "Super" file.txt
It returns all lines, because they all contain "Super"
ON,111111,TEN000812,Super,7483747483,767,Free
ON,262762,BOB747474,SuperMan,4347374,676,Free
ON,454644,FRED84848,Super Man,65757,555,Free
I have also tired with awk, and I believe I'm close, but when I do:
awk '$4==Super' file.txt
I still get output like this:
ON,111111,TEN000812,Super,7483747483,767,Free
ON,262762,BOB747474,SuperMan,4347374,676,Free
I have been at this for hours, and any help would be greatly appreciated at this point.
You were close, or I should say very close just put field delimiter as comma in your solution and you are all set.
awk 'BEGIN{FS=","} $4=="Super"' Input_file
Also one more thing in OP's attempt while comparison with 4th field with string value, string should be wrapped in "
OR in case you want to mention value to be compared as an awk variable then try following.
awk -v value="Super" 'BEGIN{FS=","} $4==value' Input_file
You are quite close actually, you can try :
awk -F, '$4=="Super" {print}' file.txt
I find this form easier to grasp. Slightly longer than #RavinderSingh13 though
-F is the field separator, in this case comma
Next you have a condition followed by action
Condition is to check if the fourth field has the string Super
If the string is found, print it
I have a file with words (1 word per line). I need to censor all letters in the word, except the first five, with a *.
Ex.
Authority -> Autho****
I'm not very sure how to do this.
If you are lucky, all you need is
sed 's/./*/6g' file
When I originally posted this, I believed this to be reasonably portable; but as per #ghoti's comment, it is not.
Perl to the rescue:
perl -pe 'substr($_, 5) =~ s/./*/g' -- file
-p reads the input line by line and prints each line after processing
substr returns a substring of the given string starting at the given position.
s/./*/g replaces any character with an asterisk. The g means the substitution will happen as many times as possible, not just once, so all the characters will be replaced.
In some versions of sed, you can specify which substitution should happen by appending a number to the operation:
sed -e 's/./*/g6'
This will replace all (again, because of g) characters, starting from the 6th position.
Here's a portable solution for sed:
$ echo abcdefghi | sed -e 's/\(.\{5\}\)./\1*/;:x' -e 's/\*[a-z]/**/;t x'
abcde****
Here's how it works:
's/\(.\{5\}\)./\1*/' - preserve the first five characters, replacing the 6th with an asterisk.
':x' - set a "label", which we can branch back to later.
's/\*[a-z]/**/ - ' - substitute the letter following an asterisk with an asterisk.
't x' - if the last substitution succeeded, jump back to the label "x".
This works equally well in GNU and BSD sed.
Of course, adjust the regexes to suit.
Following awk may help you in same.
Solution 1st: awk solution with substr and gensub.
awk '{print substr($0,1,5) gensub(/./,"*","g",substr($0,6))}' Input_file
Solution 2nd:
awk 'NF{len=length($0);if(len>5){i=6;while(i<=len){val=val?val "*":"*";i++};print substr($0,1,5) val};val=i=""}' Input_file
Autho****
EDIT: Adding a non-one liner form of solution too now. Adding explanation with it too now.
awk '
NF{ ##Checking if a line is NON-empty.
len=length($0); ##Taking length of the current line into a variable called len here.
if(len>5){ ##Checking if length of current line is greater than 5 as per OP request. If yes then do following.
i=6; ##creating variable named i whose value is 6 here.
while(i<=len){ ##staring a while loop here which runs from value of variable named i value to till the length of current line.
val=val?val "*":"*"; ##creating variable named val here whose value will be concatenated to its own value, it will add * to its value each time.
i++ ##incrementing variable named i value with 1 each time.
};
print substr($0,1,5) val##printing value of substring from 1st letter to 5th letter and then printing value of variable val here too.
};
val=i="" ##Nullifying values of variable val and i here too.
}
' Input_file ##Mentioning Input_file name here.
Personally I'd just use sed for this (see #triplee's answer) but if you want to do it in awk it'd be:
$ awk '{t=substr($0,1,5); gsub(/./,"*"); print t substr($0,6)}' file
Autho****
or with GNU awk for gensub():
$ awk '{print substr($0,1,5) gensub(/./,"*","g",substr($0,6))}' file
Autho****
It is also possible and quite straightforward with sed:
sed 's/./\*/6;:loop;s/\*[^\*]/\**/;/\*[^\*]/b loop' file_to_censor.txt
output:
explanation:
s/./\*/6 #replace the 6th character of the chain by *
:loop #define an label for the goto
s/\*[^\*]/\**/ #replace * followed by non * char by **
/\*[^\*]/b loop #then loop until it does not exist a * followed by a non * char
Here is a pretty straightforward sed solution (that does not require GNUsed):
sed -e :a -e 's/^\(.....\**\)[^*]/\1*/;ta' filename
Could you let me know how to print "user.%" string in below text by awk?
The value of 'user' is not fixed and the number of strings in '( )' are not fixed.
start user1.table% NOT (%OLD, %2016%) user.% another strings
UPDATE
It is the basis of SQL processing. $2 means schema.table but here user can use '%' and also exclude by NOT keyword. It ends with ')'. The next one is a second schema.table and that is the one I want to catch.
I think I should parse the string after ')' with a regular expression but failed.
Regular expression:
[)]\s+(\S+)
Above expression can be used to catch that string I guess.
How can I apply this one in awk script(Not one liner).
If the structure of the query keeps the same, you can use this:
awk -F'[).]' '{print $3".%"}'
I'm using the closing parenthesis or the literal dot as the delimiter. Doing so the value of interest is in field 3.
While it is simple it leaves some whitespace in front of user. We can enhance the field delimiter regex to fix this:
awk -F')[[:space:]]*|[.]' '{print $3".%"}'
Btw, you may use this sed command alternatively:
sed 's/.*)[[:space:]]*\([^.]*\).*/\1.%/'
or if you have GNU grep, use this:
grep -oP '\)\s*\K[^%]*%'
Try this (GNU awk):
awk '{match($0, /[)] +([^ ]+)/, var);print var[1];}'
You need to match first (GNU awk function).
Given your posted sample input, all you need is:
awk '{print $6}'
e.g.:
$ echo 'start user1.table% NOT (%OLD, %2016%) user.% another strings' |
awk '{print $6}'
user.%
If that doesn't work for you then your posted sample input isn't representative enough of your real input so edit your question to include a few lines of truly representative sample input and the expected output given that input.