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.
Related
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 file that contains a lot of data like this and I have to delete everything that matches this regex [-]+\d+(.*)
Input:
zxczxc-6-9hw7w
qweqweqweqweqwe-18-8c5r6
asdasdasasdsad-11-br9ft
Output should be:
zxczxc
qweqweqweqweqwe
asdasdasasdsad
How can I do this with AWK?
sed might be easier...
$ sed -E 's/-+[0-9].*//' file
note that .* covers +.*
AFAIK awk doesn't support \d so you could use [0-9], your regex is correct only thing you need to put it in correct function of awk.
awk '{sub(/-+[0-9].*/,"")} 1' Input_file
You don't need the extra <plus> sign afther [0-9] as this is covered by the .*
Generally, if you want to delete a string that matches a regular expression, then all you need to do is substitute it with an empty string. The most straightforward solution is sed which is presented by karafka, the other solution is using awk as presented by RavinderSingh13.
The overall syntax would look like this:
sed -e 's/ere//g' file
awk '{gsub(/ere/,"")}1' file
with ere the regular expression representation. Note I use g and gsub here to substitute all non-overlapping strings.
Due to the nature of the regular expression in the OP, i.e. it ends with .*, the g can be dropped. It also allows us to write a different awk solution which works with field separators:
awk -F '-+[0-9]' '{print $1}' file
I have file with Nth columns
I want to remove the 5th column from last of Nth columns
Delimiter is "|"
I tested with simple example as shown below:
bash-3.2$ echo "1|2|3|4|5|6|7|8" | nawk -F\| '{print $(NF-4)}'
4
Expecting result:
1|2|3|5|6|7|8
How should I change my command to get the desired output?
If I understand you correctly, you want to use something like this:
sed -E 's/\|[^|]*((\|[^|]*){4})$/\1/'
This matches a pipe character \| followed by any number of non-pipe characters [^|]*, then captures 4 more of the same pattern ((\|[^|]*){4}). The $ at the end matches the end of the line. The first part of the match (i.e. the fifth field from the end) is dropped.
Testing it out:
$ sed -E 's/\|[^|]*((\|[^|]*){4})$/\1/' <<<"1|2|3|4|5|6|7"
1|2|4|5|6|7
You could achieve the same thing using GNU awk with gensub but I think that sed is the right tool for the job in this case.
If your version of sed doesn't support extended regex syntax with -E, you can modify it slightly:
sed 's/|[^|]*\(\(|[^|]*\)\{4\}\)$/\1/'
In basic mode, pipes are interpreted literally but parentheses for capture groups and curly brcneed to be escaped.
AWK is your friend :
Sample Input
A|B|C|D|E|F|G|H|I
A|B|C|D|E|F|G|H|I|A
A|B|C|D|E|F|G|H|I|F|E|D|O|R|Q|U|I
A|B|C|D|E|F|G|H|I|E|O|Q
A|B|C|D|E|F|G|H|I|X
A|B|C|D|E|F|G|H|I|J|K|L
Script
awk 'BEGIN{FS="|";OFS="|"}
{$(NF-5)="";sub(/\|\|/,"|");print}' file
Sample Output
A|B|C|E|F|G|H|I
A|B|C|D|F|G|H|I|A
A|B|C|D|E|F|G|H|I|F|E|O|R|Q|U|I
A|B|C|D|E|F|H|I|E|O|Q
A|B|C|D|F|G|H|I|X
A|B|C|D|E|F|H|I|J|K|L
What we did here
As you are aware awk's has special variables to store each field in the record, which ranges from $1,$2 upto $(NF)
To exclude the 5th from the last column is as simple as
Emptying the colume ie $(NF-5)=""
Removing from the record, the consecutive | formed by the above step ie do sub(/\|\|/,"|")
another alternative, using #sjsam's input file
$ rev file | cut -d'|' --complement -f6 | rev
A|B|C|E|F|G|H|I
A|B|C|D|F|G|H|I|A
A|B|C|D|E|F|G|H|I|F|E|O|R|Q|U|I
A|B|C|D|E|F|H|I|E|O|Q
A|B|C|D|F|G|H|I|X
A|B|C|D|E|F|H|I|J|K|L
not sure you want the 5'th from the last or 6th. But it's easy to adjust.
Thanks for the help and guidance.
Below is what I tested:
bash-3.2$ echo "1|2|3|4|5|6|7|8|9" | nawk 'BEGIN{FS="|";OFS="|"} {$(NF-4)="!";print}' | sed 's/|!//'
Output: 1|2|3|4|6|7|8|9
Further tested on the file that I have extracted from system and so it worked fine.
I want to remove the decimal places in strings from a list of identifiers:
ENSG00000166224.12
ENSG00000102897.5
ENSG00000168496.3
ENSG00000010295.15
ENSG00000147533.12
ENSG00000119242.4
My desired output will be
ENSG00000166224
ENSG00000102897
ENSG00000168496
ENSG00000010295
ENSG00000147533
ENSG00000119242
I would like to do it with awk, I have been playing with printf but with no success.
UPDATE:
The awk answer setting the field separator to . works well in files with only one column, but what if the file is composed of different columns (strings and float numbers)?
Here is an example:
ENSG00000166224.12 0.0730716237772557 -0.147970450702234
ENSG00000102897.5 0.156405616866614 -0.0398488625782745
ENSG00000168496.3 -0.110396121325736 -0.0147093758392248
How can I remove only the decimal places in the first field?
Thanks
You can set the field separator to the dot and print the first element:
$ awk -F. '{print $1}' file
ENSG00000166224
ENSG00000102897
ENSG00000168496
ENSG00000010295
ENSG00000147533
ENSG00000119242
In sed you would say sed 's/\.[^\.]*$//' file, which will catch everything from the last dot on and remove it.
You would be able to do it with printf if it just was a number. Then, you would use something to not print the decimal places. However, since it is an alphanumeric string it is best to handle it as a string.
Update
Use gsub to replace everything from . in the first field:
$ awk '{gsub(/\..*$/,"",$1)}1' a
ENSG00000166224 0.0730716237772557 -0.147970450702234
ENSG00000102897 0.156405616866614 -0.0398488625782745
ENSG00000168496 -0.110396121325736 -0.0147093758392248
use sub function also.
awk '{sub(/\..*/, "")}1' file
Using cut:
$ cut -d. -f1 file
ENSG00000166224
ENSG00000102897
ENSG00000168496
ENSG00000010295
ENSG00000147533
ENSG00000119242
If you are looking for a solution in perl
perl -pne 's/\..*$//' file.txt
This eventually remove everything after the decimal point.