Replace jq predefined variable with number - jq

jq could have predefined variables and replace with --arg options. It works well with string. However, I don't know how to replace the variable as number.
$jq -n --arg number 3000 '{"number":$number}'
{
"number": "3000"
}
I would like to be able to generate something as following :
{
"number": 3000
}
Thanks in advance.

Or:
jq -n --arg number 300 '{"number": $number|tonumber}'
or (as Jeff pointed out) with jq 1.5:
jq -n --argjson number 300 '{"number": $number}'
or (with versions of jq after June 26, 2015):
jq -n --arg number 300 '{$number} | .number |= tonumber'

Here is the solution:
jq -n --argfile number <(printf '%d' 3000) '{"number":$number}'
Or
jq -n --arg number 3000 '{"number":$number|fromjson}'

Related

Why sub() can not perform a string substitution?

I don't understand why the sub() command does not replace "%s" with "my string" in the first jq command below. How to make it work?
$ jq -r --arg format '|%s|' '$format | sub("%s"; .desc)' <<< '{"desc": "my string"}'
||
$ jq -r --arg format '|%s|' '$format | sub("%s"; "my string")' <<< '{"x": "y"}'
|my string|
$ jq -r .desc <<< '{"desc": "my string"}'
my string
You have lost the input context. Save it in a variable (eg. . as $dot) to reference it later (eg. $dot.desc):
$ jq -r --arg format '|%s|' '. as $dot | $format | sub("%s"; $dot.desc)' <<< '{"desc": "my string"}'
|my string|
You can use null input -n :
jq -nr --arg format '|%s|' '$format | sub("%s"; input.desc)' <<< '{"desc": "my string"}'

Problem with jq doing a select and contains

I have limited experience with jq and am having an issue doing a a select contains for a string in a boolean. This is my json and am looking to get back just tdonn.
[
"user",
"admin"
]
[
[
"tdonn",
true
]
]
Here is what im trying. I have tried many different ways too.
jq -e -r '.results[] | .series[] | select(.values[] | contains("tdon"))[]'
With the sample JSON shown in a comment, the following filter would produce the result shown:
.results[] | .series[][] | flatten[] | select(contains("tdon")?)
Output with -r option
tdonn
You might like to consider:
jq '.. | strings | select(contains("tdon"))'

Passing Multiple Objects to jq for Recursive Filter Operation

I am trying to use jq 1.5 to develop a script that can take one or more user inputs that represent a key and recursively remove them from JSON input.
The JSON I am referencing is here:
https://github.com/EmersonElectricCo/fsf/blob/master/docs/Test.json
My script, which seems to work pretty well, is as follows.
def post_recurse(f):
def r:
(f | select(. != null) | r), .;
r;
def post_recurse:
post_recurse(.[]?);
(post_recurse | objects) |= del(.META_BASIC_INFO)
However, I would like to replace META_BASIC_INFO with one or more user inputs. How would I go about accomplishing this? I presume with --arg from the command line, but I am unclear on how to incorporate this into my .jq script?
I've tried replacing del(.META_BASIC_INFO) with del(.$module) and invoking with cat test.json | ./jq -f fsf_key_filter.jq --arg module META_BASIC_INFO to test but this does not work.
Any guideance on this is greatly appreciated!
ANSWER:
Based on a couple of suggestions I was able to arrive to the following that works and users JQ.
Innvocation:
cat test.json | jq --argjson delete '["META_BASIC_INFO","SCAN_YARA"]' -f fsf_module_filter.jq
Code:
def post_recurse(f):
def r:
(f | select(. != null) | r), .;
r;
def post_recurse:
post_recurse(.[]?);
(post_recurse | objects) |= reduce $delete[] as $d (.; delpaths([[ $d ]]))
It seems the name module is a keyword in 1.5 so $module will result in a syntax error. You should use a different name. There are other builtins to do recursion for you, consider using them instead of churning out your own.
$ jq '(.. | objects | select(has($a))) |= del(.[$a])' --arg a "META_BASIC_INFO" Test.json
You could also use delpaths/1. For example:
$ jq -n '{"a":1, "b": 1} | delpaths([["a"]])'
{
"b": 1
}
That is, modifying your program so that the last line reads like this:
(post_recurse | objects) |= delpaths([[ $delete ]] )
you would invoke jq like so:
$ jq --arg delete "META_BASIC_INFO" -f delete.jq input.json
(One cannot use --arg module ... as "$module" has some kind of reserved status.)
Here's a "one-line" solution using walk/1:
jq --arg d "META_BASIC_INFO" 'walk(if type == "object" then del(.[$d]) else . end)' input.json
If walk/1 is not in your jq, here is its definition:
# Apply f to composite entities recursively, and to atoms
def walk(f):
. as $in
| if type == "object" then
reduce keys[] as $key
( {}; . + { ($key): ($in[$key] | walk(f)) } ) | f
elif type == "array" then map( walk(f) ) | f
else f
end;
If you want to recursively delete a bunch of key-value pairs, then here's one approach using --argjson:
rdelete.jq:
def rdelete(key):
walk(if type == "object" then del(.[key]) else . end);
reduce $strings[] as $s (.; rdelete($s))
Invocation:
$ jq --argjson strings '["a","b"]' -f rdelete.jq input.json

Is there a way to ignore header lines in a UNIX sort?

I have a fixed-width-field file which I'm trying to sort using the UNIX (Cygwin, in my case) sort utility.
The problem is there is a two-line header at the top of the file which is being sorted to the bottom of the file (as each header line begins with a colon).
Is there a way to tell sort either "pass the first two lines across unsorted" or to specify an ordering which sorts the colon lines to the top - the remaining lines are always start with a 6-digit numeric (which is actually the key I'm sorting on) if that helps.
Example:
:0:12345
:1:6:2:3:8:4:2
010005TSTDOG_FOOD01
500123TSTMY_RADAR00
222334NOTALINEOUT01
477821USASHUTTLES21
325611LVEANOTHERS00
should sort to:
:0:12345
:1:6:2:3:8:4:2
010005TSTDOG_FOOD01
222334NOTALINEOUT01
325611LVEANOTHERS00
477821USASHUTTLES21
500123TSTMY_RADAR00
(head -n 2 <file> && tail -n +3 <file> | sort) > newfile
The parentheses create a subshell, wrapping up the stdout so you can pipe it or redirect it as if it had come from a single command.
If you don't mind using awk, you can take advantage of awk's built-in pipe abilities
eg.
extract_data | awk 'NR<3{print $0;next}{print $0| "sort -r"}'
This prints the first two lines verbatim and pipes the rest through sort.
Note that this has the very specific advantage of being able to selectively sort parts
of a piped input. all the other methods suggested will only sort plain files which can be read multiple times. This works on anything.
In simple cases, sed can do the job elegantly:
your_script | (sed -u 1q; sort)
or equivalently,
cat your_data | (sed -u 1q; sort)
The key is in the 1q -- print first line (header) and quit (leaving the rest of the input to sort).
For the example given, 2q will do the trick.
The -u switch (unbuffered) is required for those seds (notably, GNU's) that would otherwise read the input in chunks, thereby consuming data that you want to go through sort instead.
Here is a version that works on piped data:
(read -r; printf "%s\n" "$REPLY"; sort)
If your header has multiple lines:
(for i in $(seq $HEADER_ROWS); do read -r; printf "%s\n" "$REPLY"; done; sort)
This solution is from here
You can use tail -n +3 <file> | sort ... (tail will output the file contents from the 3rd line).
head -2 <your_file> && nawk 'NR>2' <your_file> | sort
example:
> cat temp
10
8
1
2
3
4
5
> head -2 temp && nawk 'NR>2' temp | sort -r
10
8
5
4
3
2
1
It only takes 2 lines of code...
head -1 test.txt > a.tmp;
tail -n+2 test.txt | sort -n >> a.tmp;
For a numeric data, -n is required. For alpha sort, the -n is not required.
Example file:
$ cat test.txt
header
8
5
100
1
-1
Result:
$ cat a.tmp
header
-1
1
5
8
100
So here's a bash function where arguments are exactly like sort. Supporting files and pipes.
function skip_header_sort() {
if [[ $# -gt 0 ]] && [[ -f ${#: -1} ]]; then
local file=${#: -1}
set -- "${#:1:$(($#-1))}"
fi
awk -vsargs="$*" 'NR<2{print; next}{print | "sort "sargs}' $file
}
How it works. This line checks if there is at least one argument and if the last argument is a file.
if [[ $# -gt 0 ]] && [[ -f ${#: -1} ]]; then
This saves the file to separate argument. Since we're about to erase the last argument.
local file=${#: -1}
Here we remove the last argument. Since we don't want to pass it as a sort argument.
set -- "${#:1:$(($#-1))}"
Finally, we do the awk part, passing the arguments (minus the last argument if it was the file) to sort in awk. This was orignally suggested by Dave, and modified to take sort arguments. We rely on the fact that $file will be empty if we're piping, thus ignored.
awk -vsargs="$*" 'NR<2{print; next}{print | "sort "sargs}' $file
Example usage with a comma separated file.
$ cat /tmp/test
A,B,C
0,1,2
1,2,0
2,0,1
# SORT NUMERICALLY SECOND COLUMN
$ skip_header_sort -t, -nk2 /tmp/test
A,B,C
2,0,1
0,1,2
1,2,0
# SORT REVERSE NUMERICALLY THIRD COLUMN
$ cat /tmp/test | skip_header_sort -t, -nrk3
A,B,C
0,1,2
2,0,1
1,2,0
Here's a bash shell function derived from the other answers. It handles both files and pipes. First argument is the file name or '-' for stdin. Remaining arguments are passed to sort. A couple examples:
$ hsort myfile.txt
$ head -n 100 myfile.txt | hsort -
$ hsort myfile.txt -k 2,2 | head -n 20 | hsort - -r
The shell function:
hsort ()
{
if [ "$1" == "-h" ]; then
echo "Sort a file or standard input, treating the first line as a header.";
echo "The first argument is the file or '-' for standard input. Additional";
echo "arguments to sort follow the first argument, including other files.";
echo "File syntax : $ hsort file [sort-options] [file...]";
echo "STDIN syntax: $ hsort - [sort-options] [file...]";
return 0;
elif [ -f "$1" ]; then
local file=$1;
shift;
(head -n 1 $file && tail -n +2 $file | sort $*);
elif [ "$1" == "-" ]; then
shift;
(read -r; printf "%s\n" "$REPLY"; sort $*);
else
>&2 echo "Error. File not found: $1";
>&2 echo "Use either 'hsort <file> [sort-options]' or 'hsort - [sort-options]'";
return 1 ;
fi
}
This is the same as Ian Sherbin answer but my implementation is :-
cut -d'|' -f3,4,7 $arg1 | uniq > filetmp.tc
head -1 filetmp.tc > file.tc;
tail -n+2 filetmp.tc | sort -t"|" -k2,2 >> file.tc;
Another simple variation on all the others, reading a file once
HEADER_LINES=2
(head -n $HEADER_LINES; sort) < data-file.dat
With Python:
import sys
HEADER_ROWS=2
for _ in range(HEADER_ROWS):
sys.stdout.write(next(sys.stdin))
for row in sorted(sys.stdin):
sys.stdout.write(row)
cat file_name.txt | sed 1d | sort
This will do what you want.

counting records in unix file

This was an interview question, nevertheless still a programming question.
I have a unix file with two columns name and score. I need to display count of all the scores.
like
jhon 100
dan 200
rob 100
mike 100
the output should be
100 3
200 1
You only need to use built in unix utility to solve it, so i am assuming using shell scripts . or reg ex. or unix commands
I understand looping would be one way to do. store all the values u have already seen and then grep every record for unseen values. any other efficient way of doing it
Try this:
cut -d ' ' -f 2 < /tmp/foo | sort -n | uniq -c \
| (while read n v ; do printf "%s %s\n" "$v" "$n" ; done)
The initial cut could be replaced with another while read loop, which would be more resilient to input file format variations (extra whitespace). If some of the names consist in several words, simple field extraction will not work as easily, but sed can do it.
Otherwise, use your favorite programming language. Perl would probably shine. It is not difficult either in Java or even in C or Forth.
$ cat foo.txt
jhon 100
dan 200
rob 100
mike 100
$ awk '{print $2}' foo.txt | sort | uniq -c
3 100
1 200
Its a pity you can't do a count with sort or uniq alone.
Edit: I just noticed I have the count in front ... to get it exactly the same you can do:
$ awk '{print $2}' foo.txt | sort | uniq -c | awk '{ print $2 " " $1 }'
Not very complicated in perl:
#!/usr/bin/perl -w
use strict;
use warnings;
my %count = ();
while (<>) {
chomp;
my ($name, $score) = split(/ /);
$count{$score}++;
}
foreach my $key (sort keys %count) {
print "$key ", $count{$key}, "\n";
}
You could go with awk:
awk '/.*/ { a[$2] = a[$2] + 1; } END { for (x in a) { print x, " ", a[x] } }' record_file.txt
Alternatively with shell commands:
for i in `awk '{print $2}' inputfile | sort -u`
do
echo -n "$i "
grep $i inputfile | wc -l
done
The first awk command will give a list of all the different scores (e.g. 100 and 200) which then
the for loop iterates over, counting up each separately. Not very super efficient, but simple. If the file is not to big is should not be a too big problem.

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