Echo without a newline character results a syntax error - bash

I am writing a shell script and I would like to have this code
echo $(awk '{print $1}' /proc/uptime) / 3600 | bc
without the newline character at the end.
I wanted to write it using echo -n, but this code
echo -n $(awk '{print $1}' /proc/uptime) / 3600 | bc
results a syntax error:
(standard_in) 1: syntax error
Can you help me with this?
Thank you very much!

echo $(awk '{print $1}' /proc/uptime) / 3600 | bc | tr -d "\n"
Alternatives:
echo -n $(($(cut -d . -f 1 /proc/uptime)/3600))
mapfile A </proc/uptime; echo -n $((${A%%.*}/3600))

A solution using echo -n:
echo -n $(echo $(awk '{print $1}' /proc/uptime) / 3600 | bc)
In general, if foo produces a line of output, you can print the same output without a newline using echo -n $(foo), even if foo is complicated.
A more straightforward solution using pure awk (since awk does arithmetic and output formatting, there's not much point in using both awk and bc):
awk '{printf("%d", $1 / 3600)}' /proc/uptime

Related

Get only numbers in output

I need to get only numbers from this:
release/M_0.1.0
thus, need to extract with bash to have in output this:
0.1.0.
I have tried this but cannot finish it:
echo "release/M_0.1.0" | awk -F'/' '{print $2}'
And what about if given such string? relea234se/sdf23_4Mm0.1.0.8. How to get only 0.1.0.8? Please pay attention that this can be random digits such as 0.2 or 1.9.1.
Please check if this grep command works
echo "release/M_0.1.0" | egrep -o '[0-9.]+'
You could also use general parameter expansion parsing to literally remove characters up through the last that isn't digits or dots.
$: ver() { echo "${1//*[^.0-9]/}"; }
$: ver release/M_0.1.0
0.1.0
$: ver relea234se/sdf23_4Mm0.1.0.8
0.1.0.8
With sed you can do:
echo "release/M_0.1.0" | sed 's#.*_##'
Output:
0.1.0
Considering that your Input_file will be same as shown samples.
echo "$var" | awk -F'_' '{print $2}'
OR could use sub:
echo "$var" | awk '{sub(/.*_/,"")} 1'
With simple bash you could use:
echo "${var#*_}"
echo release/M_0.1.0 | awk -F\_ '{print $2}'
0.1.0
Take your pick:
$ var='relea234se/sdf23_4Mm0.1.0.8'
$ [[ $var =~ .*[^0-9.](.*) ]] && echo "${BASH_REMATCH[1]}"
0.1.0.8
$ echo "$var" | sed 's/.*[^0-9.]//'
0.1.0.8
$ echo "$var" | awk -F'[^0-9.]' '{print $NF}'
0.1.0.8
if data in d file, tried on gnu sed:
sed -E 's/relea.*/.*([0-9][0-9.]*)$/\1/' d

Extract data between delimiters from a Shell Script variable

I have this shell script variable, var. It keeps 3 entries separated by new line. From this variable var, I want to extract 2, and 0.078688. Just these two numbers.
var="USER_ID=2
# 0.078688
Suhas"
These are the code I tried:
echo "$var" | grep -o -P '(?<=\=).*(?=\n)' # For extracting 2
echo "$var" | awk -v FS="(# |\n)" '{print $2}' # For extracting 0.078688
None of the above working. What is the problem here? How to fix this ?
Just use tr alone for retaining the numerical digits, the dot (.) and the white-space and remove everything else.
tr -cd '0-9. ' <<<"$var"
2 0.078688
From the man page, of tr for usage of -c, -d flags,
tr [OPTION]... SET1 [SET2]
-c, -C, --complement
use the complement of SET1
-d, --delete
delete characters in SET1, do not translate
To store it in variables,
IFS=' ' read -r var1 var2 < <(tr -cd '0-9. ' <<<"$var")
printf "%s\n" "$var1"
2
printf "%s\n" "$var2"
2
0.078688
Or in an array as
IFS=' ' read -ra numArray < <(tr -cd '0-9. ' <<<"$var")
printf "%s\n" "${numArray[#]}"
2
0.078688
Note:- The -cd flags in tr are POSIX compliant and will work on any systems that has tr installed.
echo "$var" |grep -oP 'USER_ID=\K.*'
2
echo "$var" |grep -oP '# \K.*'
0.078688
Your solution is near to perfect, you need to chance \n to $ which represent end of line.
echo "$var" |awk -F'# ' '/#/{print $2}'
0.078688
echo "$var" |awk -F'=' '/USER_ID/{print $2}'
2
You can do it with pure bash using a regex:
#!/bin/bash
var="USER_ID=2
# 0.078688
Suhas"
[[ ${var} =~ =([0-9]+).*#[[:space:]]([0-9\.]+) ]] && result1="${BASH_REMATCH[1]}" && result2="${BASH_REMATCH[2]}"
echo "${result1}"
echo "${result2}"
With awk:
First value:
echo "$var" | grep 'USER_ID' | awk -F "=" '{print $2}'
Second value:
echo "$var" | grep '#' | awk '{print $2}'
Assuming this is the format of data as your sample
# For extracting 2
echo "$var" | sed -e '/.*=/!d' -e 's///'
echo "$var" | awk -F '=' 'NR==1{ print $2}'
# For extracting 0.078688
echo "$var" | sed -e '/.*#[[:blank:]]*/!d' -e 's///'
echo "$var" | awk -F '#' 'NR==2{ print $2}'

error in awk of shell script

I am getting the below error ith my code.What is missing in it? My goal is to print 13.0.5.8 in $version
#!/bin/ksh
file="abc_def_APP_13.0.5.8"
if echo "$file" | grep -E "abc_def_APP"; then
echo "Version found: $file"
version1=(echo $file | awk -F_ '{print $NF}' | cut -d. -f1-3)
version2=(echo $file | awk -F_ '{print $NF}' | cut -d. -f4-)
echo $version1
echo $version2
version=$version$version2
echo $version
else
echo "Version not found"
fi
Please find below the error:
./version.sh: line 7: syntax error near unexpected token `|'
./version.sh: line 7: ` version1=(echo $file | awk -F_ '{print $NF}' | cut -d. -f1-3)'
./version.sh: line 9: syntax error near unexpected token `|'
./version.sh: line 9: ` version2=(echo $file | awk -F_ '{print $NF}' | cut -d. -f4-)'
./version.sh: line 18: syntax error near unexpected token `else'
There's no need for awk at all. Just trim every character before the last underscore, like so:
file="abc_def_APP_13.0.5.8"
version="${file##*_}"
echo "$version"
See http://mywiki.wooledge.org/BashFAQ/073 for documentation on this technique, or see "parameter expansion" in bash's own docs.
To treat the last segment separately is also straightforward:
file="abc_def_APP_13.0.5.8"
version="${file##*_}" # result: 13.0.5.8
version_end="${version##*.}" # result: 8
version_start="${version%.*}" # result: 13.0.5
echo "${version_start}/${version_end}" # result: 13.0.5/8
Because this happens internally to bash, without executing any external commands (such as awk), it should be considerably faster to execute than other approaches given.
The problem is your backticks are missing $ you need to fix the following two lines like so:
version1=$(echo $file | awk -F_ '{print $NF}' | cut -d. -f1-3)
version2=$(echo $file | awk -F_ '{print $NF}' | cut -d. -f4-)
This will fix the syntactical errors. The following line doesn't make much sense as $version hasn't been initialize yet:
version=$version$version2
Did you mean:
version="${version1}.${version2}"
A side note you are using the -E option with grep but you aren't using any extended regexp features, in fact you are doing a fixed string string search so -F is more appropriate. You probably also want to use the -q option to suppress the output from grep.
Personally I would do:
file="abc_def_APP_13.0.5.8"
echo "$file" | awk '/abc_def_APP/{print "Version found: "$0;
print $4,$5,$6;
print $7;
print $4,$5,$6,$7;
next}
{print "Version not found"}' FS='[_.]' OFS=.
If you just want the version number in the variable version then why not simply:
version=$(echo "$file" | grep -o '[0-9].*')
It can all be done in a single awk command and without additional cut command. Consider following command:
read version1 version2 < <(echo $file|awk -F "[_.]" '{
printf("%s.%s.%s ", $4, $5, $6); printf("%s", $7);
for (i=8; i<=NF; i++) printf(".%s", $i); print ""}')
echo "$version1 :: $version2"
OUTPUT
13.0.5 :: 8

No output when using awk inside bash script

My bash script is:
output=$(curl -s http://www.espncricinfo.com/england-v-south-africa-2012/engine/current/match/534225.html | sed -nr 's/.*<title>(.*?)<\/title>.*/\1/p')
score=echo"$output" | awk '{print $1}'
echo $score
The above script prints just a newline in my console whereas my required output is
$ curl -s http://www.espncricinfo.com/england-v-south-africa-2012/engine/current/match/534225.html | sed -nr 's/.*<title>(.*
?)<\/title>.*/\1/p' | awk '{print $1}'
SA
So, why am I not getting the output from my bash script whereas it works fine in terminal am I using echo"$output" in the wrong way.
#!/bin/bash
output=$(curl -s http://www.espncricinfo.com/england-v-south-africa-2012/engine/current/match/534225.html | sed -nr 's/.*<title>(.*?)<\/title>.*/\1/p')
score=$( echo "$output" | awk '{ print $1 }' )
echo "$score"
Score variable was probably empty, since your syntax was wrong.

Bash awk one-liner not printing

Expecting this to print out abc - but I get nothing, every time, nothing.
echo abc=xyz | g="$(awk -F "=" '{print $1}')" | echo $g
A pipeline isn't a set of separate assignments. However, you could rewrite your current code as follows:
result=$(
echo 'abc=xyz' | awk -F '=' '{print $1}'
)
echo "$result"
However, a more Bash-centric solution without intermediate assignments could take advantage of a here-string. For example:
awk -F '=' '{print $1}' <<< 'abc=xyz'
Other solutions are possible, too, but this should be enough to get you started in the right direction.

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