Bash Echo Date+string In The Same Line

I have the following code to extract two dates using awk, which are then read into two awk variables new and old respectively.
Each dates on the html file pulled with curl request is in this format:
2015-04-06 09:40:37
And two are being extracted
However the strings are being split on white space within the date strings. I tried changing OFS to ',', but it was still splitting incorrectly.
Code:

read dateStrNew dateStrOld < <(curl -k -q "$curl_call" | html2text | gawk '/Newest Sequence/ { new=$3" "$4 }/Oldest Sequence/ \
 {old=$3" "$4}END {OFS=","; print new,old }')  //new = date, old = date

Both parts of the date are being assigned to each variable using $3 and $4, then the space needs to be added back in so that the string can be used afterwards with a date command.

I just can't work out what is wrong, any help would be very much appreciated! Thanks!

Hi Forum members,

I'm interested to get the date on which particular command was ran.

I've below in my .bashrc

Code:

export HISTTIMEFORMAT="%d/%m/%y %T "

I ran history | grep "cp" | less

I get an output like:

Quote:
1288 27/05/15 09:12:55 sacct -j 767151 --format=elapsed,ncpus,state
27/05 is today's date. But I'm interested for the date, the above output was ran.

What am I missing?

Hi,

Below is the output of the "date" command:
Thu Apr 2 16:07:59 SGT 2015

After assigning the output to a parameter as follows:
DATE=`date`

The output has changed when "echo $DATE":
Thu Apr 2 16:09:40 SGT 2015
^^

The "2" has moved forward.

I don't want the "2" to move forward, please advise how to go about it.

Thanks in advance.

I want to get the time when program starts and finishes. But it always show the same time. Here is my shell script. But it always show that `START TIME` is same as `END TIME`.
Code:

host_list=("c15-0330-01.ad.mtu.edu" "c15-0330-02.ad.mtu.edu" "c15-0330-03.ad.mtu.edu" "c15-0330-04.ad.mtu.edu")
program=("L_1" "L_4" "L_3" "L_4")
subject="The job is finished"
START=$(date +"r")
address="/home/campus27/zwang10/Desktop/AWRR/program/power/vmodel_1/nprot/K_10"
ssh -f "${host_list[0]}" "cd '$address' && nohup Rscript '${program[0]}.R' > '${program[0]}_sh.txt';echo 'The job\n $address\n${program[0]} is finished\nSTART TIME = $START\n' END TIME =`date +"%r"` | mutt zwang10@mtu.edu -s '${host_list[0]} - Job ${program[0]}.R finished' -a '$address/${program[0]}_sh.txt';"

Hi All,

Im trying to create a script which checks the creation time of a file and if it is more than ten minutes, send out an alert.

For ex : Creation date of file is 10:30 AM
current time is 10:45 AM

Then send an alert/ message.

This is the script i wrote below :

Code:

#!/bin/ksh

filename="/apps/log/file.txt"

if [ -f "${filename}" ]
then
        createTime=`ls -lad "${filename}" | awk '{print $8}'`
        echo "$createTime"
        currentTime=`date '+%M'`
        echo "$currentTime"
        DIFF=$(( $currentTime - $createTime )) 
        echo "$DIFF"

else

        exit 1

fi

I am getting syntax error on the subraction when i try to run this script . I can understand that creation time and current time is in different format thats y this error throws, but i dont know how to rectify it .

I need to find out if the file creation time is more than ten minutes.

Please help me in achieving this output.

Hi!
I wrote a script shell about changing IP of the httpd.conf from text file and saving a new httpd.conf file for all IPs in the text file. The code is below, Im using mremote on windows (RHEL Release 5.7), and getting this error:
./script1.sh: line 19: syntax error: unexpected end of file

Please help me about that error. Thanks!

(ps: line 19 is the last line in the code => done < "$filename" )

#!/bin/bash

DATE='date +%Y%m%d_%H:%M:%S'
NAME=httpd_ip_change
EXTENSION=conf
max=10

filename="$1"
while read -r line

for ((i=2; i<=$max; ++i ))

do
touch $NAME$i.$EXTENSION.$DATE

sed "s/<VirtualHost 10.11.92.81:80>/$line/g" ./httpd.conf > $NAME$i.$EXTENSION.$DATE

done < "$filename"

I'm running LinuxLite 2.0 32bit on a Dell 3000.

I have never come across anything like this and to tell you, I am stumped.

Here are the contents of my crontab file:

Code:

# Edit this file to introduce tasks to be run by cron.
#
# Each task to run has to be defined through a single line
# indicating with different fields when the task will be run
# and what command to run for the task
#
# To define the time you can provide concrete values for
# minute (m), hour (h), day of month (dom), month (mon),
# and day of week (dow) or use '*' in these fields (for 'any').#
# Notice that tasks will be started based on the cron's system
# daemon's notion of time and timezones.
#
# Output of the crontab jobs (including errors) is sent through
# email to the user the crontab file belongs to (unless redirected).
#
# For example, you can run a backup of all your user accounts
# at 5 a.m every week with:
# 0 5 * * 1 tar -zcf /var/backups/home.tgz /home/
#
# For more information see the manual pages of crontab(5) and cron(8)
#
# m h  dom mon dow   command
*/10 * * * * /usr/bin/arecord -t wav -f cd -d 42 /home/randy/Music/lanting$(date "+\%^b\%d\%y").wav

If I run this from the terminal, everything is fine. It properly records the audio:

Code:

/usr/bin/arecord -t wav -f cd -d 42 /home/randy/Music/lanting$(date "+\%^b\%d\%y").wav

If I run this as it is shown in my crontab file, it records but there is no audio recorded.:


Code:

*/10 * * * * /usr/bin/arecord -t wav -f cd -d 42 /home/randy/Music/lanting$(date "+\%^b\%d\%y").wav

What could be causing this? I tried different cron settings for example 15 14 * * 2

This recorded at 2:15pm on Tuesday (today) but no audio. Yet if I run the code as mentioned above, from the terminal without the cron settings, the recording is fine.

Any ideas what I should do?

This is probably an easy fix, but it escapes me.

code:

D=date %Y-%m-%d
D1=`date +%s -d "$D"`
D2=`date +%s -d "2017-01-20"`
((diff_sec=D2-D1))
echo - | awk -v SECS=$diff_sec '{printf "Number of days : %d",SECS/(60*60*24)}'0
echo ""

runtime:

$ ./datediff.sh
./datediff.sh: line 1: fg: no job control
Number of days : 656
$

eek.. ./datediff.sh: line 1: fg: no job control

Code:

PREFIX=192.168.1
OCTET=1
while [ $OCTET -lt "255" ]; do
    echo -en "Pinging ${PREFIX}.${OCTET}..."
    ping -c1 -w1 ${PREFIX}.${OCTET} > /dev/null 2>&1
    if [ "$?" -eq "0" ]; then
       echo "OK"
    else
       echo "Failed"
    fi
    let OCTET=$OCTET+1
done

My question is, in the penultimate line, why is it OCTET, and not $OCTET, with a dollar before it, meaning, the variable?

What if it were written:
Code:

let $OCTET=$OCTET+1

Like in C++ (I think) or Python or whatever?

Hi all,

Been reading on stdin, stdout and stderr and encounter 2 questions, hope gurus here can advise.

0 = stdin
1 = stdout
2 = stderr

Code:

Sun Dec 21 03:53:42 SGT 2014 > cat test5.sh
#!/bin/bash

echo "Please enter value for name :"
read name
echo "Your name is $name."

echo "Next echo will be a syntax error"
ehco

Code:

Sun Dec 21 03:53:46 SGT 2014 > test5.sh 1> output.txt 2> error.txt
Noob

Sun Dec 21 03:54:56 SGT 2014 > cat output.txt
Please enter value for name :
Your name is Noob.
Next echo will be a syntax error

Sun Dec 21 03:55:23 SGT 2014 > cat error.txt
/home/alan/scripts/test5.sh: line 8: ehco: command not found
Sun Dec 21 03:55:26 SGT 2014 >

Which so far all is good and the correct way to output everything including error to a single file is

Code:

Sun Dec 21 03:59:14 SGT 2014 > test5.sh > output.txt 2>&1

Q1) How is a command being interpreted in linux , the sequence in which it is interpreted ? from left to right ? right to left ?

Shouldn't it be

Code:

test5.sh 2>&1 1>output.txt 
or 
test5.sh 2>&1>output.txt ?

Regards,
Noob