How to compare the current date with past 60 days using shell script - shell

start_time=`sed -e 's/\([0-9][0-9][0-9][0-9]-[0-9][0-9]-[0-9][0-9] [0-9][0-9]:[0-9][0-9]:[0-9][0-9]\).*/\1/' <<< "$line"`
start_time_sec=`date -d "$start_time" +%s`
now=`date +%s`
pass_time=`$now - $start_time_sec`
if [ $pass_time <=86400*60 ]
then
initial_time= $start_time
initial_time_sec=`date -d "$initial_time" +%s`
break
fi
/Here I have tried with date comparison with seconds, But I want in terms of days/

ISO dates (YYYY-MM-DD) can be compared like strings:
$ date +%Y-%m-%d
2014-01-07
$ date +%Y-%m-%d -d '-60 days'
2013-11-08
$ [[ "$(date +%Y-%m-%d -d '-60 days')" < "$(date +%Y-%m-%d)" ]]
$ echo $?
0

start_time=`echo $line | sed -e 's/^\([0-9][0-9][0-9][0-9]-[0-9][0-9]-[0-9][0-9] [0-9][0-9]:[0-9][0-9]:[0-9][0-9]\).*/\1/'`
start_time_sec=`date -d "$start_time" +%s`
now=`date +%s`
pass_time=`expr $now - $start_time_sec`
limit_time=$((60 * 60 * 24 * 60))
if [[ $pass_time -le $limit_time ]]; then
echo "in 60 days"
fi
update:
or your idea:
start_time_date=`date -d "$start_time" +%s`
past_date=`date +"%Y-%m-%d" -d "-60 day"`
if [[ $past_date -le $start_time_date ]]; then
echo 'in 60 days'
fi

Related

How to get second sunday in a Month given a date parameter in bash script

I am trying to write a bash script, to merge 24 files in a given day. The requirement changes during Day light saving time changes, where I get 23 or 25 files.
So, with further research I realized that day-light savings begins on the second Sunday of March(23) of every year and ends on first sunday of Novemeber(25).
I need more inputs to get second sunday in a given month to do the check of finding 23 or 25 files for March and November respectively.
Any inputs to help me with this will be really appreciated.
Thank you
Here is the sample code to find 24 files in a day-
if [ -z "$1" ];then
now=$(date -d "-1 days" +%Y-%m-%d);
else now=$1;
fi
load_date='load_date='$now
singlePath="$newPath/$load_date"
fileCount=$(hdfs dfs -ls -R $hdfsPath/$load_date/ | grep -E '^-' | wc -l)
path=$hdfsPath/$load_date
if [ $fileCount -eq 24 ]; then
echo "All files are available for "$load_date;
hadoop fs -cat $path/* | hadoop fs -put - $singlePath/messages.txt
else echo $fileCount" files are available for "$load_date"! Please note, few files are being missed";
fi
I wouldn't hardcode the dates of DST transistions. I would just count "how many hours did today have":
a "normal" day:
$ diff=$(( $(date -d now +%s) - $(date -d yesterday +%s) ))
$ echo $(( diff / 3600 ))
24
"spring forward"
$ diff=$(( $(date -d "2019-03-10 23:59:59" +%s) - $(date -d "2019-03-09 23:59:59" +%s) ))
$ echo $(( diff / 3600 ))
23
"fall back"
$ diff=$(( $(date -d "2019-11-03 23:59:59" +%s) - $(date -d "2019-11-02 23:59:59" +%s) ))
$ echo $(( diff / 3600 ))
25
One thing to note: since bash only does integer arithmetic, if the difference is not 86400 but 86399, you get:
$ echo $((86399 / 3600))
23
So, better to query yesterday's time first in the tiny-but-non-zero chance that the seconds tick over between the 2 date calls:
diff=$(( -$(date -d yesterday +%s) + $(date -d now +%s) ))
Here, $diff will be 86400 or 86401 (for non DST transition days), and dividing by 3600 will give 24 not 23.

How to subtract today's date with a file's modification date in unix?

For example:
echo $(date) - $(date -r sample.txt)
Output:
90 days(for example)
Use %s seconds since 1970-01-01 00:00:00 UTC as in
echo $(expr $(date +%s) - $(date -r sample.txt +%s)) #!/bin/sh
echo $(($(date +%s) - $(date -r sample.txt +%s))) #/bin/bash
One more way
$ ls -l peter.txt
-rwxrw-r--+ 1 pppp qqqq 149 Dec 15 18:39 peter.txt
$ echo "(" $(date +%s) - $(date -r peter.txt +%s) ")/" 86400 | perl -nle ' print eval, " days" '
29.254537037037 days
$

Error on backup bash script: syntax error near unexpected token `newline'

I am having problem finding error on bash script for handling backup:daily, monthly, yearly. Here is the script:
#!/bin/bash
echo > /home/alpha/folder/keep.txt
#writing dates of the backups that should be kept to the array
for i in {0..7}; do ((keep[$(date +%Y%m%d -d "-$i day")]++)); done
for i in {0..4}; do ((keep[$(date +%Y%m%d -d "sunday-$((i+1)) week")]++)); done
for i in {0..12}; do
DW=$(($(date +%-W)-$(date -d $(date -d "$(date +%Y-%m-15) -$i month" +%Y-%m-01) +%-W)))
for (( AY=$(date -d "$(date +%Y-%m-15) -$i month" +%Y); AY < $(date +%Y); AY++ )); do
((DW+=$(date -d $AY-12-31 +%W)))
done
((keep[$(date +%Y%m%d -d "sunday-$DW weeks")]++))
done
for i in {0..30}; do
DW=$(date +%-W)
for (( AY=$(($(date +%Y)-i)); AY < $(date +%Y); AY++ )); do
((DW+=$(date -d $AY-12-31 +%W)))
done
((keep[$(date +%Y%m%d -d "sunday-$DW weeks")]++))
done
#writing the array to file keep.txt line by line
for i in ${!keep[#]}; do echo $i >> /home/alpha/folder/keep.txt; done
#delete all files that not mentioned in keep.txt
cd /home/alpha/folder
ls -1 /home/alpha/folder/ | sort /home/alpha/folder/keep.txt /home/alpha/folder/keep.txt - | uniq -u | xargs rm -rf
rm /home/alpha/folder/keep.txt
When I try to run the script, throws error message:
./back.sh: line 12: syntax error near unexpected token `newline' ./back.sh: line 12: ` done'
Where did I do wrong on the script?
Your date expression seems to misbehave inside the arithmetic context. Adding temporary variables solved your issue for me :
#!/bin/bash
echo > /home/alpha/folder/keep.txt
#writing dates of the backups that should be kept to the array
for i in {0..7}; do ((keep[$(date +%Y%m%d -d "-$i day")]++)); done
for i in {0..4}; do ((keep[$(date +%Y%m%d -d "sunday-$((i+1)) week")]++)); done
for i in {0..12}; do
DW=$(($(date +%-W)-$(date -d $(date -d "$(date +%Y-%m-15) -$i month" +%Y-%m-01) +%-W)))
begin=$(date -d "$(date +%Y-%m-15) -$i month" +%Y)
for (( AY=begin; AY < $(date +%Y); AY++ )); do
((DW+=$(date -d $AY-12-31 +%W)))
done
((keep[$(date +%Y%m%d -d "sunday-$DW weeks")]++))
done
for i in {0..30}; do
DW=$(date +%-W)
begin=$(($(date +%Y)-i))
for (( AY=begin; AY < $(date +%Y); AY++ )); do
((DW+=$(date -d $AY-12-31 +%W)))
done
((keep[$(date +%Y%m%d -d "sunday-$DW weeks")]++))
done
#writing the array to file keep.txt line by line
for i in ${!keep[#]}; do echo $i >> /home/alpha/folder/keep.txt; done
#delete all files that not mentioned in keep.txt
cd /home/alpha/folder
ls -1 /home/alpha/folder/ | sort /home/alpha/folder/keep.txt /home/alpha/folder/keep.txt - | uniq -u | xargs rm -rf
rm /home/alpha/folder/keep.txt
However, I am unsure why the expression misbehaves inside the arithmetic block.

How to use DATE command with an external variable?

I'm working on a time convertion script. It is supposed to do something like this:
echo $(( ($(date -d '00:10:2.00' +%s) - $(date -d 0 +%s) ) ))
This line is working just fine giving me the result
602
But I want to put the the first part of the date string (00:10:2.00) under a) an command line argument so it could be read from $1 like:
echo $(( ($(date -d '$1' +%s) - $(date -d 0 +%s) ) ))
OR as a variable:
echo $(( ($(date -d '$myvariable' +%s) - $(date -d 0 +%s) ) ))
When I'm trying this:
foo="00:10:2.00"
echo $foo
echo $(( ($(date -d '$foo' +%s) - $(date -d 0 +%s) ) ))
All I get is:
00:10:2.00
date: invalid date `$foo'
-1417042800
So it's echoing properly but it aint working with the time command...
Variables are expanded inside double quotes, they're not expanded inside single quotes. So use
date -d "$1" +%s

How to read a file content using shell script

cat file.txt
(
while read line
do
initial_time=`head -n 1 $line| sed -e 's/\([0-9][0-9][0-9][0-9]-[0-9][0-9]-[0-9][0-9] [0-9][0-9]:[0-9][0-9]:[0-9][0-9]\).*/\1/'`
initial_time_sec=`date -d "$initial_time" +%s`
done
)
I want to take the date from every line , if it is less than 60 days I have to store it in variable
while read line ; do
initial_time=`echo $line | sed -e 's/.*\([0-9][0-9][0-9][0-9]-[0-9][0-9]-[0-9][0-9] [0-9][0-9]:[0-9][0-9]:[0-9][0-9]\).*/\1/'`
initial_time_sec=`date -d "$initial_time" +%s`
now=`date +%s`
pass_time=`expr $now - $initial_time_sec`
limit_time=$((60 * 24 * 60))
if [[ $pass_time -lt $limit_time ]]; then
echo $line
fi
done <data

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