[SOLVED] getting sub-string in a variable

tedy2808 · 07-15-2010, 04:55 AM

one more question..

i wonder if is it possible if the date can be set as : 11 Jul 2010

based on script below,

datetime=${msg##*ran on } # remove "ran on " and all proceeding
datetime=${datetime%% SST *} # remove " SST " and all following
time=${datetime##* } # remove last space and all proceeding
date=${datetime% *} # remove last space and all following

the output is : "Tue Jul 11"

is there any other easier way to make the date as 11 Jul 2010 instead of i need to do step by step and combine the all the variable for example : echo "$day" "$month" "$year" ??

in addition, can someone guide me on how to eliminate day in front of the month " Tue Jul 11" as the day is changing from monday-sunday..

thanks.

grail · 07-15-2010, 06:23 AM

This should get you going:

Code:

date -d "Tue Jul 11 23:15:00 2010" "+%d %b %Y"

The only thing removed here is SST from date and time portion of string

simon.sweetman · 07-15-2010, 07:34 PM

Well looks like it's pretty close now, what you want to do is get rid of the " is" on the end, and "ERROR: " on the font (right ?).

So replace:

Code:

name=${msg%% Failed *}

with

Code:

name=${msg%% is Failed *}
name=${name#ERROR: }

See, we are deleting from " is Failed " to the end and then removing "ERROR: " from the front.

To swap your date around use:

Code:

date=$(date -d "$date" "+%d %b %Y")

It's unfortunate that the variable is called "date" as it makes the above hard to read if your variable was renamed to logdate it would read better:

Code:

logdate=$(date -d "$logdate" "+%d %b %Y")

Using the "date -d" command grail suggested to reformat the string.

tedy2808 · 07-15-2010, 08:11 PM

sorry,my bad..

the input from test.log are:

if [success]: report_name successfully ran on Tue Jul 11 23:15:00 SST 2010.

if [failed]: ERROR: report_name is Failed on Tue Jul 11 23:15:00 SST 2010. ErrorCode : 500

the script should be like this:

#!/bin/sh
dir=/home/user/input/test.log

msg=`tail -1 $dir`
name=${msg%% Failed *}

status=Failed
if [ ${#name} -eq ${#msg} ]
then

name=${msg%% successfully *}
status=Success
fi

datetime=${msg##*ran on } # remove "ran on " and all proceeding
datetime=${datetime%% SST *} # remove " SST " and all following
time=${datetime##* } # remove last space and all proceeding
date=${datetime% *}

date=${datetime% *} # remove last space and all following
echo "$msg"
echo "$name"
echo "$status"
echo "$date"

the script output:

if success:

report_name successfully ran on Tue Jul 11 23:15:00 SST 2010.
report_name
Success
23:15:00
Tue Jul 11

if failed:

ERROR: report_name is Failed on Tue Jul 11 23:15:00 SST 2010. ErrorCode : 500

ERROR: report_name is
Failed
23:15:00
Tue Jul 11

BUT the output for name should be only "report_name" instead of "ERROR: report_name is" if failed and for date, i need to make it in dd/mm/yyyy format, thus i need to eliminate the "Tue" (in real situation, the day is changing from monday to sunday everyday)..How could i solve this problem?

grail · 07-15-2010, 09:43 PM

So how is this for a slightly different attack:

Code:

#!/bin/bash

log=/home/user/input/test.log

OLD_IFS=$IFS
IFS="|"

output($(sed -n -r -e '$ {/\[success\]/s/[^:]*: (.*) succ.*on (.*) SST ([0-9]+).*/\1|success|\2 \3/p;/ERROR/s/.*R: (.*) is.*on (.*) SST ([0-9]+).*/\1|failed|\2 \3/p}' $log))

IFS=$OLD_IFS

echo "Filename is ${output[0]}"
echo "Status is ${output[1]}"
eval echo 'Date is $(date -d "${output[2]}" "+%d %b %Y")'

tedy2808 · 07-15-2010, 10:10 PM

i edited the script and the name part works well...Really happy..Thanks!!!

But the date part still got error..

#!/bin/sh
dir=/home/input/test.log

msg=`tail -1 $dir`
type=Report
name=${msg%% is Failed *}
name=${name#ERROR: }

status=Failed
if [ ${#name} -eq ${#msg} ]
then

name=${msg%% successfully *}
status=Success
datetime=${msg##*ran on }

fi
datetime=${msg##* on }
datetime=${datetime%% SST *}
time=${datetime##* }

#date=${date -d "$date" "+%d %b %Y"}
#date=date -d "$datetime" "+%d %b %Y"

date=$(date -d "$date" "+%d %b %Y")
logdate=${date -d "$logdate" "+%d %b %Y"}

#date=${datetime% *}

#date=date -d "$datetime" "+%d %b %Y"

echo "$msg"
#echo "$datetime"
#echo "$name"
#echo "$type"
#echo "$status"
#echo "$time"
echo "$date"
echo "$logdate"

when i run the script, i got this message:

date: illegal option -- d
Usage: date [-u] [+format]
date [-u] [mmddhhmm[[cc]yy]]
date [-a [-]sss.fff]
./test2.sh[28]: logdate=${date -d "$logdate" "+%d %b %Y"}: The specified substitution is not valid for this command.

How could i solve this??

simon.sweetman · 07-15-2010, 11:05 PM

Looks like your date program dosn't support the -d option it's not POSIX so some flavours of unix dont have it. Here is an updated script that swaps the fields around but month will still be 3 letter name (not month number).

Code:

#!/bin/sh
dir=/home/input/test.log

msg=`tail -1 $dir`
status=Failed
name=${msg%% is Failed *}
name=${name#ERROR: }

if [ ${#name} -eq ${#msg} ]
then
    status=Success
    name=${msg%% successfully *}
fi
df=(${msg##* on })
logdate=${df[2]}-${df[1]}-${df[5]%.}

echo "$msg"
echo "$name"
echo "$status"
echo "$logdate"

tedy2808 · 07-15-2010, 11:24 PM

simon, i did try the script but i got the message

./simon.sh[19]: Syntax error at line 19 : `(' is not expected.

then i tried to edit the script;

df={$datetime} or df=${datetime}

but the logdate appear like above..

report_name successfully ran on Tue Jul 11 23:15:00 SST 2010.
report_name
Success
--

simon.sweetman · 07-16-2010, 12:01 AM

Change first line to:

Code:

#!/bin/bash

tedy2808 · 07-16-2010, 12:40 AM

thanks...it works great!!!

currently i'm working on another step and hoping that you're willing to guide me..
Really appreciate it!

tedy2808 · 07-16-2010, 02:11 AM

i'm trying to set all day to 2 decimal point so that if the input for date for example

if [success]: report_name successfully ran on Tue Jul _7 23:15:00 SST 2010.
if [success]: report_name successfully ran on Tue Jul 11 23:15:00 SST 2010.

the output for date is :

7 Jul 2010
11 Jul 2010

thus, i wanted the result to be:

07 Jul 2010
11 Jul 2010

so i did the script like this:

if [ ${df[2]} < "10" ]; then

logdate=`echo 0${df[2]}`\ ${df[1]}\ ${df[5]%.}

or

if [x={df[2]}; x <10 ]; then

logdate=echo"0"${df[2]}\ ${df[1]}\ ${df[5]%.}

or

df=(${msg##* on})
if [${df[2]}< 10]; then
$df[2]= 0${df[2]}
fi

but nothing changed..
someone please guide me on this..

* "_" to replace space

thanks

tedy2808 · 07-16-2010, 03:22 AM

Another problem,
Now i'm facing problem to assign AM and PM in time..

for example the input is

if [success]: report_name successfully ran on Tue Jul 11 00:00:01 SST 2010. (time = 00:00:01)
if [success]: report_name successfully ran on Tue Jul 11 03:15:00 SST 2010. (time = 03:15:00)
if [success]: report_name successfully ran on Tue Jul 11 18:15:00 SST 2010. (time = 18:15:00)
if [success]: report_name successfully ran on Tue Jul 11 24:15:00 SST 2010. (time = 23:15:00)

so, i want the output to be like this:

00:00:01 AM
03:15:00 AM
06:15:00 PM
11:15:00 PM

so,i'm thinking of breaking the time and assign the first 2 digit as hours so that i can compare the value,

hours=${time##* }
hours=${hours%% :*}

if [$hours <=12]; then
time=$time AM

else
time=$time -12 PM

fi

but it doesn't work..anyone pls guide me on how to solve this problem..
Really appreciate your help..i have no idea on this
thanks

grail · 07-16-2010, 04:44 AM

Well you would need to do more calculations than just the AM or PM because:

18:15:00 PM <- - this is incorrect

6:15:00 PM

If you then place a zero prior to the 6 that will also be incorrect as 24 hour format does not require / use the AM / PM setting.

simon.sweetman · 07-18-2010, 08:36 PM

Code:

hours=${time##* }
hours=${hours%%:*}

if [ $hours -le 12 ]
then
    time=${time:0} AM
else
    time="$((hours-12)):${time#*:} PM"
fi

Use -lt -le -gt -ge when doing numeric tests.
"[" and "]" must has space before and after.
AM/PM format should not have zero before hours.

For shell evaluation of equations (like hours - 12) put within $(( and ))

tedy2808 · 07-18-2010, 09:47 PM

I managed to solve the 2 decimal place for day but still having problem on assigning the AM and PM..

this is my latest script:

#!/bin/bash
dir=/home/devusr02/input/test.log

msg=`tail -1 $dir`

status=Failed
name=${msg%% is Failed *}
name=${name#ERROR: }

typeset -i dday

if [ ${#name} -eq ${#msg} ]
then

date=${msg##*on }
date=${date%%. *}

status=Success
name=${msg%% successfully *}
datetime=${msg##*ran on }

fi

datetime=${msg##* on }
datetime=${datetime%% SST *}
time=${datetime##* }

hours=${time}
hours=${hours%%:}
dftime=(${hours})

if [[ $dftime[1] -lt 12 ]]
then
new_time="${dftime[1]} $dftime[2] $dftime[3] AM "
else
new_time="expr ${dftime[1]-12} $dftime[2] $dftime[3] PM"
fi

date=${msg##*on }
date=${date%%. ErrorCode *}

df=(${msg##* on})
dday=${dfarray[2]}

if [[ $dday -lt 10 ]]
then
myday="0${df[2]}"
else
myday="${df[2]}"
fi

logdate="${myday} ${df[1]} ${df[5]%.}"

echo "$msg"
echo "$name"
echo "$status"
echo "$new_time"
echo "$logdate"

the output i got is:

./test2.sh: line 35: 23:15:00: syntax error in expression (error token is ":15:00")
./test2.sh: line 37: [[: [1]: syntax error: operand expected (error token is "[1]")
report_name successfully ran on Tue Jul 1 23:15:00 SST 2010.
report_name
Success
12 [2] [3] PM
03 Jul 2010

someone pls help me on this asap..really need your guidance