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for i in range ksh

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# 1  
Old 03-21-2011
for i in range ksh
Hi guys i have one handy script which is written in bash and it was using `seq ` , since we want to modified it to ksh script.

i have tried couple of tricks to work around
for i in $(x..y) like syntax it dont work .

below is the actual code can somebody help me on it

Code:
#!/bin/ksh
if [ $# != 4 ]
then
echo "Syntax  genawrs.sh dbid instanceId startsnapid endsnapid"
exit 1
fi
l_dbid=$1
l_instid=$2
l_start_snapid=$3
let l_end_snapid=$4-1
for i in "$(l_start_snapid..l_end_snapid)"     <=  not sure how to take out i out it 
do
let l_next_snapid=$i+1;
l_awr_log_file="/u01/app/awr/awrrpt_${2}_${i}_${l_next_snapid}.log"
sqlplus -s / as sysdba << EOC
set head off
set pages 0
set lines 132
set echo off
set feedback off
spool $l_awr_log_file
SELECT
output
FROM
TABLE
(dbms_workload_repository.awr_report_html
($l_dbid,$l_instid,$i,$l_next_snapid)
);
spool off
EOC
done

Moderator's Comments:
Mod Comment
Please use code tags when posting data and code samples!
Last edited by vgersh99; 03-21-2011 at 04:03 PM.. Reason: code tags, please!
# 2  
Old 03-21-2011
replace your for loop with (assuming your snapid are incremented by 1) otherwise you must consider another way to build up the list of the snap you want to get.
Code:
i=l_start_snapid
while [ $i -lt ${l_end_snapid} ]
do
      let l_next_snapid=$i+1
      l_awr_log_file="/u01/app/awr/awrrpt_${2}_${i}_${l_next_snapid}.log"
      ...
      ...
      ...
      let i+=1
done

Last edited by ctsgnb; 03-21-2011 at 04:52 PM..
# 3  
Old 03-21-2011
Thanks a lot for prompt reply

I changed the code according it.

Code:
l_start_snapid=$3
let l_end_snapid=$4-1
i=l_start_snapid
while [ $i -lt l_end_snapid ]
do
let l_next_snapid=$i+1;
l_awr_log_file="/u01/app/awr/awrrpt_${2}_${i}_${l_next_snapid}.log"
sqlplus -s / as sysdba << EOC
set head off
set pages 0
set lines 132
set echo off
set feedback off
spool $l_awr_log_file
SELECT
output
FROM
TABLE
(dbms_workload_repository.awr_report_text
($l_dbid,$l_instid,$i,$l_next_snapid)
);
spool off
EOC
let i+1
done

=======================================
but got it below error. i is not getting export. ???

Code:
(2529181125,1,l_start_snapid,7)
              *
ERROR at line 6:
ORA-00904: "L_START_SNAPID": invalid identifier

Last edited by vgersh99; 03-21-2011 at 05:10 PM.. Reason: once AGAIN - please use code tags!
# 4  
Old 03-22-2011
It looks like you forgot the dollar sign before the name of the variable
Code:
${l_start_snapid}

Code:
while [ $i -lt ${l_end_snapid} ]

Last edited by ctsgnb; 03-22-2011 at 10:28 AM..
# 5  
Old 03-22-2011
If your ksh is actually ksh93 (Korn shell 93), you can do ranges using {start..finish} syntax.

For example:
Code:
#!/bin/ksh93

for i in {11..15}; do
   echo $i
done

produces
Code:
11
12
13
14
15

Bash and zsh use the same syntax.
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