Inserting a row using shell

Shell Programming and Scripting
1

hi all,

In first run,I am inserting a row into an oracle table through shell script,
in the same shell script when I query this table the count comes as 0.

When I run the same script again then the count comes correctly,
Can anyone help how can I get the count of the row correctly in the same run when the row is inserted in table.

Thanks.

2

use

after inserting row..You will get correct count...

3

Show us what you have written

4 
     entry_data_control=`sqlplus  ${sysuser}/${syspwd} << EOSQL
                      whenever sqlerror exit sql.sqlcode;
       whenever oserror exit failure;
       INSERT INTO LSHADMIN.STUDY_LEVEL_TABLE(dataload_no,SESSION_SID,SESSION_SERIAL_NO,FILE_NAME,
                                              TIMESTAMP,session_status,STUDY_NAME,BLIND_BREAK,valid_file_naming)
                select  LSHADMIN.seq.nextval,sid,$serial, '$i' ,
                        to_char(to_date($timestamp,'YYYYMMDDHH24MISS'),
                  'YYYY/MM/DD::HH24:MI:SS'),status,'$substring1','$substring3',
                  '$valid_file'
                FROM $table 
                WHERE audsid = userenv('sessionid');
       COMMIT;
       EXIT;
       EOSQL`

this is insert stmt..
 
 

update_study_tab(){
                    echo "inside update_study_tab "
     status="$1"
     name="$2"
     echo "$status"
     echo "$name"
                   update=`nohup sqlplus  ${sysuser}/${syspwd} <<UpdateTab
               whenever sqlerror exit sql.sqlcode;
      whenever oserror exit failure;
      select count(*) from lshadmin.study_level_table where study_name = '$name';
      execute LSHADMIN.update_study_tab('${name}','${status}');
       commit;
      exit;
      UpdateTab`
       echo "update $update"
                 }

I am getting count as 0 in this update stmt.
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