Shell Programming and Scripting

Hi Everyone,

I am facing a problem regarding the select from sybase, the return with the incorrect size.

For example, field is NAME(20).

After i selected from sybase, the result is nicky.
after i assign it to another declaration variable, it will be in actual name "nicky" , what i need the result is "nicky " with the space.

Anyone, who can help me? where to set it.. then i can get the return value with correct size which is including the space.

here is my sample of coding which

FIRST::::
typeSQL()
{

CONNECT="$SYBASE/XXX-22_8/bin/isql -S$HOSTNAME -U$USERNAME -P$PASSWORD -w400 -b -s|"

data_type_SQL=$($CONNECT cat <<-__EOF__
SET NOCOUNT ON
select LOC_TYPE, WAREHOUSE, LOCATION_ID from LOCATION where
LOCATION_ID="$loc_ID"
go
__EOF__)
}

size of each filed
LOC_TYPE(1)
WAREHOUSE(4)
LOCATION_ID(6)


After return value with incorrect size which are follow the size of each table name

data_type_SQL result is --> |G |NRT |NRTW |

SECOND::
loc_TYPE=$(echo $data_type_SQL | cut -d\| -f2| sed -e 's/|//g')
ware_HOUSE=$(echo $data_type_SQL | cut -d\| -f3| sed -e 's/|//g')
local_ID=$(echo $data_type_SQL | cut -d\| -f4| sed -e 's/|//g')

result after assigned with incorrect size

loc_TYPE="NRTW "
ware_HOUSE="G "
local_ID="NRT "


Can someone help me to check it? I just hope after assign to each field with correct size, including "SPACING"

thank you

I think you just need to put your variables in double quotes "$loc_type_SQL", etc.

you might want to do

str_out=$(echo $data_type_SQL | sed -e 's/^|//g' -e 's/|$//g' )
loc_TYPE=$(echo $str_out| cut -d\| -f1)
ware_HOUSE=$(echo $str_out| cut -d\| -f2)
local_ID=$(echo $str_out| cut -d\| -f3)


cheers,
Devaraj Takhellambam

Thank you devtakh and giannicello...

the sample you teach me.. still the same.
I think in the string for data_type_SQL after selected from database already wrong with SIZE.

The error according with the TABLE name size LOC_TYPE in 8,WAREHOUSE in 9 and LOCATION_ID in 11.

so, data_type_SQL = |G*******|NRT*****|NRTW*******| <-- it is wrong, coz it is follow with the table name of size.

suppose to be in its own size

LOC_TYPE(1)
WAREHOUSE(4)
LOCATION_ID(6)

the correct one should be
data_type_SQL = |G|NRT*|NRTW**|

Do you clear what i am explain?
SOS

Thank you...

ok then try this:

loc_TYPE=$(echo $data_type_SQL | awk -F "|" '{print substr($2,1,1)}')
ware_HOUSE=$(echo $data_type_SQL | awk -F "|" '{print substr($3,1,4)}')
local_ID=$(echo $data_type_SQL | awk -F "|" '{print substr($3,1,4)}')

you should also note that even if the size in the database is defined as 4 or 5 or 6, the actual data may be lesser than the defined size.

cheers,
Devaraj Takhellambam

Hi devtakh,

still the same... the spacing is not working...

thank you

what is it you are getting now:

a typo in the lasr post, it is

local_ID=$(echo $data_type_SQL | awk -F "|" '{print substr($4,1,6)}')

this variables will give you the output with the spaces...



cheers,
Devaraj Takhellambam