capture the file name

Shell Programming and Scripting
1

I am trying to capture the file name(which is not of fixed length) and put it in a variable. It is working in unix whereas when I am running the same script in Informatica it is not giving me the desired output. But when I comment the option(find the file name) then it is working fine. It can also because I am using awk. Is there any other way(other than using awk) to find the file name? File name will be in second line of the file with 2 other items filesize numeric(10) and record count numeric(8).

Here is my code:
#!/bin/ksh

eval $(awk 'NR==2 {
printf "header_fname=\"%s\"\n", substr($0,1,length-18) ## filename
}' $1)

echo "$header_fname"

Sample file:
id_xyz20070523085554004
abcdefgh.csv000000324100000036

2 
header_fname=`awk 'NR==2 { print substr($0,1,length-18) }' $1`
echo $header_fname

or

#!/bin/ksh
{
  read line
  read line
} < $1
header_fname=${line%??????????????????}
echo $header_fname

Jean-Pierre.

3

In any BOURNE-type shell (bash, ksh, sh, etc.):

{
 read line
 IFS=0123456789 read header_fname junk
} < "$FILE"
echo "$header_fname"
4

Doesn't work if the file name contains any numeric character.

Jean-Pierre.

5

Thank you Aigles, it is working perfectly.