How to call a function with parameter in a bash script?

Beginner Bash scripting question? FASTEST CORRECT ANSWER GETS BEST ANSWER ***?

• This is my script, I am trying to write a script to take a PID as a parameter, and display the parent PID of it, and the parent PID of its parent PID, all the way until I get back to process 1. Basically, showing the lineage of any PID back to process 1. Please let me know what I'm doing wrong. #!/bin/bash if [ ! -d /proc/$1/ ]; then echo "That is not a valid PID" elif [ ! $# -eq 1 ]; then echo "I need one PID parameter" else echo $1 helper $1 fi function helper{ declare -i TEMPVAR=$(grep "PPid" /proc/$1/status | cut -c 7-) echo $TEMPVAR if [ ! $TEMPVAR -eq 1 ]; then helper $TEMPVAR fi } This is my first attempt at a function, and when I try to run the script I get line 9: helper: command not found line 13: syntax error near unexpected token 'declare' line 13: 'declare -i TEMPVAR=$(grep blah blah)' Please Help!!! First correct answer gets best answer.

• Answer:

  #!/bin/bash if [ ! -d /proc/$1/ ]; then echo "That is not a valid PID" exit fi if [ ! $# -eq 1 ]; then echo "I need one PID parameter" exit fi # specified pid: echo $1 # next pid up the tree: TEMPVAR=$(grep "PPid" /proc/$1/status | cut -c 7-) while [ ! "$TEMPVAR" = "1" ]; do TEMPVAR=$(grep "PPid" /proc/$TEMPVAR/status | cut -c 7-) echo $TEMPVAR done