Mistake in while loop? bash script

Question

My Code:

#!/bin/bash
rm screenlog.0
screen -X stuff 'X21'$(printf \\r)
while :
do
grep -i "T" $screenlog.0
if [ $? -eq 0 ];
then
FILE=/etc/passwd 
VAR=`head -n 1 $FILE` 
echo $VAR
rm screenlog.0
break
done

This script is to delete the file "screenlog.0" send a command (X21) to an screen interface. Thats the first part and it works.

The second Part is the Problem: That should test the content of "screenlog.0", is there an something with a "T" inside save the contant into a variable.

The error:

line 11: syntax error near unexpected token `done'
line 11: `done'

To the "screen": Its an screen of an usb device that recive radio messages like this:

T350B00A66E2
H34D04DE4254

The script have to scan for the incomming messages with "T" at the beginning (The first letter is a Type field behind this a hex code.

Some ideas to correct or other solutions?

I corrected my code a bit:

#!/bin/bash
>screenlog.0
screen -X stuff 'X21'$(printf \\r)
while :
do
sleep 2
grep -i "T" $screenlog.0
if [ $? -eq 0 ];
then
screenlog.0=/etc/passwd
VAR=`head -n 1 $screenlog.0`
echo $VAR
break
fi
done

The new error is:

grep: .0: No such file or directory

All 5 seconds....

The file screenlog.0 exist .. :(

Note that you can test the exit status of the grep command directly: if grep -i "T" screenlog.0. You also probably don't want the $ in front of $screenlog.0 given that the script otherwise manipulates screenlog.0 with no $. It's also better to use the $(…) notation rather than use the back-ticks notation. — Jonathan Leffler
– Jonathan Leffler, Commented Mar 29, 2014 at 12:20

Jayesh Bhoi · Accepted Answer · 2014-03-29 12:16:54Z

4

oh...you missed fi in your script :). Like syntax as follows if [ condition ];then #dosomething fi

For your script

if [ $? -eq 0 ];then
    FILE=/etc/passwd  
    VAR=`head -n 1 $FILE` 
    echo $VAR
    rm screenlog.0
    break
fi

edited Mar 29, 2014 at 12:16

answered Mar 29, 2014 at 10:28

Jayesh Bhoi

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1 Comment

Jayesh Bhoi Over a year ago

@DemonDragon change from $screenlog.0 to "screenlog.0".Also note you removed on startup rm screenlog.0. remove that line.

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