[SOLVED] Shell scripting error: head: invalid trailing option -- +

Andrew Stallard · 07-13-2019, 03:37 PM

I am trying to write a script to check if a directory exists in another directory.

Here it is:

Code:

#!/bin/bash
 
echo Enter name of directory
echo
read direct
a=`ls -l | wc -l`
for i in $a
do
k=$i+1
b=`ls -l | awk '{print $9}' | head -"$k" | tail -1`
c=`ls -l | awk '{print $1}' | head -"$k" | tail -1`
if [[ $b == $direct ]]
then
if [[ $c == d* ]]
then
echo $direct directory is found.
exit 0
fi
fi
done
echo $direct directory is not found

Now, I keep getting the following error when I run this:

Code:

head: invalid trailing option -- +
Try 'head --help' for more information.
head: invalid trailing option -- +
Try 'head --help' for more information.

This has me completely stumped. I used the exact same syntax for head and tail when running the commands outside the script and it ran just fine.

Code:

[andrew@mynewcentos ~]$ ls -l | awk '{print $9}' | head -"$k" | tail -1
Desktop
[andrew@mynewcentos ~]$ a=`ls -l | awk '{print $9}' | head -"$k" | tail -1`
[andrew@mynewcentos ~]$ echo $a
Desktop

Where I had previously set the variable k to 3. What the heck is happening here?

michaelk · 07-13-2019, 04:07 PM

See https://www.gnu.org/software/bash/ma...pressions.html

Code:

if [ -d "/path/to/$direct" ] 
then
    echo "$direct found." 
else
    echo "$direct not found."
fi

BW-userx · 07-13-2019, 07:59 PM

you can use substrings too

Code:

/home/bob/Pictures
substring=Pictures

if [[ $c =~ "$substring" ]]
then
echo $direct directory is found.
exit 0
fi

Code:

#!/bin/bash

set -xv

read -p "Please enter a (directory) name " directory

#locate $directory

#find / -type d -iname $directory

#a=`ls -l | wc -l`
#can be written like this too
a=$(ls l | wc -l)
for i in $a
do
	k=$((i+1))
	 
	b=$(ls -l | awk '{print $9}' | head -"$k" | tail -1)
	c=$(ls -l | awk '{print $1}' | head -"$k" | tail -1)
	if [[ $b == $directory ]]
	then
		if [[ $c == d* ]]
		then
			echo $directory directory is found.
			exit 0
		fi
	fi
done

your addition was wrong.

ehartman · 07-13-2019, 10:57 PM

Quote:

Originally Posted by Andrew Stallard

k=$i+1

What the heck is happening here?

When you would put an "echo $k" after the assignment, you would see what is happening, the assignment is done as a string operation, NOT as a numeric addition.
So "k" is NOT increased, but set to a string with "+1" appended.

Andrew Stallard · 07-14-2019, 12:01 AM

Quote:

Originally Posted by BW-userx

you can use substrings too

Code:

/home/bob/Pictures
substring=Pictures

if [[ $c =~ "$substring" ]]
then
echo $direct directory is found.
exit 0
fi

Code:

#!/bin/bash

set -xv

read -p "Please enter a (directory) name " directory

#locate $directory

#find / -type d -iname $directory

#a=`ls -l | wc -l`
#can be written like this too
a=$(ls l | wc -l)
for i in $a
do
	k=$((i+1))
	 
	b=$(ls -l | awk '{print $9}' | head -"$k" | tail -1)
	c=$(ls -l | awk '{print $1}' | head -"$k" | tail -1)
	if [[ $b == $directory ]]
	then
		if [[ $c == d* ]]
		then
			echo $directory directory is found.
			exit 0
		fi
	fi
done

your addition was wrong.

Thank You that worked!

MadeInGermany · 07-17-2019, 08:36 PM

You can directly test for a directory:

Code:

read -p "Please enter a (directory) name " directory
if [[ -d "$directory" ]]
then
  echo "$directory directory is found."
fi

ehartman · 07-18-2019, 02:06 AM

Quote:

Originally Posted by MadeInGermany

You can directly test for a directory:

Code:

read -p "Please enter a (directory) name " directory
if [[ -d "$directory" ]]
then
  echo "$directory directory is found."
fi

And you don't even need the double brackets for that, test also supports -d, so
if [ -d "$directory" ]
works as well.