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Logical if error

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# 1  
Old 07-04-2013
Logical if error
Hi All,

I am writing a simple script to read a file and display the lines with char count between 20 and 25. I am stuck with the if condition here. Tried a lot but still getting an error on the if condition
[: 14: unexpected operator

I am trying both the if formats below but none are working !
Code:
# if [ "$countchar" -ge "20" ] && [ "$countchar" -le "25" ]
if [ "$countchar" > "20" ] && [ "$countchar" < "25" ]

My script is very simple as below, not able to understand why its getting stuck Smilie

Code:
echo "Enter file name"
read FILENAME=$1
linecount=0
charcount=0
nu=0
linecount=`wc -l $FILENAME | cut -d " " -f7`
if [ $linecount -ge 0 ]; then
        while [ $nu -ne $linecount ]
                do
                nu=`expr $nu + 1`
                line=`sed -n ''$nu'p' $FILENAME`
                charcount=`echo $line | wc -c`
if [ [ "$countchar" > "20" ] && [ "$countchar" < "25" ] ]; then
                                        echo "Line with length between 20 and 25"
                                        echo $charcount $line
                        fi
                done
else
        echo "Error: no lines in the file"
        exit 1
fi


Regards,
# 2  
Old 07-04-2013
Quote:
Originally Posted by nss280
Code:
if [ [ "$countchar" > "20" ] && [ "$countchar" < "25" ] ]; then

Drop the outer pair of enclosing brackets.

For numeric comparison use -gt and -lt, not > and < which perform string comparison.

Regards,
Alister
# 3  
Old 07-04-2013
Hi Alister,

Thank you for the quick reply. I tried as you suggested but still its showing me the same error

Code:
$ vi prog11.sh
"prog11.sh" 22 lines, 646 characters
echo "Enter file name"
read FILENAME=$1
linecount=0
charcount=0
n=0
linecount=`wc -l $FILENAME | cut -d " " -f7`
if [ $linecount -ge 0 ]; then
        while [ $nu -ne $linecount ]
                do
                nu=`expr $nu + 1`
                line=`sed -n ''$nu'p' $FILENAME`
                charcount=`echo $line | wc -c`
if [ "$countchar" > "20" ] && [ "$countchar" < "25" ];
then
                                        echo "Line with length between 20 and 25"
                                        echo $charcount $line
                        fi
                done
else
        echo "Error: no lines in the file"
        exit 1
fi
~
~
~
~
~
~
~
~
~
~
~
~
~
~
"prog11.sh" 22 lines, 647 characters
$ sh prog11.sh
Enter file name
myfile
[: 14: unexpected operator
$

---------- Post updated at 12:57 AM ---------- Previous update was at 12:51 AM ----------

I am in bash.

Even tried without quotes, still not working Smilie

Code:
$ vi prog11.sh
"prog11.sh" 23 lines, 647 characters
echo "Enter file name"
read FILENAME=$1
linecount=0
charcount=0
n=0
linecount=`wc -l $FILENAME | cut -d " " -f7`
if [ $linecount -ge 0 ]; then
        while [ $nu -ne $linecount ]
                do
                nu=`expr $nu + 1`
                line=`sed -n ''$nu'p' $FILENAME`
                charcount=`echo $line | wc -c`
if [ $countchar > 20 ] && [ $countchar < 25 ]
then
                                        echo "Line with length between 20 and 25"
                                        echo $charcount $line
                        fi
                done
else
        echo "Error: no lines in the file"
        exit 1
fi

~
~
~
~
~
~
~
~
~
~
~
~
~
"prog11.sh" 23 lines, 639 characters
$ sh prog11.sh
Enter file name
myfile
[: 14: unexpected operator
$


Regards,
Nss280
# 4  
Old 07-04-2013
Where does "countchar"
Code:
if [ "$countchar" > "20" ] && [ "$countchar" < "25" ];

come from?

Should this be "charcount"?
# 5  
Old 07-04-2013
Hi Wisecracker, I must say that was a wonderful catch.

Code:
$ vi prog11.sh
"prog11.sh" 23 lines, 639 characters
echo "Enter file name"
read FILENAME=$1
linecount=0
charcount=0
n=0
linecount=`wc -l $FILENAME | cut -d " " -f7`
if [ $linecount -ge 0 ]; then
        while [ $nu -ne $linecount ]
                do
                nu=`expr $nu + 1`
                line=`sed -n ''$nu'p' $FILENAME`
                charcount=`echo $line | wc -c`
if [ "$charcount" > "20" ] && [ "$charcount" < "25" ]
then
                                        echo "Line with length between 20 and 25"
                                        echo $charcount $line
                        fi
                done
else
        echo "Error: no lines in the file"
        exit 1
fi

~
~
~
~
~
~
~
~
~
~
~
~
~
"prog11.sh" 23 lines, 647 characters
$ sh prog11.sh
Enter file name
myfile
[: 14: unexpected operator
$

Still the same issue Smilie

Regards,
Nss280

---------- Post updated at 01:38 AM ---------- Previous update was at 01:23 AM ----------

would redirecting to /dev/null >0 help here?

Regards,
nss280
# 6  
Old 07-04-2013
Also just noticed "n=0" should this be "nu=0"?

LBNL, make sure you have indented correctly, it does make it difficult to read...
This User Gave Thanks to wisecracker For This Post:
nss280
# 7  
Old 07-04-2013
The read statement should just use FILENAME.

Your script suffers from a pervasive lack of double quoted variables. Even the character counting is buggy because $line isn't quoted.

If after fixing those issues the script still misbehaves, enable tracing, sh -x prog11.sh, and post the output.

By the way, this is an extremely inefficient way of filtering lines. awk's length function would make this a concise and efficient one- liner.

Regards,
Alister
These 2 Users Gave Thanks to alister For This Post:
nss280 wisecracker
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