The script is supposed to print "Enter the nest number to add: " and keep doing this until the user enters a negative number. At which point it is supposed to print the sum of the positive numbers. However as is the loop asks for the next number one time, it is entered, and then does not ask again, the script just stops doing anything and doesn't even reach the next line within the loop.
#!/bin/csh -x
#
# This script adds positive numbers entered by the user, stopping
# when a negative number is added
# Usage: +#, +#, +#... -#.
#
# x=0
# sum = 0
while($x>= 0)
echo -n "Enter the next number to be added: "
# sum = $sum + $<
# x = $<
end
#
exit 0
$< reads a line from stdin. This must be assigned to a variable, else if $< is used a second time the script will expect further input before continuing.
# x=0
# sum = 0
while ($x >= 0)
echo -n "Enter the next number to be added: "
# x = $<
if ($x >= 0) then
# sum = $sum + $x
endif
end
echo $sum
exit 0
Related
I'm new to Bash and I've been having issues with creating a script. What this script does is take numbers and add them to a total. However, I can't get total to work.It constantly claims that total is a non-variable despite it being assigned earlier in the program.
error message (8 is an example number being entered)
./adder: line 16: 0 = 0 + 8: attempted assignment to non-variable (error token is "= 0 + 8")
#!/bin/bash
clear
total=0
count=0
while [[ $choice != 0 ]]; do
echo Please enter a number or 0 to quit
read choice
if [[ $choice != 0 ]];
then
$(($total = $total + $choice))
$(($count = $count + 1))
echo Total is $total
echo
echo Total is derived from $count numbers
fi
done
exit 0
Get rid of some of the dollar signs in front of the variable names. They're optional inside of an arithmetic context, which is what ((...)) is. On the left-hand side of an assignment they're not just optional, they're forbidden, because = needs the variable name on the left rather than its value.
Also $((...)) should be plain ((...)) without the leading dollar sign. The dollar sign will capture the result of the expression and try to run it as a command. It'll try to run a command named 0 or 5 or whatever the computed value is.
You can write:
((total = $total + $choice))
((count = $count + 1))
or:
((total = total + choice))
((count = count + 1))
or even:
((total += choice))
((count += 1))
when I do
$ ls | wc -l
703
It gave me the result 703, I want to print 702 (703-1)
How can I do it in bash?
You can use arithmetic expansion:
result=$(( $(ls | wc - l) - 1))
or just ignore one of the files
result=$(ls | tail -n+2 | wc -l)
Note that it doesn't work if filenames contain the newline character; use ls -q to get one filename per line in such a case. This applies to the first solution, too, if you're interested in the number of files and not the number of lines in their names.
(Cheeky answer) Remove one line from the output before counting :D
ls | sed '1d' | wc -l
How to convert result as Integer in bash
#choroba has already answered this question and it should have solved OP's problem. However, I want to add more to his answer.
The OP's wants to convert the result into Integer but Bash doesn't have any data type like Integer.
Unlike many other programming languages, Bash does not segregate its variables by "type." Essentially, Bash variables are character strings, but, depending on context, Bash permits arithmetic operations and comparisons on variables. The determining factor is whether the value of a variable contains only digits.
See this for arithmetic operation in Bash.
See this for a best example to learn the untyped nature of Bash. I have posted the example below:
#!/bin/bash
# int-or-string.sh
a=2334 # Integer.
let "a += 1"
echo "a = $a " # a = 2335
echo # Integer, still.
b=${a/23/BB} # Substitute "BB" for "23".
# This transforms $b into a string.
echo "b = $b" # b = BB35
declare -i b # Declaring it an integer doesn't help.
echo "b = $b" # b = BB35
let "b += 1" # BB35 + 1
echo "b = $b" # b = 1
echo # Bash sets the "integer value" of a string to 0.
c=BB34
echo "c = $c" # c = BB34
d=${c/BB/23} # Substitute "23" for "BB".
# This makes $d an integer.
echo "d = $d" # d = 2334
let "d += 1" # 2334 + 1
echo "d = $d" # d = 2335
echo
# What about null variables?
e='' # ... Or e="" ... Or e=
echo "e = $e" # e =
let "e += 1" # Arithmetic operations allowed on a null variable?
echo "e = $e" # e = 1
echo # Null variable transformed into an integer.
# What about undeclared variables?
echo "f = $f" # f =
let "f += 1" # Arithmetic operations allowed?
echo "f = $f" # f = 1
echo # Undeclared variable transformed into an integer.
#
# However ...
let "f /= $undecl_var" # Divide by zero?
# let: f /= : syntax error: operand expected (error token is " ")
# Syntax error! Variable $undecl_var is not set to zero here!
#
# But still ...
let "f /= 0"
# let: f /= 0: division by 0 (error token is "0")
# Expected behavior.
# Bash (usually) sets the "integer value" of null to zero
#+ when performing an arithmetic operation.
# But, don't try this at home, folks!
# It's undocumented and probably non-portable behavior.
# Conclusion: Variables in Bash are untyped,
#+ with all attendant consequences.
exit $?
I have a shell script that runs a test repeatedly :
#!/bin/tcsh
set x = 1
while ($x <= 10000)
echo $x
./test
# x += 1
end
I am trying to adapt it to break the loop and stop running if the test failed, i.e. the test executable returned with a non-zero status. I thought the following change would work.
#!/bin/tcsh
set x = 1
set y = 0
while ($x <= 10000 && $y == 0)
echo $x
# y = ./test
# x += 1
end
But, I get error #: Expression syntax
Can you please tell me what did I do wrong, and how to capture the return value of ./test in a variable so I can break the loop, or some other way to break the loop upon encountering the test failure
I'm not a fan of scripting in csh, and I highly advise against it. However, in this case, csh seems to do the right thing, and you can simply do:
#!/bin/tcsh
set x = 1
while ($x <= 10000)
echo $x
./test || break
# x += 1
end
I have a bash program which extracts marks from a file that looks like this:
Jack ex1=5 ex2=3 quiz1=9 quiz2=10 exam=50
I want the code to execute such that when I input into terminal:
./program -ex1 -ex2 -ex3
Jack does not have an ex3 in his data, so an output of 0 will be returned:
Jack 5 3 0
how do I code my program to output 0 for each unrecognized argument?
If I understand what you are trying to do, it isn't that difficult. What you need to do is read each line into a name and the remainder into marks. (input is read from stdin)
Then for each argument given on the command line, check if the first part matches the beginning of any grade in marks (the left size of the = sign). If it does, then save the grade (right side of the = sign) and set the found flag to 1.
After checking all marks against the first argument, if the found flag is 1, output the grade, otherwise output 0. Repeat for all command line arguments. (and then for all students in file) Let me know if you have questions:
#!/bin/bash
declare -i found=0 # initialize variables
declare -i grade=0
while read -r name marks; do # read each line into name & marks
printf "%s" "$name" # print student name
for i in "$#"; do # for each command line argument
found=0 # reset found (flag) 0
for j in $marks; do # for each set of marks check for match
[ $i = -${j%=*} ] && { found=1; grade=${j#*=}; } # if match save grade
done
[ $found -eq 1 ] && printf " %d" $grade || printf " 0" # print grade or 0
done
printf "\n" # print newline
done
exit 0
Output
$ bash marks_check.sh -ex1 -ex2 -ex3 < dat/marks.txt
Jack 5 3 0
I want to extract lines before and after a matched pattern.
eg: if the file contents are as follows
absbasdakjkglksagjgj
sajlkgsgjlskjlasj
hello
lkgjkdsfjlkjsgklks
klgdsgklsdgkldskgdsg
I need find hello and display line before and after 'hello'
the output should be
sajlkgsgjlskjlasj
hello
lkgjkdsfjlkjsgklks
This is possible with GNU but i need a method that works in AIX / KSH SHELL WHERE NO GNU IS INSTALLED.
sed -n '/hello/{x;G;N;p;};h' filename
I've found it is generally less frustrating to build the GNU coreutils once, and benefit from many more features http://www.gnu.org/software/coreutils/
Since you'll have Perl on the machine, you could use the following code, but you'd probably do better to install the GNU utilities. This has options -b n1 for lines before and -f n1 for lines following the match. It works with PCRE matches (so if you want case-insensitive matching, add an i after the regex instead using a -i option. I haven't implemented -v or -l; I didn't need those.
#!/usr/bin/env perl
#
# #(#)$Id: sgrep.pl,v 1.7 2013/01/28 02:07:18 jleffler Exp $
#
# Perl-based SGREP (special grep) command
#
# Print lines around the line that matches (by default, 3 before and 3 after).
# By default, include file names if more than one file to search.
#
# Options:
# -b n1 Print n1 lines before match
# -f n2 Print n2 lines following match
# -n Print line numbers
# -h Do not print file names
# -H Do print file names
use warnings;
use strict;
use constant debug => 0;
use Getopt::Std;
my(%opts);
sub usage
{
print STDERR "Usage: $0 [-hnH] [-b n1] [-f n2] pattern [file ...]\n";
exit 1;
}
usage unless getopts('hnf:b:H', \%opts);
usage unless #ARGV >= 1;
if ($opts{h} && $opts{H})
{
print STDERR "$0: mutually exclusive options -h and -H specified\n";
exit 1;
}
my $op = shift;
print "# regex = $op\n" if debug;
# print file names if -h omitted and more than one argument
$opts{F} = (defined $opts{H} || (!defined $opts{h} and scalar #ARGV > 1)) ? 1 : 0;
$opts{n} = 0 unless defined $opts{n};
my $before = (defined $opts{b}) ? $opts{b} + 0 : 3;
my $after = (defined $opts{f}) ? $opts{f} + 0 : 3;
print "# before = $before; after = $after\n" if debug;
my #lines = (); # Accumulated lines
my $tail = 0; # Line number of last line in list
my $tbp_1 = 0; # First line to be printed
my $tbp_2 = 0; # Last line to be printed
# Print lines from #lines in the range $tbp_1 .. $tbp_2,
# leaving $leave lines in the array for future use.
sub print_leaving
{
my ($leave) = #_;
while (scalar(#lines) > $leave)
{
my $line = shift #lines;
my $curr = $tail - scalar(#lines);
if ($tbp_1 <= $curr && $curr <= $tbp_2)
{
print "$ARGV:" if $opts{F};
print "$curr:" if $opts{n};
print $line;
}
}
}
# General logic:
# Accumulate each line at end of #lines.
# ** If current line matches, record range that needs printing
# ** When the line array contains enough lines, pop line off front and,
# if it needs printing, print it.
# At end of file, empty line array, printing requisite accumulated lines.
while (<>)
{
# Add this line to the accumulated lines
push #lines, $_;
$tail = $.;
printf "# array: N = %d, last = $tail: %s", scalar(#lines), $_ if debug > 1;
if (m/$op/o)
{
# This line matches - set range to be printed
my $lo = $. - $before;
$tbp_1 = $lo if ($lo > $tbp_2);
$tbp_2 = $. + $after;
print "# $. MATCH: print range $tbp_1 .. $tbp_2\n" if debug;
}
# Print out any accumulated lines that need printing
# Leave $before lines in array.
print_leaving($before);
}
continue
{
if (eof)
{
# Print out any accumulated lines that need printing
print_leaving(0);
# Reset for next file
close ARGV;
$tbp_1 = 0;
$tbp_2 = 0;
$tail = 0;
#lines = ();
}
}
I had a situation where I was stuck with a slow telnet session on a tablet, believe it or not, and I couldn't write a Perl script very easily with that keyboard. I came up with this hacky maneuver that worked in a pinch for me with AIX's limited grep. This won't work well if your grep returns hundreds of lines, but if you just need one line and one or two above/below it, this could do it. First I ran this:
cat -n filename |grep criteria
By including the -n flag, I see the line number of the data I'm seeking, like this:
2543 my crucial data
Since cat gives the line number 2 spaces before and 1 space after, I could grep for the line number right before it like this:
cat -n filename |grep " 2542 "
I ran this a couple of times to give me lines 2542 and 2544 that bookended line 2543. Like I said, it's definitely fallable, like if you have reams of data that might have " 2542 " all over the place, but just to grab a couple of quick lines, it worked well.