help! please simplify this shell program!

Pacific Yan

Now my program will grow bigger and bigger, but I still missing how to take 2 arguments,


#!/usr/bin/ksh
     
fflag=
sflag=
while getopts hf:s: name
do
            case $name in
            f)  fflag=1
                fval="$OPTARG";;
            s)  sflag=1
                sval="$OPTARG";;
            h)  printf "Usage: p2\n p2 year\n p2 -f firstyear -s second year\n"
                exit 2;;        
            ?)  printf "Usage: %s: [-f] [-s value] args\n" $0
                exit 2;;
            esac
done


# if sflag is missing, take 1st year and proceed to main computation block
if [   ! -z "$fflag"  -a  -z "$sflag"    ]; then
            year=$fval
           if ((year<1)) || ((year>9999))
           then
             echo "year must be 1 to 9999"
             exit 1
           fi            

else
{
da=0
for st in Sunday Monday Tuesday Wednesday Thursday Friday Saturday;do
        eval str_$da=$st
        da=$(($da + 1))
done

mon=1
while [ $mon -le 12 ];do
        da=`cal $mon $1 |awk 'NR==3{loc=index($0,"1";print (loc-2)/3;exit}'`
        eval "count_$da=\$((\$count_$da+1))"
        mon=$(($mon + 1 ))
done

da=0
while [ $da -le 6 ];do
        eval dcn=\$count_$da
        if [ $dcn -eq 1 ];then
                eval "print \"\$str_$da \""|awk 'BEGIN {OFS=" "; ORS=" "} {print $1}
'
fi
        da=$(($da + 1))
done
echo
}
fi


# if fflag is missing,
if [    -z "$fflag"  -a  ! -z "$sflag"    ]; then
           PS3="choose:
           1 quit
           2 change the beginning of the range
           3 change end of the range
           :"
           select indx in "quit" "change beginning" "change end"
             do
               case $indx in
                 "quit" break;;
                 "change beginning"
                                   print -n "Enter first year "
                                   read firstyear
                                   year1=$firstyear
                                   print -n "Enter second year "
                                   read secondyear
                                   year2=$secondyear                                 
                                   break
                                   ;;     
                 "change end"  
                                   print -n "Enter first year "
                                   read firstyear
                                   year1=$firstyear
                                   print -n "Enter second year "
                                   read secondyear
                                   year2=$secondyear                                 
                                   break
                                   ;;                                
                 *) echo  default;;
                esac               
            done
fi

夜未眠

不用那么复杂的.写脚本我觉得实用为主,写的速度越快越好.加点注释就行了，简单功能不用考虑什么usage(),下面是另一个script [ hh.sh ]，它会调用上次的test.sh。另外写只是为了结构更清楚。你可以自己试着把两个脚本合在一起成为一个.good luck!

1. 
   
2. #!/bin/sh
   
3. #shell script [ hh.sh ]
   
4. if [ $# -ne 2 ];then if [ $# -eq 1 ];then ./test.sh $1;exit 0;else exit 1;fi;fi
   
5. if [ $1 -ge 1 ] && [ $1 -le 9999 ] && \
   
6.         [ $2 -ge 1 ] && [ $2 -le 9999 ] && [ $1 -ne $2 ];then
   
7. 
   
8.         if [ $1 -gt $2 ];then from=$2;to=$1;
   
9.         else from=$1;to=$2;fi
   
10. 
    
11.         while [ $from -le $to ];do
    
12.                 echo -n "$from " #这条指令你机器上不行，你自己改.
    
13.                 ./test.sh $from    #在这儿调用上次的test.sh
    
14.                 from=$(($from + 1))
    
15.         done
    
16. fi
    

复制代码

测试:

1. 
   
2. $ ./hh.sh 2002
   
3. Wednesday Thursday Saturday
   
4. $ ./hh.sh 1999 2003
   
5. 1999 Sunday Tuesday Saturday
   
6. 2000 Sunday Monday Thursday
   
7. 2001 Tuesday Wednesday Friday
   
8. 2002 Wednesday Thursday Saturday
   
9. 2003 Sunday Thursday Friday
   
10. $ ./hh.sh 2003 1999
    
11. 1999 Sunday Tuesday Saturday
    
12. 2000 Sunday Monday Thursday
    
13. 2001 Tuesday Wednesday Friday
    
14. 2002 Wednesday Thursday Saturday
    
15. 2003 Sunday Thursday Friday
    
16. $
    

复制代码

Pacific Yan

I know C is a really powerful and faster program.  But my major is biology and even the shell and PERL is difficult for me to learn.  

Your program is exercent, but the output format isn't what I want.  I want it will give me the same format as you input one argument.

That mean:

$test.sh 2003
Sunday Thursday Friday

$test.sh 2003 2009  or $test.sh 2009 2003
Wednesday Thursday Saturday   (I guess it)

夜未眠

只要删除上面[ hh.sh ]的下面这一行就行了.
"echo -n "$from " #这条指令你机器上不行，你自己改. "

Pacific Yan

That only solve me one problem, but how can I add all the years together like from 2003 to 2009, then give just one line output, I don't want it list all the years.  For example,

I should have
$ hh.sh 2003 2009
$ Wednesday Thursday Saturday (one line output, I guess it)

rather than
$ hh.sh 2003 1999
Sunday Tuesday Saturday
Sunday Monday Thursday
Tuesday Wednesday Friday
Wednesday Thursday Saturday
Sunday Thursday Friday

夜未眠

Sunday Tuesday Saturday
Sunday Monday Thursday
Tuesday Wednesday Friday
Wednesday Thursday Saturday
Sunday Thursday Friday
我不太明白你的意思，从这些行里输出哪一行的依据是什么？

Pacific Yan

for example ( all the number is my guess):

2003 (Sunday 2 Friday 2 Thursday 1 Saturday 3 Tuesday 2  Monday 1 Wednesday 1)   

2004  (Sunday 1 Friday 1 Thursday 2 Saturday 1 Tuesday 2 Monday 3 Wednesday 2)

2005 (Sunday 1 Friday 2 Thursday 1 Saturday 3 Tuesday 2  Monday 1 Wednesday 2)

add all sunday, monday, tuesday, ..... together, after you get the numbers, then you compare to get the output.


$hh.sh 2003 2005
Sunday Thursday

Note: Sunday and Thursday only appears 4 times, then both of them is the lowest.

夜未眠

ftp://bu.dns0755.net/newtest.sh

1. 
   
2. #!/bin/sh
   
3. # shell script [ newtest.sh ]
   
4. 
   
5. if [ $# -ne 2 ];then exit 1;fi
   
6. if [ $1 -ge 1 ] && [ $1 -le 9999 ] && \
   
7.         [ $2 -ge 1 ] && [ $2 -le 9999 ];then
   
8. 
   
9. if [ $1 -gt $2 ];then from=$2;to=$1;else from=$1;to=$2;fi
   
10. 
    
11. da=0
    
12. for st in Sunday Monday Tuesday Wednesday Thursday Friday Saturday;do
    
13.         eval str_$da=$st
    
14.         da=$(($da + 1))
    
15. done
    
16. 
    
17. while [ $from -le $to ];do
    
18.         mon=1
    
19.         while [ $mon -le 12 ];do
    
20.                 da=`cal $mon $from |awk 'NR==3{loc=index($0,"1");print (loc-2)/3;exit}'`
    
21.                 eval "count_$da=\$((\$count_$da+1))"
    
22.                 mon=$(($mon + 1 ))
    
23.         done
    
24.         from=$(($from + 1))
    
25. done
    
26. 
    
27. 
    
28. min=$count_0
    
29. 
    
30. da=1
    
31. while [ $da -le 6 ];do
    
32.         eval dcn=\$count_$da
    
33.         if [ $dcn -lt $min ];then min=$dcn;fi
    
34.         da=$(($da + 1))
    
35. done
    
36. 
    
37. da=0
    
38. while [ $da -le 6 ];do
    
39.         eval dcn=\$count_$da
    
40.         if [ $dcn -eq $min ];then
    
41.                 eval "echo -n \"\$str_$da \""
    
42.         fi
    
43.         da=$(($da + 1))
    
44. done
    
45. echo
    
46. fi
    

复制代码

夜未眠

确认一下你是不是想要累加和最小的     
你仍然要把echo好儿改成适合你系统的。

Pacific Yan

Thanks, 夜未眠

I will test it after I do my shopping.  It is almost 10 clock, wish you have a good sleep.  If you need any any softwares, books, please let me know.