Source file (containing list of variables) to be run in bash loop?

Question

I've got a file containing 2 variables, start and end that I want invoked in a for loop of a bash script but I can't seem to get the result I want

outputfile2.text 

START=26 ; END=47
START=48 ; END=69
START=70 ; END=91
START=92 ; END=113
START=114 ; END=135
...

The loop:

range.bash

#!/bin/bash
rm range.text -f

while read line
do
    for (( c=$START; c<=$END; c++ ))
    do
            echo -n "$c ";
    done >> range.text
done < "outputfile2.text"

Desired output:

1 2 3 ... 19 20 21
26 27 28 ... 45 46 47
48 49 50 ... 67 68 69
70 71 72 ... 89 90 91
... etc

How can this be done?

I'm currently getting:

./range.bash: line 6: ((: c=: syntax error: operand expected (error token is "=")

So I'm assuming it's not reading outputfile2.text as I has hoped.

END GOAL: What I really want is it to print a grid that is 15 columns wide with each entry being 6 characters long, separated by a space and titled with something, but I have no clue how to code that:

line 1
xxxx1 xxxx2 xxxx3 xxxx4 xxxx5 xxxx6 xxxx7 xxxx8 xxxx9 xxx10 xxx11 xxx12 xxx13 xxx14 xxx15
xxx16 xxx17 xxx18 xxx19 xxx20 xxx21

line 2
xxx48 xxx49 xxx50 xxx51 xxx52 xxx53 xxx54 xxx55 xxx56 xxx57 xxx58 xxx59 xxx60 xxx61 xxx62
xxx63 xxx64 xxx65 xxx66 xxx67 xxx68

line 3
... etc

UPDATE (thanks to Mark Setchell):

#!/bin/bash

while IFS="=;" read a START c END e
do
   ((j++))
   echo "[ L${j} ]"
   for (( i=$START; i<=$END ; i++ ))
   do
      printf "%6s" $i
   done
   echo ; echo                          # Just a newline
done < outputfile2.text >> range.text

Produces:

[ L1 ]
    26    27    28    29    30    31    32    33    34    35    36    37    38    39    40    41    42    43    44    45    46    47

[ L2 ]
    48    49    50    51    52    53    54    55    56    57    58    59    60    61    62    63    64    65    66    67    68    69

almost there!! :)

Mark Setchell · Accepted Answer · 2014-06-20 07:52:58Z

2

You can do something like this:

#!/bin/bash
COLS=15
line=1
while IFS="=;" read a START c END e
do
   echo Line: $line
   col=1
   for (( i=$START; i<=$END ; i++ ))
   do
      printf "%6s" $i
      if [[ $col -eq 15 ]]; then
         col=1
         echo
      fi
      (( col++ ))
   done
   echo
   (( line++ ))
done < file

The trick is to set the input field separator (IFS) to separate variables when a = is seen or a ;.

edited Jun 20, 2014 at 7:52

answered Jun 20, 2014 at 7:18

Mark Setchell

210k32 gold badges309 silver badges503 bronze badges

Sign up to request clarification or add additional context in comments.

4 Comments

Patrick Over a year ago

Thanks! Works a charm ... now to figure out how to get my end goal working :)

Mark Setchell Over a year ago

Ooops! I didn't see that bit. What are all the x's in the output - spaces or zeroes?

Patrick Over a year ago

Not a problem. They're spaces :)

Patrick Over a year ago

Thanks! You're awesome! :) just need it to start new line every 15 columns printed. I've edited my original question

Collectives™ on Stack Overflow

Source file (containing list of variables) to be run in bash loop?

1 Answer 1

4 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

4 Comments

Your Answer

Sign up or log in

Post as a guest

Related