how to print CSV values from bash script when values have spaces

Question

I have the following CSV files ( I have linux red-hat 6.2 version )

# more  SIMPLE_FILES.CSV
  FILE1,FILE2,FILE3
  /etc/hosts,/etc/info.txt,/var/log.txt
  /etc/some_file,/var/collect.txt,/etc/INFO.txt
  /sbin/ls,/sbin/awk,/sbin/sed

 # more COMPLEX_FILES.CSV
   FILE1,FILE2,FILE3
   /etc/config/net ip.txt,/var/log/summary snap/LOG OF global/info.txt
   /etc/hosts files hosts.info/etc/hosts,var/log/messages,/sbin/collect file/all.info
   /etc/old/TEXT INFO/info.txt/etc/OLD FILES/info.txt,/root/customor select/info.txt

the following bash script should read the CSV file and print the files PATH for FILE1,FILE2,FILE3

remark - I set the param CSV_LINE=2 only for example ( the second values in CSV )

 #!/bin/bash


 CSV=SIMPLE_FILES.CSV

 CSV_LINE=2

 eval $(awk -v A=$CSV_LINE -F, 'NR==1 { for (i=1; i<=NF; i++) sn[i]=$i }
            NR==A { for (i=1; i<=NF; i++) print sn[i] "=" $i; exit }' $CSV  )


 echo $FILE1
 echo $FILE2
 echo $FILE3

so if I run the bash script ( when I set CSV=SIMPLE_FILES.CSV ) get the following

 /read_path_from_csv.bash


 /etc/hosts
 /etc/info.txt
 /var/log.txt

until now its fine,

but if I run the bash script ( when I set CSV=COMPLEX_FILES.CSV ) get the follwoiojng

  ./read_path_from_csv.bash: line 20: ip.txt: command not found

My conclusion - its seems that the spaces in the PATH cause for this error

please advice how to update my code in order to print the PATH as defined in the COMPLEX_FILES.CSV ,

How to add one " before PATH and one " after the PATH ?

Example what I should to get:

/read_path_from_csv.bash

/etc/config/net ip.txt,/var/log/summary snap/LOG OF global/info.txt
/etc/hosts files hosts.info/etc/hosts,var/log/messages,/sbin/collect file/all.info
/etc/old/TEXT INFO/info.txt/etc/OLD FILES/info.txt,/root/customor select/info.txt

I don't see the difference between COMPLEX_FILES.csv input and output, apart from the fact 1st line is not printed. Could you indicate so? — fedorqui
– fedorqui, Commented Jul 16, 2014 at 9:51
the first line shuld not printed because they the parameters , we want to print only the values from the second line until end — maihabunash
– maihabunash, Commented Jul 16, 2014 at 9:58
Then why not just awk 'NR>1' file? It is still not clear what you want and definitely using eval looks a bit odd. — fedorqui
– fedorqui, Commented Jul 16, 2014 at 10:01
awk 'NR>1' print only the values , but what I need is to set the values in to the parameters that defined in the first line in the CSV ( parameters shuld be print in side the bash script ) — maihabunash
– maihabunash, Commented Jul 16, 2014 at 10:03

WKPlus · Accepted Answer · 2014-07-16 10:03:40Z

2

Yes, it the spaces in the PATH that causing this problem, and you can add quotes to avoid it.

eval $(awk -v A=$CSV_LINE -v q='"' -F, 'NR==1 { for (i=1; i<=NF; i++) sn[i]=$i }
            NR==A { for (i=1; i<=NF; i++) print sn[i] "=" q $i q; exit }' $CSV)

Explanation:

If you run eval 'FILE1=/path/file name' in bash, it will give an error:

-bash: name: command not found

And you can quotes the /path/file name part to avoid this error:

> eval 'FILE1="/path/file name"'
> echo $FILE1
/path/file name

edited Jul 16, 2014 at 10:03

answered Jul 16, 2014 at 9:51

WKPlus

7,3152 gold badges38 silver badges55 bronze badges

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

1 Comment

maihabunash Over a year ago

YES GOOD SOLUTION I AGREE ( +1 and UPVOTE )

Collectives™ on Stack Overflow

how to print CSV values from bash script when values have spaces

1 Answer 1

1 Comment

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

1 Comment

Your Answer

Sign up or log in

Post as a guest

Related