如何在Unix shell脚本中从字符串中剪切空格

Question

I am reading a file and cutting a column based on some logic. My issue is I am not able to cut a column with space.

This is the code for testing -

st="1|alalhabad|up|tushar|kesarwani|90|   mls   k|19990|india|420|24|m"
HardErrorCheckColumnValue=`echo $st | cut -d'|' -f7`
echo $HardErrorCheckColumnValue

The output should be -

   mls   k

But I am getting-

mls   k

How do I make it not trim the leading or trailing spaces? It should give space, even if it contains only space.

Answer 1

You must use quotes around your variables:

HardErrorCheckColumnValue=$(echo "$st" | cut -d'|' -f7)
echo "$HardErrorCheckColumnValue"
   mls   k

Better to use $(...) instead of old fashioned back-tick for command substitution.

Answer 2

awk Will help you with that

$ cat file.dat
1|alalhabad|up|tushar|kesarwani|90|  mls ki |19990|india|420|24|m

$ awk -F"|" '{print "|"$7"|"}' file.dat
|  mls ki |
||

EDIT 2

If you echo the st variable there is a problem there, where some spaces disapear:

check the difference:

$ st="1|alalhabad|up|tushar|kesarwani|90|   mls   k|19990|india|420|24|m"

$ echo $st
1|alalhabad|up|tushar|kesarwani|90| mls k|19990|india|420|24|m  <-- ONE SPACE GONE

$ cat file.dat
1|alalhabad|up|tushar|kesarwani|90|  mls ki |19990|india|420|24|m

$ awk -F"|" '{print "|"$7"|"}' file.dat
|  mls ki |  <---- SPACE OK
||

Answer 3

Protect your variable before printing it:

echo "$HardErrorCheckColumnValue"

Without the quote it gets expanded to echo mls k , while with it get's expanded to echo " mls k"

But as pointed out @Mat, damage already done. So refer to @anubhava answer :)