6

I'm reading the last line of a file into a variable. Then I want to get the last X characters of the string variable:

#!/bin/bash
someline="this is the last line content"
echo ${someline}
somepart=${someline: -5}
echo ${somepart}

Run with: sh lastchars.sh

Result:

this is the last line content
line 4: Bad substitution

What might be wrong here?

user285259
  • 262
membersound
  • 1,460

2 Answers2

20

It sounds like you're not using bash at all. I can only reproduce the error you show if I use dash instead of bash:

  1. bash:

    $ line="someline content"
$ echo ${line}
someline content
$ lastchars=${line: -5}
$ echo ${lastchars}
ntent

  2. dash:

    $ line="someline content"
echo ${line}
lastchars=${line: -5}
echo ${lastchars}
$ someline content
$ dash: 3: Bad substitution

Your shebang line is pointing to bash, but you are running the script with sh, so the shebang is ignored. /bin/sh on Ubuntu systems is actually dash, a minimal shell that doesn't support the syntax you are trying to use.

When using a shebang line, there's no reason to explicitly call a shell for the script, just make it executable (chmod a+x /path/to/script.sh) and run it without specifying an interpreter:

/path/to/script.sh

Alternatively, just use the right one:

bash /path/to/script.sh
Dan
  • 14,180
terdon
  • 104,119
1

Obviously using the build-in functions of a certain shell are nice, but you can also accomplish the task using standard UNIX commands, so it will work in any shell:

String="This is some text"
StrLen=`echo ${String} | wc -c`
From=`expr $StrLen - 5`
echo $String | cut -c${From}-${StrLen}