How can I extract the last X chars from a string variable?

Question

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?

What gives the actual error? The echo? If so, please show us the contents of the $line variable. Do you also get the error if you use quotes (echo "${lastchars}")? — terdon, Nov 20 '17 at 11:11
Your code works exactly like it's supposed to for me, please add input and full output to your question. — dessert, Nov 20 '17 at 11:13
the ${line} content prints just fine. anyways the substitution error is shown. @terdon the echo ${lastchars} is not reached at all, because getting the last 5 characters from the variable already produces the error. — membersound, Nov 20 '17 at 11:22
I can't reproduce what you describe. Are you sure you're using bash? Are you running this in a script or from the command line? If a script, how are you running that script? — terdon, Nov 20 '17 at 11:30
I added the complete example above. I cannot help, still getting the error of this... — membersound, Nov 20 '17 at 11:36
That's because you're using the wrong shell. You're not running bash at all but instead are using sh which is a different shell: dash. See my answer. — terdon, Nov 20 '17 at 11:37

score 20 · Accepted Answer · edited Nov 20 '17 at 15:53

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:

bash:

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

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

edited Nov 20 '17 at 15:53

Dan

13,119

answered Nov 20 '17 at 11:34

terdon

100,812

I have the #!/bin/bash and still getting the error. Interestingly running the script with bash lastchars.sh works! Do you know why? – membersound Nov 20 '17 at 11:37
1

@membersound yes, because you are running it as sh lastchars.sh. The #!/bin/bash is ignored if you run it with a specific shell. So either run it directly (lastchars.sh) or run it with bash (bash lastchars.sh). – terdon Nov 20 '17 at 11:38
Ok so that's probably the solution. – membersound Nov 20 '17 at 11:43
4

@membersound For details about why the #!/bin/bash line is ignored when you run sh lastchars.sh, see Is #!/bin/sh read by the interpreter? (which applies to any "hashbang" line, including #!/bin/bash, and not just to #!/bin/sh). – Eliah Kagan Nov 20 '17 at 13:53

score 1 · Answer 2 · answered Nov 20 '17 at 22:56

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}

How can I extract the last X chars from a string variable?

2 Answers2