Simple read line_ function hangs indefinitely
 

I would like to use the function read line{line number here} in linux bash programming language. Unfortunately, today that function does not work. I would like to read commented out lines 6-10. My attempt below:
The result is that the terminal just hangs indefinitely. Both read -r and read don't work.


What am I doing wrong?

Ok, so first off, use / and not \ in the closing code tag to make them work :)

As for your example, you do realise that the read command will wait until you enter something before moving to the next line of the script?

So we can ignore all but the first read:
After this line in the code your terminal will wait for you to enter for the read command to read :) Only once the enter key has been hit (or maybe ctrl-d too) will you get the next line of code executed

questions formated to 'human readable' form important
 

Thanks, Pardon me.

Yes, and since this is a homework assignment, get your teacher involved and keep him there. That's what he's there for.

Probably the most important skill in programming is to learn how to see what a program is (not) doing, and to devise a way to figure it out. You can short-change yourself of that if you simply throw your hands up, post out to a forum, and ask someone else to give you the answer. A much better strategy would be to ask Mr. Cooper to give you a nudge in the right direction. Don't have any reluctance to do that!

something strange going on
 

Even though I have used "read -r Var" and "read Var" before, I was under the impression that the "line" in "read line#" & "read -r line#" was an dashless option, which would do something along the lines that "$0" would do, almost like a preset variable.

I gained this false notion from when Example 15-4. What happens when read has no variable uses the code block:
and the second line of the script is "# read-novar.sh", which is echoed.

Looking at this practice exercise I still really think something special exists about "line2", more specifically the line "read line2" above. If you look at this entire program you will see that the variable "line2" was never set.

Actually I have no teacher
 

This could be a misunderstanding, but when I wrote
what I meant by "Teacher: Mendell Cooper" I meant that the author of the book I was reading and learning shellscripting from is Mendell Cooper. I'll probably never get a chance to meet Mendell Cooper, but certainly his eBook Advanced Bash Scripting is the most useful thing I have used/discovered so far. Not everybody has a non-poverty income thus we can only use open source materials/resources online to learn. However even though I can't pay/donate $ to Mendel Cooper now, I can pay my respects/thanks.

___________________________________

Actually I did give significant effort to try to solve this problem on my own, more than just the three attempts that were "presentable". I simply just don't know why
works the way it does exactly how it is written but then when you replace the "read line2" with another number, e.g.
Still this will echo "line 2"??!

In the example it is set by reading from the input of $0 and seeing that $0 refers to the script itself it is now reading lines from itself. I can assure you that those variables could have just as easily been
foo and bar and bar would contain the same data as line2. Of course you can try this and also change the digit to be say line10 but you will still only get the second line of the file.

Solution
 

I GOT IT!!!!!
If you want the code block to read line "N" of the script $0, than the "read" command must be executed "N-1" times before line "N" is read. E.g.
Too bad read doesn't have a number option, but that's besides the point.

Now I understand how to echo the choice line of the script $0.

Thanks a-lot grail!!

Original example program correction
 

Correction of original program below:
Where I finally got the desired results:

Alternative:

PEDAGOGICAL SYNOPSIS for grail's alternative method
 

______________________ PEDAGOGICAL ANALYSIS 01 _________________________
At first I did not understand
Even though i could not find this technique on the web, I pondered and hacked around with it, then I realized that this is a way to read a line not starting with the first word. More specifically:

Read a line starting with the Nth word
syntax: $ read _(write " _" N times) word(N+1)

Examples
1) Read a line skipping the first 5 words.
2) Read first 4 words of a line, skip 5th word and then read rest of sentence.
_______________________ PEDAGOGICAL ANALYSIS 02 ________________________
Then I did not quite get
I returned to the Advanced Bash scripting textbook, and I reviewed the chapter on tests, which led me to realize that:
and thus
_____________________ PEDAGOGICAL ANALYSIS 03 __________________________

Finally I also had some problems with
With the above principles covered we can immediately see the
pattern present. Concerning the
statement: This merely says that if variable "$out" already exists, then define variable "out" as its own previous meaning, plus the string value of '$word'; if "out" does not already exist, than "out=$word", e.g.:

Your findings are correct and here is the same in more layman terms:

1. Variable names may start with [a-zA-Z_], so our _ could be replaced by a simple 'a', but the generally practised rule is that _ is treated as what you wish to throw away

2. It is also important to remember that read uses the IFS variable to advise the delimiter for information, so when multiple variables are provided, read will separate using the IFS information and if there
are less variables than items on the line read, all remaining items are placed in the last variable

3. (()) is specifically for arithmetic in bash, either testing or setting and calculating

4. [[ $var ]] is the same as [[ -n $var ]], which of course is your findings that if the variable is empty / null then the statement is true

Here is an additional site that i think is a must have to slowly work through after completing ABS