Why does " $Namefile* " give me such a result?

Question

In my code , there is such a part ;

Namefile=$1
for n in $Namefile*
do
echo $n
done

and just for this part's result is ;

$ sh example.sh hw1
hw1_evening_sun.txt
hw1_morning_sun.txt
hw1_out_si_wire.txt
hw1_script.sh
hw1_script2.sh
hw1_script3.sh
hw1_sun1.txt
hw1_sun2.txt

which are the files whose name starting with "hw1" , but I don't understand why $Namefile* takes these names.As far as I know * calls the the argument which the script receives.

Why does this method work like that ? Could someone explain the logic behind it ?

Answer 1

I don't understand why $Namefile* takes these names.As far as I know * calls the the argument which the script receives.

$Namefile expands to the value that you passed in (probably "hw"). This means that $Namefile* after variable expansion becomes "hw*".

When this string is used in a location where it is interpreted as a filename, the * is special. See "pathname expansion" in the man page.

... bash scans each word for the characters *, ?, and [. If one of these characters appears, then the word is regarded as a pattern, and replaced with an alphabetically sorted list of file names matching the pattern.

Answer 2

Here you are doing a substitution in you for loop. So star will match every character after $Namefile string (which I guess it is "hw1"). If you want to match all the arguments provided to the script use the internal variable $*. In your script you are saving $1 (positional parameter number 1) which is wh1 to the variable Namefile. So using * after $Namefile has a different meaning than $*. For more info see here Internal Variables (Positional Parameters section)

Answer 3

It's simply because of -l option of ls. In this case, because $Namefile* would expand to hw1_evening_sun.txt ..snip.. hw1_sun2.txt:

ls -l $Namefile*

is effectively equals to

ls -l hw1_evening_sun.txt ..snip.. hw1_sun2.txt

For more information, you can run man ls. Here I quote:

...  
   -l     use a long listing format  
...

Welcome to *nix.

Answer 4

It is because your do loop is so. For the output to be as expected for -l you would need to execute your that part of script inside the backticks as below.

---

tmp.sh

Namefile=$1
for n in $Namefile*
do
echo $n
done

---

Run that particular part as,

ls -l `./tmp.sh "d"`

-bash-3.2$ ls -l `./again d`
-rw-r--r-- 1 me me 0 Jun  3 05:16 dat
-rw-r--r-- 1 me me 0 Jun  3 05:16 dis