Re: bash array
> ===========================================================
> #!/bin/sh
> exec 3< "user.list"
> exec 4< "pass.list"
>
> while IFS= read -r user <&3 && IFS= read -r pass <&4
> do
> exampleprogram "$user" "$pass"
> done
>
> exec 3<&-
> exec 4<&-
I arranged according to this example. Very thanks.
On Thu, Jan 25, 2018 at 4:34 PM, Greg Wooledge <wooledg@eeg.ccf.org> wrote:
> On Thu, Jan 25, 2018 at 03:35:35PM +0300, Gokan Atmaca wrote:
>> The problem is: I can pull the users one by one. I can not do the same
>> for passwords.
>
> Why not?
>
>> Do I need to use "Mapfile" for this?
>
> How would we know? You haven't told us what is in the files, or what
> you want to DO with the contents of the files.
>
>> passwords=(cat $pass.list)
>
> Is wrong.
>
>> mapfile -t pass < "$passwords"
>
> Would be OK if $passwords contains the NAME of the file. Is WRONG if
> $passwords is a string containing the CONTENTS of the file.
>
> Is also WRONG if passwords is an array containing the word "cat" and the
> expansion of "$pass" with the string .list appended to it, which is what
> your previous line created.
>
>> example-script:
>> -- user.list
>> a
>> b
>> c
>>
>> -- pass.list
>> xxx
>> yyy
>> zzz
>
> With all that whitespace? Or should I just assume that the "xxx" is
> NOT indented as shown, but is instead an entire line?
>
> What do you want to DO with these files?
>
>> script:
>>
>> #!/bin/bash
>>
>>
>> pass=$(cat pass.list)
>
> Now, you see, this is ENTIRELY DIFFERENT from what you showed up above.
>
> Do you see the difference? Look at the two lines again:
>
>> passwords=(cat $pass.list)
>
>> pass=$(cat pass.list)
>
> If you can't see the difference, you can't write computer programs.
>
> Moving on....
>
>> for i in user=$(cat user.list); do exampleporagram $i $pass; done
>
> Syntax errors. Logic errors. Cannot even make SENSE of this.
>
> Are you trying to read ONE USERNAME and ONE PASSWORD at a time, and
> pass each PAIR to your "exampleprogram"?
>
>
>> On Mon, Jan 15, 2018 at 5:39 PM, Greg Wooledge <wooledg@eeg.ccf.org> wrote:
>> > What kind of array do you want? Why do you even want an array?
>> >
>> > What is your program supposed to do with these users and passwords?
>
> Can't you just write ONE SENTENCE to tell us the objective of the script?
>
> Let's go out on a limb and assume "I want to read one username and
> one password at a time, and pass them as a pair of arguments to
> 'exampleprogram'."
>
> The way you would do that is to open the two files, and then read a
> line at a time from each file, and then close the files.
>
> Like this:
>
> ===========================================================
> #!/bin/sh
> exec 3< "user.list"
> exec 4< "pass.list"
>
> while IFS= read -r user <&3 && IFS= read -r pass <&4
> do
> exampleprogram "$user" "$pass"
> done
>
> exec 3<&-
> exec 4<&-
> ===========================================================
>
>
> See the following pages:
>
> http://mywiki.wooledge.org/Quotes
> http://mywiki.wooledge.org/FileDescriptor
> http://mywiki.wooledge.org/BashFAQ/001
>
> Then read this page again, because I know you didn't read it the
> first time:
>
> http://mywiki.wooledge.org/Quotes
>
> Then keep reading it over and over until you UNDERSTAND it.
>
>
> You do not need an array for this. You don't even need bash features.
> I used pure POSIX sh up there, hence the #!/bin/sh shebang.
>
>
> P.S. read http://mywiki.wooledge.org/Quotes
>
> Also, be sure you quote properly.
>
> I'm serious.
>
> Quotes.
>
> They are NOT optional.
>
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