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Using an array value as another array name

Posted by shubb (shubb), 6 January 2006
Hi, I'm trying to use the values from one array as names of other arrays for excecuting commands.

I create several data arrays with values.
set testArr1('name') "test1"
set testArr1('val1') "value11"
etc...

set testArr2('name') "test2"
set testArr2('val1') "value21"
etc...

Then I create another array with the names of each data array.
set arrayNames(0) "testArr1"
set arrayNames(1) "testArr2"
etc...

I want to use the arrayNames in a foreach loop to get the name of each of the data arrays, so I can use the data in those arrays for commands.  The problem is that when I use the foreach loop, it reads the names correctly, but it doesnt use the value to look up the data array by the same name.

foreach arrData [array names arrayNames] {
    send_user "data array's name is $arrayNames($arrData)('name')\n"
}


What this returns is:
data array's name is testArr1('name')
data array's name is testArr2('name')

when what I want to get is the value of testArr1('name'), which is "test1" and "test2"

Any Idea how I to do this?
As you can probably guess, I'm a beginner with Expect.
Is there a better way to accomplish this?

Thanks for any help you can offer.

Posted by admin (Graham Ellis), 7 January 2006
I would tackle it rather differently - the way you try to store one name in another is a bit obtuse and you'll do better with just one array:

Code:
set abc(def,name) hello
set abc(def,value) "a way to greet people"
set abc(ghi,name) goodbye
set abc(ghi,value) "a way to conclude a conversation"

foreach gname [array names abc *,name] {
       set subname [lindex [split $gname ,] 0]
       puts "$abc($subname,name) IS $abc($subname,value)"
       }


runs

Code:
earth-wind-and-fire:~/jan06 grahamellis$ expect eee
hello IS a way to greet people
goodbye IS a way to conclude a conversation
earth-wind-and-fire:~/jan06 grahamellis$


Posted by shubb (shubb), 8 January 2006
Awesome!  That works perfect.

Thanks!  I knew there had to be a better way to do that.



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