Hi Alex
Thanks, but a newbie here so that was "drinking at a firehose". Let me
parse it to better understand.
First can you really invoke on two separate enumerable objects with that
[n,n1,n2] syntax? That is tremendous.
My problem is I need to first check the presence of a key before I unset it
b/c I found out unset-ing a non-existent key seems to zero the struct; also
it is more courteous, grin. So how would I construct the and parameterize
the annonymous function to do the check and if present "do it to itself".
Ideally I would like to pass in the enumerable, call the function and return
the modified(or not) enumerable in the function's return.
On Thu, Apr 8, 2010 at 4:41 PM, Alex Wallace <alexmlwall...@gmail.com>wrote:
> Ah, I see. You can handle this using Enumerable's invoke. This code should
> make it clear:
>
> a = new Hash({ x : "foo", y : "bar" });
> b = new Hash({ x : "zam", y : "moof" });
> a.get("x");
> "foo"
> b.get("x");
> "zam"
> [a,b].invoke("get","x");
> ["foo", "zam"]
> [a,b].invoke("unset","x")
> ["foo", "zam"]
> a.get("x")
> (undefined)
>
> Cheers,
> Alex
>
> On Thu, Apr 8, 2010 at 3:24 PM, chrysanthe m <chrysant...@gmail.com>wrote:
>
>> Sorry sudden send resume in this reply
>>
>> Hello
>> I am having a difficult time trying to enumerate a hash to determine if a
>> give key is in the hash and if so delete it and its value.
>> If I could approach it index it would be
>> function remove(valueToTest, hashToBeTested){
>> for(i=0;i<hashToBeTested.length;i++){
>> if(valueToTest==hashToBeTested[i])
>> hashToBeTested.unset(valueToTest);
>> }
>>
>> Would I do it like
>>
>> hashToBeTested.each(function(valueToTest){
>> if(valueToTest==this)hashToBeTested.unset(valueToTest);
>> },hashToBeTested);
>> return hashToBeTested;
>>
>>
>> which I am sure is wrong syntactically if not semantically. Can someone
>> guide the proper way and more deeply the use of this?
>>
>>
>> On Thu, Apr 8, 2010 at 3:06 PM, chrysanthe m <chrysant...@gmail.com>wrote:
>>
>>>
>>>
>>
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