Displaying results with comma

Hi i have two tables that have a one-to-many relationship and i want to display the output as comma delimitters. The two tables are
TABLE TABLE1
OVERSION_ID NUMBER NOT NULL,
PREVIOUS_V_ID NUMBER NOT NULL
TABLE TABLE2
OVERSION_ID NUMBER NOT NULL,
so for example if i run the query
select * from table1
the result is
OVERSION_ID PREVIOUS_V_ID
12456 25677
12456 25702
12456 27897
how can i write an sql query that will display all previous_v_id seperated by a comma like below
25677, 25702, 27897
Thank you very much.

i have used the wm_concat function and it worked
select oversion_id, wm_concat(previous_v_id) from table1 group by oaversion_id;
however i have 2 other tables
TABLE TABLE3
OVERSION_ID NUMBER NOT NULL,
OID NUMBER NOT NULL
TABLE TABLE4
OID NUMBER NOT NULL
ONUMBER VARCHAR2 NOT NULL,
You can see the relation between table 3 and table 4
how can I change the query
"select oversion_id, wm_concat(previous_v_id) from table1 group by oaversion_id;"
so that i get the output as onumber and not the previous_v_id
in other words how can I join these tables so the output is not the numbers (previous_v-id) , but the varchar2 (onumber) from table4

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Please Mark This As Answer if it helps to solve the issue Visakh ---------------------------- http://visakhm.blogspot.com/ https://www.facebook.com/VmBlogs

Display a number with commas

Here's a way to display a number with commas. See attached
code.
I had searched here for help on how to do it, and couldn't
find anything. So, then I wrote this. I'm a hack, self-taught
coder, so there may be a more eloquent way, but this does
work.

Nice, I'm always happy to be humbled.
One more point. If you're going ot use a return statement I
like to keep it
at the very end of the function. So, in a conditional
statement, setting a
local variable to the value to be returned allows you to
return that local
variable and always keep the return statement as the last
line. It just
saves a little searching around for returns in longer
functions. Just a
style choice not necessarily better.
Craig Wollman
Word of Mouth Productions
phone 212 928 9581
fax 212 928 9582
159-00 Riverside Drive West #5H-70
NY, NY 10032
www.wordofmouthpros.com
"duckets" <[email protected]> wrote in
message
news:[email protected]...
>
quote:
Originally posted by:
FasterPastor
> I had searched here for help on how to do it, and
couldn't find
> anything.
>
> You should have asked! here's one I have had sitting
around for a while.
> (code
> attached below). It handles lots of cases which the
other handlers posted
> so
> far do not. Specifically:
>
> Negative numbers - input: -123456
> - craig's: -,123,456
> - dougs's: -,123,456
> - mikes's: -,123,456
> - duck's: -123,456
>
> Floating point numbers - input: pi
> - craig's: 3
> - dougs's: 3
> - mikes's: 3.1,416
> - duck's: 3.1416
>
> Numbers larger than the maxinteger - input: 149668992000
> - craig's: -,654,863,360
> - dougs's: -,654,863,360
> - mikes's: 14,966,899,200,0.0,000
> - duck's: 149,668,992,000
>
> Negative floats - input: -123456.7890
> - craig's: -,123,457
> - dougs's: -,123,457
> - mikes's: -12,345,6.7,890
> - duck's: -123,456.789
>
> Floats larger than the maxinteger - input:
2233445566.7788
> - craig's: -2,061,521,729
> - dougs's: -2,061,521,729
> - mikes's: 223,344,556,6.7,788
> - duck's: 2,233,445,566.7788
>
> Yes.. the code is longer, but if you want to print
something like the
> distance
> in meters to the sun, you need to handle floats
properly!
>
> enjoy ;-)
>
> - Ben
>
>
>
> on stringNumber n
>
> outputString = ""
> inputString = string(n)
>
> if inputString.char[1] = "-" then
> negative = true
> delete inputString.char[1]
> else
> negative = false
> end if
>
> fraction = ""
>
> if inputString contains "e" then
>
> mantissa = inputString.char[1..(offset("e",
inputString)-1)]
> exponent = value(inputString.char[(offset("e",
> inputString)+1)..inputString.length])
>
> decimalChar = mantissa.char[2]
> mantissa = mantissa.char[1] &
mantissa.char[3..mantissa.length]
>
> if mantissa.length < exponent+1 then
> plainNumber = mantissa
> repeat while plainNumber.length < exponent+1
> put "0" after plainNumber
> end repeat
> else
> plainNumber = mantissa.char[1..(exponent+1)]
> fraction = mantissa.char[(exponent+2)..mantissa.length]
> end if
>
> else
>
> if offset(".", inputString)>0 then
> decimalChar = "."
> end if
> if offset(",", inputString)>0 then
> decimalChar = ","
> end if
>
> if offset(decimalChar, inputString)>0 then
> plainNumber = inputString.char[1..(offset(decimalChar,
> inputString)-1)]
> fraction = inputString.char[(offset(decimalChar,
> inputString)+1)..inputString.length]
> else
> plainNumber = inputString
> fraction = ""
> decimalChar = string(1.2).char[2]
> end if
>
> end if
>
> if decimalChar = "." then
> separatorChar = ","
> else
> separatorChar = "."
> end if
>
>
> repeat while plainNumber.char[1] = "0"
> delete plainNumber.char[1]
> end repeat
>
>
> repeat while plainNumber.length > 0
> if plainNumber.length > 3 then
> nextDigits =
>
separatorChar&plainNumber.char[plainNumber.length-2..plainNumber.length]
> delete
plainNumber.char[plainNumber.length-2..plainNumber.length]
> else
> nextDigits = plainNumber
> plainNumber = ""
> end if
> put nextDigits before outputString
> end repeat
>
> repeat while fraction.char[fraction.length] = "0"
> delete fraction.char[fraction.length]
> end repeat
>
> if fraction.length > 0 then
> put decimalChar&fraction after outputString
> end if
>
>
> if negative then
> put "-" before outputString
> end if
>
> return outputString
>
> end
>

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Is not a PL/SQP functiion??
What do I do wrong?
thanks,
Roseline Paquin
Montreal, Canada

1 error has occurred
Invalid query, no select statement foundWhat if you try
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Mac mini -> HDMI cable -> monitor (result: fuzzy and blurry image)
Mac mini -> displayport –HDMI adapter-> HDMI cable -> monitor (result: fuzzy and blurry image)
Mac mini -> displayport –DVI adapter-> DVI-to-HDMI cable -> monitor (result: it gives me a blank screen, no image at all)
I’m running out of options. Don’t know what to do. The 24” monitor produces beautiful, crisp images when hooked to my Windows 7 PC. It looks horrible when connected to my brand new Mac Mini.
Any alternatives you may have come across? Please let me know, thank you.

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Hi All,
I have a table with 10 records(employee name) and i have to make a string
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Thanks in advance.

// I believe the following example gives you the answer.
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Here dont assume that I asked you to get all the results and putting
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The half-way coding below answers your question, I hope.
vempname = "";
while (emprs.next()) {
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trace(result);
This function displays: [object Object].
I know that "result" is an ArrayCollection type but i don`t know how to get rows and columns from this. I know that my data is there but i have no idea how to get it.
Clarify: I don`t know how to get to Arrays and simple data variables which are in ArrayCollection.
Could anyone help me with that problem. I would be gratefull
P.S. I tried also change query type in service.PHP for mysql_fetch_query but in that case i`ve got only one row (not all data).

Thanks for fast reply,
arr_coll:ArrayCollection = new ArrayCollection ({col1:"data1",col2:"data2"}, {col1:"data3",col2:"data4"});
you would get the data like
var resultstr:String = arr_coll[1][1].col2;
trace(resultstr);
//results in data4
could you explain me how it was happen (arr_coll[1][1].col2)? It`s not clear to me. I thought in this case rather something like this :
var resultstr:String = arr_coll[1]['col2'];
It should give me "data4". I know it wasn`t but i don`t understand ArrayCollection in level which is needed to use your advice in my case. Could you clarify "arr_coll[1][1].col2" a bit?
What would it look like when you would have something like this:
arr_coll:ArrayCollection = new ArrayCollection ({col1:"data1",col2:"data2"}, {col1:"data3",col2:"data4"},{col1:"data5",col2:"data6"},{col1:"data7",col2:"data8"});
and you would want know f.e. position in ArrayCollection of "data6". How would you code this? arr_coll[1][2].col2?

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