Listing row length and count of rows

Hi,
I am trying to write a piece of code to fetch number of rows grouped monthly and running out of ideas, please help .. here is my requirement.
I have a flat file which contains table names and I need to query for the date column from those table, group it by month, number rows in that month. The generated output should contain the table_name, month, rowcount, and avg_row_length of that table.
Please advice.
Thanks.

Hi
Good, I understand a lot better now. Sorry, I mis-read your first message.
You need to do some kind of dynamic SQL, that is, you want to run a query like:
select    '&table_name'
,         to_char(update_date, 'Mon-yyyy')
,         count(*) row_count
from      &table_name
group by  TRUNC (update_date, 'MM')
ORDER BY  TRUNC (update_date, 'MM');Does every table have a column called update_date, or is it called by different names in different tables? Either way, you can create a script (let's call it one_table.sql) that accepts an input parameter (the table name), and contains the query above.
Then, to call that for each of your tables, you vould create another script that contains:
-- Adjust SQL*Plus output settings
SET  FEEDBACK    OFF
SET  PAGESIZE    0
SET  VERIFY      OFF
SPOOL  combined_output.lst
-- Write common header line for the whole file
PROMPT  table_name   Month     Rowcount    avg_row_len
-- Run query on each table
@@one_table  tab1
@@one_table  tab2
@@one_table  tab3
-- Finished.
SPOOL OFFThis will produce an output file like this:
table_name   Month     Rowcount    avg_row_len
tab1         Jan-2008       20      100
tab1         Feb-2008       30      100
tab1         Mar-2008       40      100
tab2         Jan-2008       20      200
tab2         Feb-2008       50      200If you want to print the table name only on the first row, like this:
table_name   Month     Rowcount    avg_row_len
tab1         Jan-2008       20      100
             Feb-2008       30      100
             Mar-2008       40      100
tab2         Jan-2008       20      200
             Feb-2008       50      200You can make the query in one_table.sql a little more complicated: use ROW_NUMBER or ROWNUM to number the rows, and use CASE to print the table name on row #1, and nothing on the others.
Are you generating the list of tables yourself? If so, then instead of generating just the names, you might generate
@@one_table  tab1
@@one_table  tab2
@@one_table  tab3or even the entire main script.
Sorry, I have to go now. See how far you can get with this, and post again if you have problems.

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    lbl14.Text = "The number of 14 year olds is "
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    lbl13.Visible = True
    lbl14.Visible = True
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    I think I should use a For..Next loop but I don't know how to identify and match using this database, and then count how many repeated 12, 13, 14 years old there are.
    Any help would be really appreciated.

    Hello,
    Conceptually speaking you would group the data and count. Beings this is school work the demo below is a static example and not suitable for your assignment, its to show a point. Now if you have learned about LINQ and Lambda this logic can apply to your
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    Module Module1
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    .Select(Function(group) New With
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    Function(x) x.Field(Of String)("Name"))}) _
    .OrderBy(
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    dt.Columns.Add(New DataColumn With {.ColumnName = "Name", .DataType = GetType(String)})
    dt.Columns.Add(New DataColumn With {.ColumnName = "Age", .DataType = GetType(Int32)})
    dt.Rows.Add(New Object() {Nothing, "Bill", 13})
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    dt.Rows.Add(New Object() {Nothing, "Jim", 13})
    dt.Rows.Add(New Object() {Nothing, "Paul", 15})
    dt.Rows.Add(New Object() {Nothing, "Mike", 14})
    dt.Rows.Add(New Object() {Nothing, "Jill", 13})
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    End Function
    End Module
    Output in the IDE output window
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    6 Jill
    1 Bill
    Group: 14 count: 2
    5 Mike
    2 Karen
    Group: 15 count: 1
    4 Paul
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    13 3
    14 2
    15 1
    Please remember to mark the replies as answers if they help and unmark them if they provide no help, this will help others who are looking for solutions to the same or similar problem.

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