When calculating totals itterater run twice

Hello,
=> when I check the log file I see that the itterater runs twice
=> I have a view Wedstrijden with 2 transient (Totaal & TotaalMin)
=> I have made an object E_Som
package oracle.foot.userinterface;
import oracle.jbo.domain.Number;
public class E_Som {
private Number som1;
private Number som2;
public E_Som(
Number som1,
Number som2) {
this.som1=som1;
this.som2=som2;
public E_Som() {
public void setSom1(Number som1) {
this.som1 = som1;
public Number getSom1() {
return som1;
public void setSom2(Number som2) {
this.som2 = som2;
public Number getSom2() {
return som2;
=> I have created a getSumOfTotal in the WedstrijdenImpl.java
public E_Som getSumOfTotal() {
System.out.println("in get SumofTotal ");
E_Somtest = new E_Som();
sumOfSal = new Number(0);
sumOfSal2 = new Number(0);
RowSetIterator rsi = createRowSetIterator(null);
while (rsi.hasNext()) {
System.out.println("in de itterater "+ sumOfSal);
System.out.println("in de itterater2 "+ sumOfSal2);
Row r = rsi.next();
Number curSal = (Number)r.getAttribute("Mntrecette");
if (curSal != null) {
sumOfSal = sumOfSal.add(curSal);
Number curSal2 = (Number)r.getAttribute("Mntfrarb");
if (curSal2 != null) {
sumOfSal2 = sumOfSal2.add(curSal2);
System.out.println("Na de itterater "+ sumOfSal);
rsi.closeRowSetIterator();
E_Somtest.setSom1(sumOfSal);
E_Somtest.setSom2(sumOfSal2);
return E_Somtest;
=> In the WedstrijdenRowImpl.java if have change the get's of Totaal & TotaalMin in
public Number getTotalMin() { 
return ((WedstrijdenImpl)getViewObject()).getSumOfTotal().getSom2();
public Number getTotaal() {
System.out.println("voor de return van get Totaal ");
return ((WedstrijdenImpl)getViewObject()).getSumOfTotal().getSom1();
=> In the jspx I have
#{bindings.WedstrijdenlijstTotaal.inputValue} &
#{bindings.WedstrijdenlijstTotalMin.inputValue}
=> When I call the detail jspx I see in the log that the getSumOfTotal is used twice.

Yes, So you have to figure out what's causing that. The only part of your code that I can see that does any parsing is this:fltPrice = Float.parseFloat(txtPrice.getText()); so that's probably what's throwing the exception.
So basically, you need to put some code around there to check the return value of txt.getText and see if it's in valid floating-point number format. If it's not in valid format, then when you call parseFloat, it will throw that exception.
and I am new to programming , so I'm not quite
for sure what you mean by code tags.It has nothing to do with programming. These forums have special markup tags. You can make text bold or italic or underlined. When you post code, put "[code]" before it and "[/code]" after it, and the forum software will display the code legibly (like it did above).

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    not sure how that would work.
    I'm not sure how you managed to subtract 9pm from 2pm and get 7 hours. When I try 2am - 9pm, I get #VALUE!, as AppleWorks reads both 'times' as text, and cannot use Text values in a formula that requires numbers. AppleWorks is a bit fussy on what it considers to be a number.
    Using the default alignment, you can recognize what AppleWorks thinks is a number—it will be aligned to the right of the cell—and what AW considers to be Text—it will be aligned to the left of the cell. In the case of Time, you're better off using the 24 hour clock, and entering one minute betore 2pm as 13:59.
    The Mac clock's unit of time is the (24 hour) Day, and all times entered are regarded as the portion of one day that has elapsed up to that time. When you do the subtraction, the result is also a time of day. For start and end times within the same calendar day, simply doing the subtraction will give a result that can be read as the (correct) elapsed time. For the example times you gave, the 'correct' result is also visible:
    02:00 - 21:00 = 05:00
    But the actual numbers behind that 05:00 result and the two times used to get it are the portion of the current day that has passed up to each of the times entered, and the difference obtained when you do the subtraction:
    0.0833 - 0.8750 = -0.7917
    If you only want the (readable) result for a single day, that's not a problem. But if you want to add the results to get a total time for a week, you need to (make AppleWorks) do some additional math. In simplest terms, you need to add 1 to the end time if the end time is on the calendar day after the start time. There's an explanation of this below, which you should read for a better understanding of Time and the Mac, but which isn't necessary to use the formula.
    The formula assumes your start time is in column B, your end time in column C, your formula reporting time worked is in column D, and that you will enter times in rows 2..6, and calculate the week's sum in D7
    Enter in D2, and fill down to D6: =IF(C2,C2-B2+(C2<B2),"")
    Format these cells to display as Time.
    Enter in D7: =24(SUM(D2..D6)) —to report the result in hours (eg. 25.678)
    Format the cell to display as Number, General.
    or
    Enter in D7: =INT(24(SUM(D2..D6))&":"MINUTE(SUM(D2..D6)) —to report the result as hours and minutes
    Note that the second option for D7 produces a text string which will not be useable in further calculations.
    This is Text, so no Number formatting is necessary.
    Regards,
    Barry
    Time and the Mac
    The Mac clock tracks time in Days and Fractions of a Day, starting at midnight, January 1, 1904.
    Nine PM today, May 8, 2006, represented as a decimal number, is 37383.875. Two AM tomorrow, May 9, 2006, is 37384.0833333333.
    An AppleWorks spreadsheet cell (or database field) formatted to display as Time will ignore the whole number part of those numbers, and interpret the decimal fraction part as Time of Day, and display the result in the Time format you have chosen for the cell.
    A cell (or field) formatted to display as a Date will ignore the fractional part of the number, and display the whole number part in the Date format you have chosen.
    When you enter a time, the result is the time of day on January 1, 1904. When calculating elapsed times where start and end times cross the midnight boundary, you need to add 1 (day) to the result to compensate for the change in day and get a result that can be used in further calculations. That's what the "+(C2<B2)" part of the first formula above does—adds 1 if "(C2<B2)" is TRUE or adds 0 it it's false.
    If you add times, the result is 'correct', but if the total reaches 24 hours, the displayed result will appear incorrect as a Time formatted cell ignores the whole number part of the actual result (eg. 27 hours, or 1.25 days), will display as 03:00, the same as 3 hours, or 0.25 days. You can convert the result to hours and fractions of hours by simply multiplying the result by 24, and displaying it as a Number rather than as Time.
    B

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