How to capture the value of type column report

It would be a great help that you might suggest the following

(i) Java script that I wrote on the standard status column (in a tabular report) fails.
This happens because of the standard status column, if I use the text field then called Javascript function and showing correct results.

Now, here, I want to know how to capture the value of the column standard report, since javascript fails due to
the column values of standard report unrecognized.

Example:

Columns of the report as a table:
```````````````````
Standard report column: salary
TextField report column: new treatment
Report column standards: difference

I've written the javascript on column New_salary function which returns the value of the salary column to calculate the difference.

For this column of report types, the javascript function does not work, that is to say, it does not show the difference., since it is not read
the value of the column Salary (which is the Standard type of report column)
Once I have change the Salary column to text (display as saved state) javascript field works fine.

All entries on how to capture the standard value of the column, so that I can access the value of the javascript function.

Thanks in advance
Vijay

Hi Vijay - I had problems with the forum as well (and my workspace OTN so)!

You have two questions, I think.

First of all, that the old wage is not an input element, you can not get to it by referring to an ENTRY tag. you will need to change that to something like:

var s1 = d[k-1].firstChild.nodeValue;

This means that there is nothing else in the cell apart from the text - as in < td > < table > 123. If there is something else there that stops you getting the value using the above, you can go through these nested tags referring to objects further firstChild-

var s1 = d[k-1].firstChild.firstChild.nodeValue;

As I don't see your page, you need to check yourself to see firstChild how much you need. This example assumes that you had something like: < td > < b > 123 < /b > < table >

Second, the values of s and s1 are strings. You must convert these numbers to be able to perform calculations. You have two javascript functions to do this: ('string') parseFloat and parseInt ('string'). One contains decimals, the other is not - but know that one is not very accurate (it can change "123.456' in ' 123.455999999995' or something similar - not very good! '").

So, your calculation should be:

var diff = parseInt(s) - parseInt(s1);

Andy

Tags: Database

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