| how to display all datas in line chart [message #551676] |
Mon, 09 August 2010 06:58  |
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Hi!
I create a line chart. Data of X axis is connected to timestamps which are string type like "1100804032005000". I converted them to date type like 10-8-4 3:20:05 in category X series expression build using expression:
new Date( parseInt("1"+row["Timestamp"].substring(1,3)) - 100 + 2000,
parseInt("1"+row["Timestamp"].substring(3,5)) - 100 - 1,
parseInt("1"+row["Timestamp"].substring(5,7)) - 100,
parseInt("1"+row["Timestamp"].substring(7,9)) - 100,
parseInt("1"+row["Timestamp"].substring(9,11)) - 100,
parseInt("1"+row["Timestamp"].substring(11,13)) - 100 )
The timestamp ranges from 10-8-4 3:20:05 to 10-8-7 3:20:05 by step 5 minutes.
When I preview my chart, I find only datas from 10-8-4 3:20:05 to 10-8-4 19:20:05 are displayed but other datas not. I don't know what is wrong and how to solve the broblem. Any help will be appreciated!
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| Re: how to display all datas in line chart [message #551804 is a reply to message #551676] |
Mon, 09 August 2010 15:26  |
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Any chance you can convert your report to use a scripted dataset so we
can run it and test it?
Jason
On 8/9/2010 2:58 AM, Rong Yi wrote:
> Hi!
> I create a line chart. Data of X axis is connected to timestamps which
> are string type like "1100804032005000". I converted them to date type
> like 10-8-4 3:20:05 in category X series expression build using expression:
>
> new Date( parseInt("1"+row["Timestamp"].substring(1,3)) - 100 + 2000,
> parseInt("1"+row["Timestamp"].substring(3,5)) - 100 - 1,
> parseInt("1"+row["Timestamp"].substring(5,7)) - 100,
> parseInt("1"+row["Timestamp"].substring(7,9)) - 100,
> parseInt("1"+row["Timestamp"].substring(9,11)) - 100,
> parseInt("1"+row["Timestamp"].substring(11,13)) - 100 )
>
> The timestamp ranges from 10-8-4 3:20:05 to 10-8-7 3:20:05 by step 5
> minutes.
>
> When I preview my chart, I find only datas from 10-8-4 3:20:05 to 10-8-4
> 19:20:05 are displayed but other datas not. I don't know what is wrong
> and how to solve the broblem. Any help will be appreciated!
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