How to add Time field in models database in odoo 15? - time

In Py
time = fields.Float('Time')
in views
<field name="time" widget="float_time"/>
the above method is not working as when I try to record time format such 10:50,It shows invalid error message. In order to save the window after editing, I have to record time format in float like 10.50 but when saving, it appears in the report 10:39 in time format.
Another problem when I import excel sheet consists of time, It show me an error that i can not import
second error imagethird error image
error image]3

You got that error because you used the wrong field (you didn't define the float_time widget).
You just need to use the float_time widget in the XML definition.
Example:
<field name="field_name" widget='float_time'/>

Related

How can we get the value of textbox in QTP?

I am executing the automated test scripts in UFT 12.5 I am new to UFT. Not very familiar with codes.There is an edit box wherein i have to type the value "S05292". Example:
Browser(Browsername").Page("Pagename").WebEdit("ctl00$ConBody$txtPDNumber").Set "S05292"
The problem is my script fails at this step and does not type the value. Can somebody provide me with a solution which is easy to understand. I tried the below two methods
Method (1)
a=Browser().page().webedit(ctl00$ConBody$txtPDNumber).getroproperty("value")
if a=="S05292" then
msgbox ("displayed message is S05292")
else
msgbox ("msg is not S05292")
end if
Method (2)
x = Browser("Browsername").Page("Pagename").Webedit("ctl00$ConBody$txtPDNumber").GetROProperty("value")
msgbox x
The error message that displays is
Cannot identify the object "ctl00$ConBody$txtPDNumber" (of class WebEdit).
Verify that this object's properties match an object currently displayed in your application.
Use the Object Spy to get the properties of that text box at run time and then make sure they match up to the properties of that text box in your object repository that you defined. Perhaps that don't match up or you didn't uniquely identify that text box.
If you don't want to use an object repository then you have to pass it a property at run time to uniquely identify it. Some thing like:
Browser().page().webedit("developer name:=PDNumber").
Instead of a .set you can do a .type to set/type the value into the text box

I can't import com.parse.ParseImageView

I've been trying to import com.parse.ParseImageView so i can display image queried from parse. But I get this error in my xml file
The following classes could not be found:
- com.parse.ParseImageView
Is ParseImageView no longer supported ? What is the alternative way to display image from parse database? Thanks

Coded UI test failing in Visual Studio 2013 when entering a blank value from a csv into a WPF data grid

I am running a coded ui test in Visual Studio 2013 into a WPF data grid using values from a csv file. When I have a blank value in the csv file eg ,, it is working fine for input fields but when it comes to entering the empty string into a field on the data grid the coded ui test fails with the following error:
An exception of type 'System.ArgumentNullException' occurred in Microsoft.VisualStudio.TestTools.UITesting.dll but was not handled in user code
Additional information: Value cannot be null.
When I run the test manually I can submit the form without this value so I know it is not mandatory on the UI, the code just seems to be falling over if a value is not sent. If I enter a value on the csv the test will run but I deliberately want the field to be empty.
Has anyone come across this problem before and if so is there a way I could either adapt the csv or the code to get this to work? I have also tried ,"", and this did not work either.
Thanks
I think the way you're doing it (using the isNullOrWhiteSpace method to determine if you should skip entering the value) is the right way. If you don't want to write that each time you're entering values into the field, you could write an extension method instead:
public static void EnterValue(UITestControl control, string inputString)
{
if (!String.IsNullOrWhiteSpace(inputString)
Keyboard.SendKeys(control, inputString);
}
And then just call that when you want to enter text:
string csvValue = /*value from the .csv file*/
StaticUtilityClass.EnterValue(myControl, csvValue);
Not a ground breaking change, but it would cut down on the number of times you have to write that if-statement.

when using remapColumns in different event, sort icon doesn't show when clicking a header

I saved my column permutation info into a table. This information can be reloaded in beforeRequest event:
mynewperm = {....};
myGrid.jqGrid("remapColumns", mynewperm, true);
The columns are reordered correctly. However I lost the header icon. Now if I click any column header, I can not see
the sort icon anymore, then can not sort any column. How can I get it back?
Thank you,
yh
if you are able to change the code, you could test for an undefined of a.grid.headers[a.p.lastsort].
In the source file it could look like this:
// old
var previousSelectedTh = ts.grid.headers[ts.p.lastsort].el
// new:
var previousSelectedTh = ts.grid.headers[ts.p.lastsort] ? ts.grid.headers[ts.p.lastsort].el : null
Indeed if you look at the jquery.jqGrid.src.js source, the line is:
var previousSelectedTh = ts.grid.headers[ts.p.lastsort].el, newSelectedTh = ts.grid.headers[idxcol].el;
Line #1982 in my version. I fixed it by modifying the file and added this just before that line:
if (ts.p.lastsort < 0) // johnl.
ts.p.lastsort = 0;
The problem was that ts.p.lastsort was -1.
I've just managed to fix this issue myself but not using the methods described above. I was receiving the following error message when trying to sort columns in a jqgrid:
TypeError: a.grid.headers[a.p.lastsort] is undefined js/jqgrid/jquery.jqGrid.min.js?1.4:86
I should note that it was Firebug that produced this error message. Our company develops web applications for Chrome but Chrome's Javascript console produced a very uninformative error message:
Uncaught TypeError: Cannot read property 'el' of undefined
After stripping out all but the jqGrid declaration on the page causing the issue, it transpired that removing the "multiSelect" option declaration for the jqGrid solved the issue. Apparently, declaring this option causes an additional hidden column to be added into the grid rendered which enables users to select multiple grid rows at a time. I'm not exactly sure why this caused an issue but after consultation with the programming director here our best guess is that there is a for loop somewhere in the jqGrid library code which is called when column sorting is applied and the loop is not taking into account this extra column which results in it not being defined.
Strange answer to a strange issue but hopefully this will help somebody out in future and save them around 3 hours of debugging!
I've the same issue :
It append to me since i apply the "remapColumns" method in the "loadComplete" event (i get back user column configuration from a cookie).
So when I try to sort a column nothing happen. I got this error in firebug :
a.grid.headers[a.p.lastsort] is undefined -> jquery.jqGrid.min.js (line 93)
maybe it will be helpful to find what the problem is
Thank you

BIRT: Specifying XML Datasource file as parameter does not work

Using BIRT designer 3.7.1, it's easy enough to define a report for an XML file data source; however, the input file name is written into the .rptdesign file as constant value, initially. Nice for the start, but useless in real life. What I want is start the BIRT ReportEngine via the genReport.bat script, specifying the name of the XML data source file as parameter. That should be trivial, but it is surprisingly difficult...
What I found out is this: Instead of defining the XML data source file as a constant in the report definition you can use params["datasource"].value, which will be replaced by the parameter value at runtime. Also, in BIRT Designer you can define the Report Parameter (datasource) and give it a default value, say "file://d:/sample.xml".
Yet, it doesn't work. This is the result of my Preview attempt in Designer:
Cannot open the connection for the driver: org.eclipse.datatools.enablement.oda.xml.
org.eclipse.datatools.connectivity.oda.OdaException: The xml source file cannot be found or the URL is malformed.
ReportEngine, started with 'genReport.bat -p "datasource=file://d:/sample.xml" xx.rptdesign' says nearly the same.
Of course, I have made sure that the XML file exists, and tried different spellings of the file URL. So, what's wrong?
What I found out is this: Instead of defining the XML data source file as a constant in the report definition you can use params["datasource"].value, which will be replaced by the parameter value at runtime.
No, it won't - at least, if you specify the value of &XML Data Source File as params["datasource"].value (instead of a valid XML file path) at design time then you will get an error when attempting to run the report. This is because it is trying to use the literal string params["datasource"].value for the file path, rather than the value of params["datasource"].value.
Instead, you need to use an event handler script - specifically, a beforeOpen script.
To do this:
Left-click on your data source in the Data Explorer.
In the main Report Design pane, click on the Script tab (instead of the Layout tab). A blank beforeOpen script should be visible.
Copy and paste the following code into the script:
this.setExtensionProperty("FILELIST", params["datasource"].value);
If you now run the report, you should find that the value of the parameter datasource is used for the XML file location.
You can find out more about parameter-driven XML data sources on BIRT Exchange.
Since this is an old thread but still usefull, i ll add some info :
In the edit datasource, add some url to have sample data to create your dataset
Create your dataset
Then remove url as shown
add some script

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