How to get the caller menuitemName in custom workflow submit button.

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Posted on by Vishu Beniwal

135

Hello Everyone,

I have created a custom workflow which is working fine then I got a requirement to change the caption of formContol of submit workflow button, so I write code on run method of WorkflowSubmitDialog but the thing is I want to change the caption only for one legal entity so & just for that custom workflow, I tried to get the menuItemName of caller but I'm not able to achieve that, Pl suggest me a way though that I can change the caption only for one legal enity and only for that custom workflow.

[PostHandlerFor(formStr(WorkflowSubmitDialog), formMethodStr(WorkflowSubmitDialog, run))]

public static void WorkflowSubmitDialog_Post_run(XppPrePostArgs args)

{

FormRun formRun = args.getThis();

if(curExt() == 'UK30' && formRun.args().menuItemName() == menuItemActionStr(LSSalesOrderReturnWFTypeSubmitMenuItem)

{

FormCommandButtonControl okBtn = formRun.design(0).controlName('Submit');

okBtn.text('@LSWorkFlow:sbmtctrol');

}

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Answers (1)

Vishu Beniwal 135 on at

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RE: How to get the caller menuitemName in custom workflow submit button.

Hi I already tried this but didn't work.
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WillWU 22,350 on at

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RE: How to get the caller menuitemName in custom workflow submit button.

Hi Vishu,

Please try to change

formRun.args().menuItemName() == menuItemActionStr(LSSalesOrderReturnWFTypeSubmitMenuItem)

to

formRun.args().menuItemName() == menuitemdisplaystr(YourFormMenuItem)