How to know if opened form is quick create form from JavaScript?

(0) Share

Report

Posted on by PoTaTo

I used Xrm.Utility.getGlobalContext().getQueryStringParameters().pageType to check form type, but getQueryStringParameters() looks like deprecated. I don't see the needed in formContext.data.attributes. I've seen ways with additional boolean entity field, checking if there is confirm button that is on quick create form or if there is no some field. Are there any other ways that more like using build-in function?

I have the same question (0)

All responses (4)

Answers (2)

Suggested answer

Community Member on at

Like (0)

Report

RE: How to know if opened form is quick create form from JavaScript?

Hi,

You can use this code to get it: formContext.context.getQueryStringParameters().pageType .

Result:

Was this reply helpful? Yes No
Ram Prakash 2,287 on at

Like (0)

Report

RE: How to know if opened form is quick create form from JavaScript?

Hello,

You can use formContext.context.getQueryStringParameters().pageType

Link for your reference :

docs.microsoft.com/.../gg334511(v=crm.7)

Was this reply helpful? Yes No
Verified answer

Pradeep Rai 5,489 Moderator on at

Like (0)

Report

RE: How to know if opened form is quick create form from JavaScript?

Hi,

I would suggest two ways.

1. CRM Function - In our script we can write below code to get formType:

Code:

function onLoad(executionContext)

{

var contextObj=executionContext.getContext();

var pageType=contextObj.getQueryStringParameters().pageType

}

This is best way to achieve this.

2. Add one more parameter in script function and while registering the script pass the formName as:

function onLoad(executionContext,formType){

}

Was this reply helpful? Yes No
Verified answer

meelamri 13,216 User Group Leader on at

Like (0)

Report
RE: How to know if opened form is quick create form from JavaScript?

Hi,

you can get this information by performing a retrieve on the systemForm table using the GUID of the current form.

To get the GUID of the current form, you can use the following code:

formItem = formContext.ui.formSelector.getCurrentItem();

formItemId = formItem.getId();

You can retrieve the form details by using the following code:

Xrm.WebApi.retrieveRecord('systemform', "f7d3b417-6220-42ac-9aec-228621129f3b", "?$select=type").then( function success(result) { console.log(result) }, function (error) { console.log(error.message); // handle error conditions } );

Result:

Good luck !

Was this reply helpful? Yes No