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

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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?

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Community Member on at

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Hi,

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

Result:

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Ram Prakash Duraisamy 2,289 on at

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Hello,

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

Link for your reference :

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

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Pradeep Rai 5,489 Moderator on at

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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){

}

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meelamri 13,218 User Group Leader on at

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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 !

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