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9.0.3 How do you make toastr work on postback?
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Joe Payne2  
#1 Posted : Friday, March 5, 2021 4:36:32 AM(UTC)
Joe Payne2

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I really like how the toastr notification works. But I cannot figure out how to make it work on my own admin page.

I have a simple form with one button to trigger some server-side code when the button is clicked:

Code:

                        @using (Ajax.BeginForm("LateShipments", "Automation", new AjaxOptions() { HttpMethod = "Post", OnSuccess = "ShowSavedSuccessMessage" }))
                        {
                            @Html.HiddenFor(model => model.JobName)
                            @Html.AntiForgeryToken()
                            <input type="submit" name="FireJob" value="Fire Job" class="btn btn-sm btn-primary" />
                        }


Here is my controller action:

Code:

        [HttpPost]
        [ValidateAntiForgeryToken]
        [ValidateAjax]
        public ActionResult LateShipments(LateShipmentsSettingsModel model)
        {
            // Enqueue the job
            BackgroundJob.Enqueue(() => LateShipmentHelper.ProcessLateShipments());

            // update user with result
            return Json(new { Success = true, msg = model.JobName + " job has been initiated.", model = model, Reload = false });
        }


and at the bottom of my view I have this javascript:

Code:

@section Scripts {
    <script type="text/javascript">
        /*FORM SUBMIT SUCCESS CALLBACK*/
        function ShowSavedSuccessMessage(data) {
            if (data == "success") {
                toastr.success(msg, "Success");
            }
        }
    </script>
    
}


My understanding is the Ajax form will call ShowSavedSuccessMessage() when the controller action is finished. And pass it my string data as a variable called 'msg'. So then toastr would then pop up using the msg string I passed in the action return json data.

But it just won't work and I can't figure out why not. It seems so damn simple. These MVC forms are so explicit in how they want to work.
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shaharyartiwana25816656  
#2 Posted : Friday, March 5, 2021 4:43:01 AM(UTC)
shaharyar

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Try using this code in your view file. Replace this with your code.

Code:
@section Scripts {
    <script type="text/javascript">
        /*FORM SUBMIT SUCCESS CALLBACK*/
        function ShowSavedSuccessMessage(data) {
            if (data.Success) {
                toastr.success(data.msg, "Success");
            }
        }
    </script>
    
}


You can see the response data object in the console by writing the following code in your callback function. Just for testing/debugging.

Code:
console.log(data)
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thanks 1 user thanked shaharyar for this useful post.
Joe Payne2 on 3/5/2021(UTC)
Joe Payne2  
#3 Posted : Friday, March 5, 2021 5:28:40 AM(UTC)
Joe Payne2

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Holy cow that worked perfectly thank you!!
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Joe Payne2  
#4 Posted : Monday, March 8, 2021 4:51:32 AM(UTC)
Joe Payne2

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So what if you want to return a partial view AND fire toastr? Is that possible?
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shaharyartiwana25816656  
#5 Posted : Monday, March 8, 2021 6:36:37 AM(UTC)
shaharyar

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Yes, we have handled this case. Please see the following line in the post method of BuyProductDialog method in retail side product controller.

Code:
return Json(new { status = "HTML", html = RenderRazorViewToString("_BuyProductDialog", model) });


You have to then handle accordingly in your ajax onSuccess method.
Search the onBuyProductDialogSuccess method in the retail app.js file.
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thanks 1 user thanked shaharyar for this useful post.
Joe Payne2 on 3/8/2021(UTC)
Joe Payne2  
#6 Posted : Monday, March 8, 2021 11:43:50 AM(UTC)
Joe Payne2

Rank: Advanced Member

Groups: HelpDesk, Developers
Joined: 11/9/2018(UTC)
Posts: 564

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Was thanked: 26 time(s) in 25 post(s)
Awesome. I'll give that a try and report my results here to share with others. Thanks again!!
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