Deserialize form from JSON data using jQuery

Written by  on July 10, 2011 

Recently, I got task to make feature which will make possible for user to update form data based on some unique “search” field (like VIN or SSN). I wanted to make this short and simple, and above all reusable. So, I made some investigation this weekend and came up with simple solution to use same model object that is used as type of view in asp.net mvc. It is simple javascript function:

 

var formPrefix = 'myFormPrefix';
$('sometextbox').change(function () {
 var myValue = $(this).val()
 $.ajax({ type: 'POST',
  url: 'ajaxcontroller/getcarbysomething',
  data: { something: myValue },
  success: function (data) {
  if (data){
   $("input[name*='" + formPrefix +"'],select[name*='" + formPrefix + "']").each(function () {
    var itemName = $(this).attr("name");
    var prop = itemName.substring(itemName.indexOf(formPrefix) + formPrefix.length);
    var itemValue = data.person[prop];
    if (itemValue && itemValue.substring &&
      itemValue.substring(0, 1) == "/" &&
      itemValue.substring(itemValue.length - 1) == "/") {
         itemValue = "new" + itemValue.substring(1, itemValue.length - 1);
         itemValue = eval(itemValue);
         $(this).datepicker("setDate", eval(itemValue));
    }
    else if (itemValue) {
     $(this).val(itemValue);
    }
   });
 
   }
  }
 });
});

So, to use this you need to make controller action that returns JsonResult and Json(model) instead of View(model). If form is partial view (preffered) or has explicit prefix, formPrefix is place to set it. Each form input element’s name is used to index json object returned from controller and value (if any) is set back into input element.

Category : ASP.NET MVCjQuery

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