如何在jQuery AJAX中使用参数调用Rails表单助手？（超级简单）

Question

I'm new on Rails, and i think this is super simple, just not used to the syntax.

this is the jQuery part,

$.ajax({
    type: "get",
    contentType: "application/json; charset=utf-8",
    url: "/users/search_users",
    data: {
        name: name
    },
    dataType: "text",
    success: function (result) {
        var test = "<%=get_user_list_html(result)%>";
        if (result == "User not found") {
            alert("User not found");
        } else {
            //console.log(result);
            var peopleData = jQuery.parseJSON(result);
            var resultHTML = "<tr>";
            resultHTML += "<th></th><th style='display:none'>User ID</th>" + "<th>First Name</th><th>Last Name</th><th>Email Address</th>" + "<th style='display:none'>Time Zone</th>";
            resultHTML += "</tr>";
            $.each(peopleData, function (index, obj) {

                resultHTML += "<tr>";
                resultHTML += "<td><input type='checkbox'></td>" + "<td style='display:none;'>" + obj.id + "</td>" + "<td>" + obj.firstname + "</td>" + "<td>" + obj.lastname + "</td>" + "<td>" + obj.email + "</td>" + "<td style='display:none;'>" + "Etc/GMT+9 (Japan)" + "</td>";
                //consider now
                resultHTML += "</tr>";
            });
            $("#internal_table").html(resultHTML);
        }
    },
    error: function () {
        window.alert("something wrong!");
    }
});

in here, i'm going to call get_user_list_html, which is in the helper.

But my problem is, how can i use the result from the respone, i have to send this as a param?

if i just put it like that, it says undefined var.

Answer 1

Ruby code must be executed on the server side, this line: var test = "<%=get_user_list_html(result)%>"; won't work after an ajax call, and it should be in a .erb file.

You can have a ruby + javascript view template with .js.erb extension but ruby code is always executed before your document loads.

I don't know what the helper does but this is not the way to do it.

Answer 2

Solution 1: Render a HTML partial instead of JSON from server. That's make things much easier in you case.

Solution 2: Use a template to render JSON response, say jQuery template or Underscore template.

The usage in your case is not good. You missed better part in both server and client side.