javascript弹出窗口不会弹出,因为已经分配了绑定

问题描述 投票:0回答:1

c#'/ Clients / ClientSelectDataTable'被调用并返回。

      self.LookupClient = function () {
          $.ajax({
              type: 'POST',
              url: '/Clients/ClientSelectDataTable'
          }).done(function (msg) {
              //grab the partial from the controller
              $("#SelectClient").html('');
              $("#SelectClient").html(msg);
              //initialize the js
              theClientPicker = new ClientSelectModel('ClientSelectDataTable', '/API/GetAllClients', AssignClient, self.CancelClientSelect, "Select");
              ko.applyBindings(theClientPicker, $("#SelectClient")[0]);

              $.fancybox("#SelectClient", { //launch ClientSelect widget
                  modal: true,
                  afterClose: function () {
                      theClientPicker.Dispose();
                      ko.cleanNode($("#SelectClient")[0]);
                  }
              });
          });
      }

这里抛出异常ko.applyBindings(theClientPicker,$(“#SelectClient”)[0]);

...

    if (!sourceBindings) {
        if (alreadyBound) {
            throw Error("You cannot apply bindings multiple times to the same element.");

我可以捕获到异常,但是弹出窗口不起作用,因为未分配按钮上的事件。

这是一个新的theClientPicker,所以我看不到如何分配它。

javascript jquery knockout-3.0
1个回答
0
投票

发现了问题。这是3.4.2和3.5.0之间引入的缺陷或重大缺陷。恢复并冻结3.4.2可以修复它。

是否有适当的方式来举报?

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