Question

我在我的多国师协会申请中,有一个令人信服的问题需要解决。我没有 Java言,我有3个意见,每个步骤都有一个意见。

The scenario is: I have a 3 step process. S1 posts to S2 and S2 posts to S3. If i have a model validation error, the error is catched on the next view because the data was already posted there. My code looks like: return(View, modeldata) if ok ELSE return ("PreviousView", modeldata). The problem is my URL that remains in the post url requested

Answer 1

在所有情况下,我都会把每个表格张贴到原来的控制人员/行动上,然后转入适当的下一个步骤......即。

[HttpGet]
public ActionResult StepOneNameHere() {
    StepOneModel model = new StepOneModel();

    // Populate any bits on the model...
    return View(model);
}

[HttpPost]
public ActionResult StepOneNameHere(StepOneModel model) {
    // Validation

    if (ModelState.IsValid) {
        // You can add logic to choose the next step, for example if they have
        // selected an option that requires an additional step in the wizard or
        // if you can skip a step in the wizard
        return RedirectToAction("StepTwoNameHere");
    } else {
        return View(model);
    }
}

友情链接