Wicket 中的测试表单响应 [英] Test form response in Wicket
本文介绍了Wicket 中的测试表单响应的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我正在使用 Wicket,我想测试提交我的表单是成功还是找不到页面"错误.我怎么能做到这一点?下面是我的代码:
I am using Wicket and I want to test whether submitting my form result in a success or "page not found" errors. How could I achieve this? Below is my code:
HTML 代码
<form wicket:id="form" enctype='multipart/form-data'>
<div wicket:id="feedback"></div>
<input type="file" wicket:id="file"></input>
<br></br>
<span wicket:id="progress"></span>
<input wicket:id="save" type="submit" value="Save"></input>
<input wicket:id="cancel" type="submit" value="Cancel"></input>
</form>
Java 代码
public class TestPage extends WebPage {
private static final long serialVersionUID = 1L;
private FeedbackPanel feedback;
// TODO Add any page properties or variables here
/**
* Constructor that is invoked when page is invoked without a session.
*
* @param parameters
* Page parameters
*/
public TestPage(final PageParameters parameters) {
Form<?> form = new Form<String>("form");
add(form);
feedback = new FeedbackPanel("feedback");
feedback.setOutputMarkupPlaceholderTag(true);
form.add(feedback);
FileUploadField file = new FileUploadField("file");
file.setRequired(false);
form.add(file);
UploadProgressBar progress = new UploadProgressBar("progress", form);
form.add(progress);
AjaxFallbackButton cancel = new AjaxFallbackButton("cancel", form) {
private static final long serialVersionUID = 1L;
@Override
protected void onSubmit(AjaxRequestTarget target, Form<?> form) {
Session.get().getFeedbackMessages().add(this,
"Everything is ok", FeedbackMessage.INFO);
target.addComponent(form);
}
@Override
protected void onError(AjaxRequestTarget target, Form<?> form) {
super.onError(target, form);
target.addComponent(feedback);
}
};
cancel.setDefaultFormProcessing(false);
form.add(cancel);
AjaxFallbackButton save = new AjaxFallbackButton("save", form) {
private static final long serialVersionUID = 1L;
@Override
protected void onSubmit(AjaxRequestTarget target, Form<?> form) {
Session.get().getFeedbackMessages().add(this,
"Everything is ok", FeedbackMessage.INFO);
target.addComponent(form);
}
@Override
protected void onError(AjaxRequestTarget target, Form<?> form) {
super.onError(target, form);
target.addComponent(feedback);
}
};
form.add(save);
}
}
以下是我正在尝试编写的测试用例:
Below is the test-case that I am trying to write:
//start and render the test page
tester.startPage(TestPage.class);
//assert rendered page class
tester.assertRenderedPage(TestPage.class);
FormTester formTester = tester.newFormTester("form");
formTester.submit();
tester.assertNoErrorMessage();
我想点击提交表单并阅读提交的回复以进一步处理它.
I want to click to submit the form and read the response of the submission to further process it.
推荐答案
您可以发送指定使用哪个提交者的表单,例如:
You can send form specifying which submitter to use like:
formTester.submit("save");
然后你可以检查从测试者那里得到的响应:
Then you can check the response getting it from tester:
String responseTxt = tester.getLastResponse().getDocument();
有关详细信息,请参阅用户指南.
See user guide for more details.
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