I want to write class codes to open form. How can I make changes the codes to work?
My Class Codes:
public void formac(Form frm, Form formName)
{
formName frm = new formName();
frm.Show();
}
Button Click Codes:
openFormClass MyClass = new openFormClass();
MyClass.formac(frm,Form1);
Hello and welcome to StackOverflow!
Unfortunately your question is not very concise on what you want to achieve. Maybe have a look at How do I ask a good question?.
From what you've shown, I'm guessing that you're just want to show another instance of Form1 on a button click. Then the following should be sufficient:
private void button1_Click(object sender, System.EventArgs e)
{
var frm = new Form1();
frm.Show();
}
i want to write code to open form in class (parameter will be "Form Name")
Related
I have a user control, I want to make if conditions in the ascx, and upon that call a function in the user control and return data from the method to be displayed as html in the ascx. Any ideas? I am still newbie to asp.
Just drop a literal onto your user control. Then in your code behind, it'll look something like this:
private void Page_Load(object sender, System.EventArgs e)
{
// Make your IF conditions here
MyFunction();
}
private void MyFunction()
{
string myHtml = "<div>My Html Code!</div>";
literal1.Text = myHtml;
}
That should render your HTML.
I need some help. I've been reading this site for days, and have read a lot of tips about controlling for example a button's property from another form. There's even a video on Youtube, which works to me as stand alone, but when I implenet it in my application it throws a NullReferenceException.
Let's say that I have a toolstrip menu on Form1. A click on the Kalibracio option opens the second form (also called Kalibracio - not Form2). Then, click on Proba in the menu should disable an ordinary button on the Kalibracio form which propery is set to public. The code on Form1 is as follows:
private void kalibracioToolStripMenuItem_Click(object sender, EventArgs e)
{
Kalibracio Kalibr = new Kalibracio(this);
Kalibr.Owner = this;
Kalibr.Show();
}
private void probaToolStripMenuItem_Click(object sender, EventArgs e)
{
if (Application.OpenForms.OfType<Kalibracio>().Any())
(this.Owner as Kalibracio).button1.Enabled = false;
// the above line throws a NullReferenceExcteption if Kalibracio form is open (Kalibracio is null)
}
What am I missing?
LoL just had to declare globaly an instance of Kalibracio, open it and then access it's properties from all other methods.
I tried this approach at first, but my problem was that I was creating and instance locally, then I had to create another one in some other method because I couldn't address the former one created locally, and ofc it didn't work...
I am using GWT 2.4. I have a Suggestbox and I have a requirement to hide the suggestion list under certain cases. The context is as below.
After user selects a suggestion from suggestion list, I am populating two other text box fields, with values corresponding to the selection. For example, suppose the suggestbox contains user-names, and user selects a user-name from suggestions, then other two fields, say user address and email are populated in two other text boxes. These two fields are read only now. Then user clicks on an 'Edit' button. Now the user can edit either user- name ( ie edit in suggestion box), user address and email. It doesn't make sense to show the suggestions again when the user is editing the user-name, since the user has already selected the user and decided to edit it. In a nutshell my SuggesBox should behave as a normal text box. I tried following code, (I know hideSuggestionList() is deprecated) but its not working.
display.getSuggestBox().hideSuggestionList();
Reading the javadoc for hideSuggestionList() it is said that, "Deprecated. use DefaultSuggestionDisplay.hideSuggestions() instead". I don't know how to use DefaultSuggestionDisplay, and I'm using SuggestBox with 'MultiWordSuggestOracle'.
Thanks for helping me out!!
What you can do is simply swap the SuggestionBox with a normal TextBox when the user clicks edit and back when edit is closed. Also because if you would hide the suggestions list, it still queried from the server. By swapping the widget you don't have to care about side effects. SuggestionBox itself uses also a TextBox and thus for the user it's not visible the widget has changed.
If you don't use your own SuggestionDisplay, then this should Just Workâ˘:
((DefaultSuggestionDisplay) suggestBox.getSuggestionDisplay()).hideSuggestions();
Here is the Solution
My Entry Point Class
public class SuggestionEntryPoint implements EntryPoint {
#Override
public void onModuleLoad() {
SuggestBoxWidget suggestBoxWidget = new SuggestBoxWidget();
RootPanel rootPanel = RootPanel.get();
suggestBoxWidget.createOracle();
suggestBoxWidget.createWidgetAndShow(rootPanel);
rootPanel.add(suggestBoxWidget);
DOM.getElementById("loader").removeFromParent();
}
}
And here is my Widget
public class SuggestBoxWidget extends Composite {
private TextBox textSuggestBox = new TextBox();
private SuggestBox suggestBox = null;
DefaultSuggestionDisplay suggestionDisplay = new DefaultSuggestionDisplay();
MultiWordSuggestOracle suggestOracle = new MultiWordSuggestOracle();
private static SuggestBoxWidgetUiBinder uiBinder = GWT
.create(SuggestBoxWidgetUiBinder.class);
interface SuggestBoxWidgetUiBinder extends
UiBinder<Widget, SuggestBoxWidget> {
}
public SuggestBoxWidget() {
initWidget(uiBinder.createAndBindUi(this));
}
public void registerEvents(){
suggestBox.addKeyUpHandler(new KeyUpHandler() {
#Override
public void onKeyUp(KeyUpEvent event) {
if(suggestBox.getText().equalsIgnoreCase("1")){
suggestionDisplay.hideSuggestions();
}
}
});
}
public void createWidgetAndShow(HasWidgets container){
suggestBox = new SuggestBox(suggestOracle,textSuggestBox,suggestionDisplay);
container.clear();
container.add(suggestBox);
registerEvents();
}
public void createOracle(){
for(int i=1;i<=100;i++){
suggestOracle.add(i+"");
}
}
}
Actually you have to create a SuggestBox with 3 Parameters to the Constructor.
I'm using GWT 2.4 with JUnit 4.8.1. When writing my class that extends GWTTestCase, I want to simulate clicking on a button on the page. Currently, in my onModuleLoad method, this button is only a local field ...
public void onModuleLoad() {
final Button submitButton = Button.wrap(Document.get().getElementById(SUBMIT_BUTTON_ID));
...
// Add a handler to send the name to the server
GetHtmlHandler handler = new GetHtmlHandler();
submitButton.addClickHandler(handler);
How do I simulate clicking on this button from the GWTTestCase? Do I have to expose this button as a public member accessor is there a more elegant way to access it? Here is what I have in my test case so far ...
public class GetHtmlTest extends GWTTestCase {
// Entry point class of the GWT application being tested.
private Productplus_gwt productPlusModule;
#Override
public String getModuleName() {
return "com.myco.clearing.productplus.Productplus_gwt";
}
#Before
public void prepareTests() {
productPlusModule = new Productplus_gwt();
productPlusModule.onModuleLoad();
} // setUp
#Test
public void testSuccessEvent() {
// TODO: Simulate clicking on button
} // testSuccessEvent
}
Thanks, - Dave
It can be as easy as buttonElement.click() (or ButtonElement.as(buttonWidget.getElement()).click(), or ButtonElement.as(Document.get().getElementById(SUBMIT_BUTTON_ID)).click())
But remember that a GWTTestCase doesn't run in your own HTML host page, but an empty one, so you'll first have to insert your button within the page before simulating your module's load.
gwt-test-utils seems to be the perfect framework to answer your need. Instead of inheriting from GWTTestCase, extend the gwt-test-utils GwtTest class and implement your click test with the Browser class, like shown in the getting starting guide :
#Test
public void checkClickOnSendMoreThan4chars() {
// Arrange
Browser.fillText(app.nameField, "World");
// Act
Browser.click(app.sendButton);
// Assert
assertTrue(app.dialogBox.isShowing());
assertEquals("", app.errorLabel.getText());
assertEquals("Hello, World!", app.serverResponseLabel.getHTML());
assertEquals("Remote Procedure Call", app.dialogBox.getText());
}
If you want to keep your button private, you'd be able to retrieve it by introspection. But my advice is to make you view's widgets package protected and to write your unit test in the same package so it could access them. It's more convinent and refactoring-friendly.
gwt-test-utils provide introspection convinence. For example, to retrieve the "dialogBox" field which could have been private, you could have do this :
DialogBox dialogBox = GwtReflectionUtils.getPrivateFieldValue(app, "dialogBox");
But note that using GwtReflectionUtils is not mandatory. gwt-test-utils allows you to use ANY java classes in GWT client side tests, without restriction :)
You can do it like this:
YourComposite view = new YourComposite();
RootPanel.get().add(view);
view.getSubmitButton.getElement().<ButtonElement>cast().click();
Here is what I'm doing, I have a large amount of reports, rather than copy and pasting the repeated code, I have the following..
public class ReportPage : System.Web.UI.Page
{
// Code
}
My report Pages have
public class MyReportPage : ReportPage
{
}
Problem I'm having is, I want to be able to access a control inside MyReportPage within ReportPage, for instance, I want all pages grid to have a specific property.
Inside ReportPage I tried
protected void Page_PreRender(object s, EventArgs e)
{
var obj = this.FindControl("reportGridName");
}
But unable to find the control, went through all the controls to see if I could it in, no luck..
Any ideas how to get the grid access on the inherit?
Page page = (Page)HttpContext.Current.Handler;
var obj = page.FindControl("reportGridName");
Does that work? I'm having the exact same problem and stumbled upon this link