link to url

[Former Member]

hi,

In a webdynpro application I have placed an ui element link to url.The reference property of this ui element is mapped with a context variable of type string.In the init method I have set the context variable with the corresponding url.

Now when i deploy the webdynpro application and click the link , the link is opened in another new window.But i want the link to be opened in the same window of my application .How can i do this?what value should i give for the target property of the ui element?

please help..

Thanks in advance,

shami.

[Former Member · Accepted solution]

Regards,ANilkumar

[Former Member · Accepted solution]

hi,

I have given all the four options like blank ,self,_parent,_top to the target property that was specified . But still the link is opening in a new window. I am using NWDS7.0.06.I think here we cannot use _self ..... in 2004s.

In the following link it is specified that self is not supported

<a href="http://help.sap.com/saphelp_nw2004s/helpdata/en/9f/656442a1c4de54e10000000a155106/frameset.htm">http://help.sap.com/saphelp_nw2004s/helpdata/en/9f/656442a1c4de54e10000000a155106/frameset.htm</a>

In linktourl api page...

How can make the link open in the same window?

please help...

Thanks in advance.

shami

[Former Member · Accepted solution]

You are right.

Check this

Regards,Anilkumar

[Former Member · Accepted solution]

hi,

It is given that <b>self is no longer supported</b>. So <b>Use exit plugs instead and specify the URL there.</b>

Where will i specify the url in the exit plug?

I mean exit plug means outboundplug of a view.I am right....

Please help..

Thanks n advance,

shami.

[abhimanyu_lagishetti7 · Accepted solution]

Hi Shami

1.Define an exit plug in the interface view of your Web Dynpro component:

a. Enter a name and check the checkbox Exit Plug.

b. Enter the name Url for the parameter and select string as the type.

This parameter is case-sensitive and must be entered exactly this way.

2. In your view, define the action Exit and add the event handler onActionExit().

3. In the view layout, define a button or a link and bind its onAction event to the Exit action.

4. In your view, define the use of the interface view controller of your Web Dynpro component under Properties.

wdThis.wdGetTimeCompInterfaceViewController().wdFirePlugExit("your URL");

which will redirect to the page with in the same window

Regards

Abhimanyu L

[Former Member · Accepted solution]

hi,

Does exit plug mean outbound plug?

Thanks in advance,

shami.

[abhimanyu_lagishetti7 · Accepted solution]

Hi Shami,

Exit plug is the outbound Plug for Interface View,

for every window you create there will be an Interface View under Component Interface.

double click the Interface View, go to plugs tab.

press new button adjacent to outbound plugs,

a dialog appears enter the name of the outbound plug and select exit radio button

Regards,

Abhimanyu L

[Former Member · Accepted solution]

hi,

Thanks I got it.

I have linked an action to the button, and in the action i have written the following code.

wdThis.wdGetTestURLWindowInterfaceViewController().wdFirePlugUrl("http://www.google.co.in/");

but when i click on the button it gives application expired error

please help.

Thanks in advance,

shami.

[abhimanyu_lagishetti7 · Accepted solution]

Hi Shami

On clicking the button, you are redirected to google's page?

if your answer is No

Your Parameter name of the exit plug is "Url" <b>U is capital here</b>.

if Yes

Exit plug exits from the applicaiton you no longer can access the page.

Regards,

Abhimanyu L

[Former Member · Accepted solution]

Hi,

Still do u have the problem .

Better to see the Inter navigationapplication in examples that explais the same requirement.

But the change is only there using the Button instead of the LinktoURL.

https://www.sdn.sap.com/irj/sdn/downloaditem?rid=/webcontent/uuid/f3b0ad90-0201-0010-4995-9bdb1849eb... [original link is broken]

Try this,

Thanks,

Lohi.

[Former Member · Accepted solution]

Hai,

If you are unable to use exit plug.

I will give you alternate solution.

create a out bound plug to your view

that create a new view -->create frame ui element.bind url to that frame ui element.

create a inbound plug to this view.

connect these two plugs

in the action of the button

fire the out bound plug of this view. it will open the newly created view with frame.

make sure that frame height and width are to be fit to Browser.

regards,

Naga

[manuel-jeremy_schaffner · Accepted solution]

Hi Shami,

how did you solve the problem?

Using ExitPlugs does not seem to work inside the SAP Portal:

com.sap.tc.webdynpro.services.exceptions.WDRuntimeException: Exit-Plug must no be triggered with an URL when running in portal. Use portal navigation instead to navigate to another application!

Looks like I have to create an iView with URL-Parameters and then use PortalNavigation (EPCM) or the IFrame-Solution.

Regards,

Manuel

[Former Member]

hi!!

keep the target property as none. don't select any of the 4 options. keep it empty.... do let me know if the problem is solved....

[former_member189631]

Hi Shami,

Please implement This Logic:

1)Use the UI element called Link To Action.

2)In the onAction Method U can call the Exit Plug to Display ur Paticular Url.

<b>Procedure :</b>

i.In the Component Interface View you have to define an exit plug called GotoUrl, together with parameter Url of type String.

ii.Write the following code in You Link to Action UI's Action wdThis.wdGetYourInterfaceViewController().wdFirePlugGotoUrl("http://www.sap.com");

iii)Goto Ur View Where ur using the Link to Action Ui element.

Click Properties---> Add ---> select Ur component Interface View

This will Solve Ur Problem

Regards,

Ramganesan K.