Hi, I am trying formspider for while and I like it very much but I could not find asnwer for my problem. :) I created a small app and created an action "onLeftClick" which should open a specific URL. Is there any API how to open URL in the same window as my app running in target="_top"? I have found component htmlRenderer which opens URL in a specific formspider panel. I would like to open that URL outside the scope of my formspider app. Thanks a lot. :)

asked 30 Jul '12, 03:07

vojto's gravatar image

vojto
94
accept rate: 0%


Hi,

You can use API api_application.run with browserWindowName parameter:

api_application.run(in_weblink_tx => 'http://....', in_browserWindowName_tx => '_top')
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answered 30 Jul '12, 03:20

Ugur%20Kocak's gravatar image

Ugur Kocak ♦
5616
accept rate: 23%

Thanks a lot. :) exactly what I wanted :)

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answered 30 Jul '12, 04:41

vojto's gravatar image

vojto
94
accept rate: 0%

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Asked: 30 Jul '12, 03:07

Seen: 1,580 times

Last updated: 30 Jul '12, 04:41


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