PeterE
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| Posted: 03/25/2004, 8:32 AM |
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This doesn't work:
$sActionFileName = "http://www.hotspring.nl/html/contact_succes.html target=_top";
This does:
$sActionFileName = "http://www.hotspring.nl/html/contact_succes.html";
So what is wrong here?
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peterr
Posts: 5971
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| Posted: 03/25/2004, 11:53 AM |
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Nothing is wrong.
The $sActionFileName is a variable that must contain the file name, nothing else.
_________________
Peter R.
YesSoftware Forums Moderator
For product support please visit http://support.yessoftware.com |
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PeterE
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| Posted: 03/26/2004, 2:06 AM |
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One mistake that I made is that it should in fact look like this (with " in the right places):
$sActionFileName = "http://www.hotspring.nl/html/contact_succes.html" target="_top";
But then I get this error:
Parse error: parse error, unexpected T_STRING
So there is definetely something wrong with it! Another suggestion?
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PeterE
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| Posted: 03/26/2004, 2:12 AM |
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Ah, I suddenly realised that you cannot combine a http://www.etcetera with a target=etcetera! That only works for a href type link!
So how I can break out of a frame with $sActionFileName is till a puzzle for me, so if anyone knows how to do that, please react!
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peterr
Posts: 5971
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| Posted: 03/26/2004, 2:39 AM |
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You cannot do this on the server, meaning that the sever cannot open a new page in user's browser. You'd need to do this on the client (HTML or JavaScript). For example you can create a page that does nothing else but automatically opens a new Window via JavaScript, then you can redirect users to that special page from the server.
The exact implementation depends on the specific case and I'm not exactly sure what and why you're trying to do. Hopefully the above information can help though.
_________________
Peter R.
YesSoftware Forums Moderator
For product support please visit http://support.yessoftware.com |
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Michael Davidson
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| Posted: 02/02/2005, 6:54 AM |
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Scouring the archives, I put together this solution.
Creating links that open to a new window
PHP4 with templates
Fields:
site_url_c = contains the URL
site_name_c = contains the name of the site
1. Create a new field (Field1), set it's type in the fields list as Label.
2. Open the Field Properties editor and set Data Type=Text and check the
Html box.
3. Add the DB fields "site_url_c" and "link_name" in the fields list and set both their types in the fields list to Hidden.
4. Go to the Form Properties, Events tab. In the Before Show event put the
following code:
$fldField1='<a href="' .$fldsite_url_c. '" target="blank">' .$fldsite_name_c. '</a>';
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