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building a menu ?
Posted: 15 Apr 2018 11:13
by yusef88
hi , my problem is the menu will only return the first extension
if I choose video menu. the echo will return mp4
any suggestion to fix this please?
Code: Select all
$menu = <<<MENU
Video;*.mp4;*.avi
Audio;*.mp3;*.wav
MENU;
$s = popupmenu("$menu");if !($s) { end 1==1; };
echo $s
Re: building a menu ?
Posted: 15 Apr 2018 11:34
by PeterH
In the expression Video;*.mp4;*.avi ";" is the separator "sep_item", so:
Video = caption
*.mp4 = data
*.avi = icon
You didn't mean it this way, I think.
Re: building a menu ?
Posted: 15 Apr 2018 12:26
by yusef88
PeterH wrote:In the expression Video;*.mp4;*.avi ";" is the separator "sep_item", so:
Video = caption
*.mp4 = data
*.avi = icon
You didn't mean it this way, I think.
you are right. I guess quoting "*.mp4;*.avi" will make it work
thanks
Re: building a menu ?
Posted: 15 Apr 2018 12:33
by highend
Then you get a quoted string...
The better way (as Peter already indicated) is:
Code: Select all
$menu = <<<MENU
Video|*.mp4;*.avi
Audio|*.mp3;*.wav
MENU;
$s = popupmenu($menu, 7:="|");
if !($s) { end true; };
echo $s;
or you use a different separator for the extensions in the first place...
Re: building a menu ?
Posted: 15 Apr 2018 12:50
by yusef88
dear highend
Do you know how wonderful you are?
Thank you
