Hi,
I want to make shortcut of my application in start-programs. For this I
written following code
<Directory Id="TARGETDIR" Name="SourceDir" >
<Directory Id="ProgramFilesFolder" Name="PFiles" >
<Directory Id="INSTALLDIR" Name="ALM6">
</Directory>
<Directory Id="ProgramMenuFolder">
<Directory Id="ApplicationProgramsFolder" Name="ALM6"/>
</Directory>
</Directory>
</Directory>
<Shortcut Id="ApplicationStartMenuShortcut" Name="Bancware ALM 6.0"
Description="Bancware ALM 6.0"
Target="[INSTALLDIR]ALMGUI\bwr.ALM.exe"
WorkingDirectory="[INSTALLDIR]ALMGUI "/>
However I am facing problem with setting WorkingDirecotry.
My working directory is [INSTALLDIR]ALMGUI. However it gives me error l
" Error 3 The Shortcut/@WorkingDirectory attribute's
value, '[INSTALLDIR]ALMGUI', is not a legal identifier. Identifiers may
contain ASCII characters A-Z, a-z, digits, underscores (_), or periods
(.). Every identifier must begin with either a letter or an underscore.
d:\WiX_prog\WixProject1\WixProject1\Product.wxs 83 1
WixProject1
"
So I tried
<Property Id="StartinDir" Value ='[INSTALLDIR]ALMGUI'></Property>
<Shortcut Id="ApplicationStartMenuShortcut" Name="Bancware ALM 6.0"
Description="Bancware ALM 6.0"
Target="[INSTALLDIR]ALMGUI\bwr.ALM.exe"
WorkingDirectory="StartinDir"/>
However it sets "start in" - [INSTALLDIR]ALMGUI' value in shortcut
property.
How can I set the directory [INSTALLDIR]ALMGUI to WorkingDirectory
Thanks,
Rahul Ekbote
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