Bernhard Seckinger
2017-01-20 10:21:48 UTC
Hi,
I've got a program, that contains some php-script frontend (cli not web)
(and other stuff, which is irrelevant here). I've put the php-scripts into
$pkgdatadir. Now I'd like to have a link from $bindir to the main script i.e.
ln -s ${pkgdatadir}/croco.php ${bindir}/croco
To do this I've added to the Makefile.ac the following:
install-exec-local:
mkdir -p ${bindir}
ln -s ${pkgdatadir}/croco.php ${bindir}/croco
When using "make install" this works. But when I run "make distcheck" I get an
error, telling that I'm not allowed to create the ${bindir}. I've allready
tried to replace the mkdir command with
${srcdir}/../install-sh -d ${bindir}
which is probably architecture-independend, but I still get a similar error.
Does anyone know, how to do this?
Thanks, Berni
I've got a program, that contains some php-script frontend (cli not web)
(and other stuff, which is irrelevant here). I've put the php-scripts into
$pkgdatadir. Now I'd like to have a link from $bindir to the main script i.e.
ln -s ${pkgdatadir}/croco.php ${bindir}/croco
To do this I've added to the Makefile.ac the following:
install-exec-local:
mkdir -p ${bindir}
ln -s ${pkgdatadir}/croco.php ${bindir}/croco
When using "make install" this works. But when I run "make distcheck" I get an
error, telling that I'm not allowed to create the ${bindir}. I've allready
tried to replace the mkdir command with
${srcdir}/../install-sh -d ${bindir}
which is probably architecture-independend, but I still get a similar error.
Does anyone know, how to do this?
Thanks, Berni
--
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