in reply to SCALAR ref error help!
means"${foo}"
but anything more complex and $... / ${ ... } becomes a derefence operator. Solutions:"$foo"
print qq|...contatc=|. uri_escape( $address ) .qq|...|;
printf qq|...contatc=%s...|, uri_escape( $address );
my $address_uri = uri_escape( $address ); print qq|...contatc=$address_uri...|;
print qq|...contatc=${\( uri_escape( $address ) )}...|;
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