When I try to compile PHP and apache v.1.3.4 with mysql I found a compiling error in the folowing statement:
Step 1:
cd apache_1.3.4
./configure --prefix=/<localpath>
result: OK
Step 2:
cd ../php-3.0.x
./configure --with-mysql --with-apache=../apache_1.3.4 --enable-track-vars
result: OK
Step 3:
make
result:
 gcc -g -O2 -O2   -I. -I.   -I../apache_1.3.4/src/include -I../apache_1.3.4/src/os/unix -I/usr/local/include/mysql <...>
bn=language-parser.tab.o: Command not found.
*** Error code 1 (bu21)
----
list from Makefile:
.c.o:
        $(CC) $(CFLAGS) -c $< -o $@
        @bn=`echo $@ | sed -e 's#functions/##'`; test -f $@ || \
          (test "$@" != "$$bn" && test -f "$$bn" && mv $$bn $@)

functions/number.o: functions/number.c
        $(CC) $(CFLAGS) -w$(WARNING_LEVEL) -c $< -o $@
        @bn=`echo $@ | sed -e 's#functions/##'`; test -f $@ || \
          (test "$@" != "$$bn" && test -f "$$bn" && mv $$bn $@)
----
I think the problem is with @bn=`...  , but I don't know how to solve it. Please, would you be so kind as to help me?
Thanks,
JV (vanek@czu.cz)