我正在尝试自己在Mac OS X Mavericks上编译PHP。
这是configure命令:
./configure --prefix=/opt/local/php --with-config-file-path=/opt/local/php/conf
--with-apxs2=/opt/local/apache/bin/apxs --with-mysql=/opt/local/mysql
--with-mysql-sock=/tmp --with-libedit --with-pcre-dir=/opt/local
--enable-bcmath --enable-mbstring --enable-sockets --enable-zip --with-bz2
--with-curl --with-jpeg-dir=/opt/local --with-png-dir=/opt/local
--with-freetype-dir=/opt/local --with-gd=/opt/local
当我尝试d
sudo make
时,出现此错误:Undefined symbols for architecture x86_64:
"_XpmLibraryVersion", referenced from:
_zm_info_gd in gd.o
ld: symbol(s) not found for architecture x86_64
clang: error: linker command failed with exit code 1 (use -v to see invocation)
PHP版本是5.5.5 ,而 libgd 是使用以下配置编译的:
./configure --prefix=/opt/local --with-png=/opt/local --with-jpeg=/opt/local
--with-tiff=/opt/local --with-freetype=/opt/local
没有
--with-gd
标志,PHP编译就很好。此错误是什么意思,我该如何解决?
最佳答案
您可以传递--with-xpm-dir = no来配置哪些应禁用xpm支持,如果不能解决该问题,请尝试为Mac OS X安装X11 Update。而且,您可能应该在安装该更新后重新编译libgd(包括libXpm)
关于php - PHP make失败,出现错误: Undefined symbols _XpmLibraryVersion,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/19895312/