Php 在没有完整路径的情况下将内容添加到ZIP不起作用

Php 在没有完整路径的情况下将内容添加到ZIP不起作用,php,linux,zip,Php,Linux,Zip,我正在创建Epub包,我需要按照以下顺序进行打包 $dest = "ebooks/pack/"; exec("zip -DX0 ".$dest."/book.zip ".$dest."/mimetype "); exec("zip -DrX9 ".$dest."/book.zip ".$dest."/META-INF ".$dest."/OEBPS"); 所以当我打包时,Zip存档结构如下所示 ebooks/pack/mimetype ebooks/pack/META-INF ebooks/

我正在创建Epub包,我需要按照以下顺序进行打包

$dest = "ebooks/pack/";
exec("zip -DX0 ".$dest."/book.zip ".$dest."/mimetype ");
exec("zip -DrX9 ".$dest."/book.zip ".$dest."/META-INF ".$dest."/OEBPS");
所以当我打包时,Zip存档结构如下所示

ebooks/pack/mimetype 
ebooks/pack/META-INF
ebooks/pack/OEBPS
mimetype 
META-INF
OEBPS
但我需要文件在家中,例如,当我解压缩时,它应该如下所示

ebooks/pack/mimetype 
ebooks/pack/META-INF
ebooks/pack/OEBPS
mimetype 
META-INF
OEBPS
我还试图通过
cd电子书/pack/
导航,这会引发错误

zip warning: name not matched: mimetype 
我甚至使用了完整路径,比如
home/user/public\u html/ful

exec("cd home/sam/public_html/domain.com/".$dest."/; zip -DX0 ".$dest."/book.zip  mimetype 2&1",$output); 
我甚至尝试了
chdir
,它显示了文件所在的正确路径

echo getcwd() . "\n";
chdir('home/sam/public_html/domain.com/".$dest."/');
echo getcwd() . "\n";