Php 在没有完整路径的情况下将内容添加到ZIP不起作用
我正在创建Epub包,我需要按照以下顺序进行打包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/
$dest = "ebooks/pack/";
exec("zip -DX0 ".$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";