Linux 如何压缩没有完整路径的文件夹
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How to zip folder without full path
提问by
firstly I don't which is the correct stackExchange site to post this question if the question is for other stack-site please remove my question.
首先，我不知道哪个是发布此问题的正确 stackExchange 站点，如果问题是针对其他堆栈站点的，请删除我的问题。
Now let talk about the question:This situation is I have a file which is located in: /home/user/public_html/folder-one/folder-two/folder-three/file.phpand the file must create archive of folder /folder-onewith all files and sub-folders of /folder-one. I create the archive with system function (exec(), shell_exec() or system()) and it works perfect. My code is:
现在让我们谈谈这个问题：这种情况是我有一个文件位于：/home/user/public_html/folder-one/folder-two/folder-three/file.php并且该文件必须创建文件夹的存档，/folder-one其中包含/folder-one. 我用系统函数（exec()、shell_exec() 或 system()）创建了存档，它工作得很好。我的代码是：
<?php
$output = 'zip -rq my-zip.zip /home/user/public_html/folder-one -x missthis/\*';
shell_exec($output);
?>
But when I download and open the archive the archive include the sub-folders as /home; /user; /public_htmlbut this folders are totally unneeded and I wanna know how to create zip without them.
但是当我下载并打开档案时，档案包括子文件夹为/home；/user; /public_html但是这个文件夹完全不需要，我想知道如何在没有它们的情况下创建 zip。
When I try with something like this $output = 'zip -rq my-zip.zip ../../../folder-one -x missthis/\*';but then when I open the archive (on Windows 7 based OS) the name of folder-one is ../folder-one
当我尝试使用这样的东西$output = 'zip -rq my-zip.zip ../../../folder-one -x missthis/\*';但是当我打开存档（在基于 Windows 7 的操作系统上）时，文件夹一的名称是../folder-one
Postscript: It will be better if somebody gives me correct$outputthe make zips on windows based hosting plans.
后记：如果有人给我更正基于 Windows 的托管计划上的 make zips会更好$output。
Best regards, George!
最好的问候，乔治！
采纳答案by vkorchagin
By default, zip will store the full path relative to the current directory. So you have to cdinto public_htmlbefore running zip:
默认情况下，zip 将存储相对于当前目录的完整路径。所以你必须在运行之前cd进入：public_htmlzip
$output = 'cd /home/user/public_html; zip -rq my-zip.zip folder-one -x missthis/\*';
shell_exec($output);
回答by pgmank
There is a better way, without doing a cd. As stated herethere is an option to ignore full paths. This is -jor --junk-paths. So you could do:
有一种更好的方法，无需执行cd. 如前所述这里有忽略完整路径的选项。这是-j或--junk-paths。所以你可以这样做：
<?php
$output = 'zip -jrq my-zip.zip /home/user/public_html/folder-one -x missthis/\*';
shell_exec($output);
?>
