blocks|key|606688|text|改用shell_exec()。exec不应该调用任何shell来执行您的程序。或者，您可以使用exec调用bash，如|type|unstyled|depth|inlineStyleRanges|entityRanges|data|606689|exec("/bin/bash+-c+\"echo+$foo+>+bar.txt'\"")|code-block|syntax|javascript|606690|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Try shell_exec() instead. exec should not invoke ANY shell to execute your program. Alternately, you can invoke bash with exec like 

<pre><code>exec("/bin/bash -c \"echo $foo &gt; bar.txt'\"")
</code></pre>

blocks|key|1263957|text|我认为问题是没有$PATH设置。尝试使用指向二进制文件的完整路径(即/bin/echo+)|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1263958|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

I think the problem is that there is no $PATH setup. Try using full paths to your binaries ie 
 /bin/echo

blocks|key|1263981|text|如果你想做的是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1263982|/usr/bin/mysql+--user=asdf+--password=asdf+mydb+<+./dump.sql|offset|length|style|CODE|1263983|然后，我想这是可行的(不管是什么shell)：|1263984|/usr/bin/mysql+--user=asdf+--password=asdf+mydb+<+/full/path/to/dir/dump.sql|1263985|entityMap^0|0|0|1O|0|0|0|24|0^^$0|@$1|2|3|4|5|6|7|N|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|O|8|@$D|P|E|Q|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|R|8|@]|9|@]|A|$]]|$1|J|3|K|5|6|7|S|8|@$D|T|E|U|F|G]]|9|@]|A|$]]|$1|L|3|-4|5|6|7|V|8|@]|9|@]|A|$]]]|M|$]]

If what you want to do is:

<code>/usr/bin/mysql --user=asdf --password=asdf mydb &lt; ./dump.sql</code>

Then I imagine this would work (regardless of shell):

<code>/usr/bin/mysql --user=asdf --password=asdf mydb &lt; /full/path/to/dir/dump.sql</code>

blocks|key|606710|text|我认为PHP在内部(或cmd.exe上)调用/bin/sh，而不管您使用的是哪个函数。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|606711|在Linux系统中的终端中运行ls+-l+/proc/$$/exe+%7C+sed+'s%25.*/%25%25'应该告诉您您使用的是什么shell，默认情况下这是我的Ubuntu终端中的bash。如果在Ubuntu中运行以下代码：|offset|length|style|CODE|606712|<?php

exec('ls+-l+/proc/$$/exe+%7C+sed+\'s%25.*/%25%25\'',+$res);
//+or:
//$res+=+shell_exec('ls+-l+/proc/$$/exe+%7C+sed+\'s%25.*/%25%25\'');
echo(json_encode($res));|code-block|syntax|javascript|606713|无论您使用哪个函数，结果都会是“破折号”。如果将/bin/sh改为bash而不是dash，则PHP的结果将更改为bash，但我不建议明显地这样做。|606714|长话短说，如果不修改整个服务器的默认shell或从源代码编译PHP，就不可能改变这一点。|606715|如果您想进一步解开这个谜团，它可能与ext/standard/proc_open.c和ext/standard/exec.c有关(所有这些功能都在其中、系统、密码等等)。在PHP源代码中，但老实说，它比我更难接受。|606716|entityMap^0|0|F|Y|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|V|8|@$D|W|E|X|F|G]]|9|@]|A|$]]|$1|H|3|I|5|J|7|Y|8|@]|9|@]|A|$K|L]]|$1|M|3|N|5|6|7|Z|8|@]|9|@]|A|$]]|$1|O|3|P|5|6|7|10|8|@]|9|@]|A|$]]|$1|Q|3|R|5|6|7|11|8|@]|9|@]|A|$]]|$1|S|3|-4|5|6|7|12|8|@]|9|@]|A|$]]]|T|$]]

I think PHP calls /bin/sh internally (or cmd.exe on Windows) regardless of which function you use.

Running <code>ls -l /proc/$$/exe | sed 's%.*/%%'</code> in your terminal in a Linux system should tell you what shell you're using, which is bash in my Ubuntu terminal by default. If you run the following in Ubuntu:

<pre><code>&lt;?php

exec('ls -l /proc/$$/exe | sed \'s%.*/%%\'', $res);
// or:
//$res = shell_exec('ls -l /proc/$$/exe | sed \'s%.*/%%\'');
echo(json_encode($res));
</code></pre>

you will get ["dash"] as the result, regardless of which function you use. If you change /bin/sh to point to bash rather than dash, the result from PHP changes to bash, but I don't recommend doing that obviously.

So long story short it's probably not possible to change this without modifying your entire server's default shell or compiling PHP from source.

If you want to solve the mystery further, it's probably related to ext/standard/proc_open.c and ext/standard/exec.c (all these functions are in there, system, passthru, etc.) in the PHP source, but to be honest it beats me.

blocks|key|1296480|text|我对Debian+Docker+PHP映像有问题，这帮助了我：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1296481|https://www.reliableembeddedsystems.com/wiki/index.php?title=Make_bash_default_shell|offset|length|1296482|/bin/sh是破折号的一个符号链接，这可能更好，但不适用于我的脚本。|style|CODE|1296483|++++++++echo+"dash+dash/sh+boolean+false"+%7C+debconf-set-selections
++++++++dpkg-reconfigure+-f+noninteractive+dash|code-block|syntax|javascript|1296484|entityMap|0|LINK|mutability|MUTABLE|url^0|0|0|2C|0|0|0|7|0|0^^$0|@$1|2|3|4|5|6|7|V|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|W|8|@]|9|@$D|X|E|Y|1|Z]]|A|$]]|$1|F|3|G|5|6|7|10|8|@$D|11|E|12|H|I]]|9|@]|A|$]]|$1|J|3|K|5|L|7|13|8|@]|9|@]|A|$M|N]]|$1|O|3|-4|5|6|7|14|8|@]|9|@]|A|$]]]|P|$Q|$5|R|S|T|A|$U|C]]]]

I had a problem with Debian Docker PHP image, this helped me:
<a href="https://www.reliableembeddedsystems.com/wiki/index.php?title=Make_bash_default_shell" rel="nofollow noreferrer">https://www.reliableembeddedsystems.com/wiki/index.php?title=Make_bash_default_shell</a>
<code>/bin/sh</code> was a symlink to dash, which may be better but doesnt work with my scripts.
<pre><code> echo &quot;dash dash/sh boolean false&quot; | debconf-set-selections
 dpkg-reconfigure -f noninteractive dash
</code></pre>

I want to use php's exec() function on an ubuntu server.
The problem is, I alway get an error, that the command is not found. For example using 

<pre><code>exec("echo 123");
</code></pre>

prints

<blockquote>
 sh: /echo: not found
</blockquote>

To me, it looks like php is using the sh shell, when I want to be using bash.
I tried changing the shell for www-data in /etc/passwd, that didn't help either.

Does anybody have an idea where else the problem might be coming from or how I can change the shell for php's ubuntu user.

Thanks, Timo

<hr>

[EDIT]

Maybe this helps:

I call a bash script from ssh as timo, this script calls a php script, which uses exec. I know, it sounds weird, but it's part of a bigger development environment...

The point is, I'm not ever certain, as which user the script inside exec is executed.

<hr>

[EDIT]

By now I figured out that there must be another rights problem involved.
Even if I try calling a bash script test.sh (by it's full path!) from within exec, php test.php will just say.

<blockquote>
 sh: /test.sh: not found
</blockquote>

How do I change the shell for php's exec()

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我想在ubuntu服务器上使用php的exec()函数。问题是，我总是得到一个错误，命令没有找到。例如，使用exec("echo 123");版画 sh: /echo:找不到在我看来，当我想要使用bash时，php似乎在使用shell。我尝试在/etc/passwd中更改www-数据的shell，这也没有帮助。有没有人知道问题可能来自哪里，或者我如何为php的ubuntu用户更改shell。谢谢你

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