drk7700
2018-07-16 11:57
浏览 122
已采纳

使用shell_exec写入Python中的文本文件

So first off, sorry if this question has already been answered elsewhere, but I couldn't find the answer myself.

I've got a LAMP server running in Ubuntu 16.04 and I'm trying to use Python 2.7.12 to write to a text file from PHP.

I've got the following two files (note, these two files are in the same directory):

index.php

<?php
    $command = escapeshellcmd('python test.py a b c');
    $output = shell_exec($command);
    echo $output;
?>

test.py

import sys

print "argv1: ", sys.argv[1]
print "argv2: ", sys.argv[2]
print "argv3: ", sys.argv[3]

f = open("test-" + sys.argv[1] + ".txt","w+")
f.write("test")
f.close()

print "Text added."

When I navigate to the index.php file in my browser, I get the following output:

argv1: a argv2: b argv3: c

However, when I execute this code in the terminal the file test-a.txt is created and I get the following output:

layer8@alpha:/var/www/html/$ python test.py a b c
argv1:  a
argv2:  b
argv3:  c
Text added.

I can't seem to understand and work out why the Python file isn't executing any code past the third print argv3 statement?

I've done a lot of research into this and I've been unable to find a solution. If you need any further information from me, please do not hesitate to ask; any responses are appreciated.

Thanks.

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首先，对不起，如果这个问题已在其他地方得到解答，但我自己找不到答案。

我在 Ubuntu 16.04 中运行了一个LAMP服务器，我正在尝试使用 Python 2.7.12 来写入 来自PHP的文本文件。

我有以下两个文件（注意，这两个文件 在同一目录中）： \ n

index.php 

 ＆lt;？php 
 $ command = escapeshellcmd（'python test.py ab c'）  ; 
 $ output = shell_exec（$ command）; 
 echo $ output; 
？＆gt; 
   
 
 
  test.py    
 
 
  import sys 
 
print“argv1：”，sys.argv [1] 
print“argv2：”，sys.argv [2] 
print“argv3：”  ，sys.argv [3] 
 
f = open（“test-”+ sys.argv [1] +“。txt”，“w +”）
f.write（“test”）
f.close（）  
 
print“已添加文字。”
   
 
 
当我在浏览器中导航到index.php文件时，我得到以下输出： \  ñ
 
 <代码> argv1  ：a argv2：b argv3：c   
 
 
但是，当我在终端中执行此代码时，会创建文件 test-a.txt ， 我得到以下输出： 
 
 
  layer8 @ alpha：/ var / www / html / $ python test.py abc 
argv1：a 
argv2：b 
argv3：c \  nText添加。
   
 
 
我似乎无法理解并弄清楚为什么Python文件没有执行超过第三个print argv3语句的任何代码？ 
 
 
我已经对此进行了大量研究，但我一直无法找到解决方案。 如果您需要我的任何进一步信息，请不要犹豫; 任何回复都表示赞赏。 
 
 
谢谢。
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1条回答 默认 最新

  • dongzhuo3059 2018-07-16 12:16
    已采纳

    shell_exec will only return what get's printed to stdout. If an exception arises in the Python script, the whole exception message and traceback are printed on stderr, so they won't be part of the output. Note that theoretically, shell_exec should return NULL if an error happens but well, this is PHP so it wouldn't be much fun if the builtin funcs behaved as documented...

    Anyway: the problem is very probably with opening the file, and is very probably a permission issue.

    First, you're using a relative path, and relative path are resolved against the current working directory, not against the directory where your script is located, so if you expect a sane and predictable behaviour always use an absolute path (either pass it as argument or build it against the script's current location).

    Also your Apache server may not (this is an understatement) run with the same env and user as you so it might just not be allowed to open a file for wrting in whatever-his-own-current-working-directory is at that time.

    And finally, even it's PYTHONPATH might be different than yours so it might even be executing another (leftover) test.py OR a stale test.pyc file.

    To make a long story short: it's impossible to tell for sure what's wrong without having access to your system, but at least now you know where to start debugging.

    The first thing I'd personnaly do would be to make sure any error in the python script is printed to stdout:

    import sys

print "argv1: ", sys.argv[1]
print "argv2: ", sys.argv[2]
print "argv3: ", sys.argv[3]

try:
    fname = "test-%s.txt" % sys.argv[1]
    with open(fname,"w+") as f:
        f.write("test")

    print "Text added."
except Exception as e:
    print "oops : got %s" % e
    raise
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