Bash腳本從文件中選擇一個Python函數

Question

對於git alias problem，我希望能夠通過名稱從文件中選擇一個Python函數。例如：

... 
    def notyet(): 
     wait for it 

    def ok_start(x): 
     stuff 
     stuff 
     def dontgettrickednow(): 
     keep going 
    #stuff 
     more stuff 

    def ok_stop_now(): 

在算法方面，以下就足夠了近：當你發現一個匹配/^(\s*)def $1[^a-zA-Z0-9]/

保持匹配的行，直到找到一條線是不

1. 開始過濾^\s*#或^/\1\s]（即，可能是縮進的註釋，或比上一個更長的縮進）

（我並不在乎是否拾取了以下函數之前的裝飾器。結果是爲人類閱讀。）

我試圖用awk（我幾乎不知道）做到這一點，但它比我想象的有點困難。對於初學者，我需要一種方法來存儲原始def之前的縮進長度。

Answer 1

單向使用awk。代碼很好評論，所以我希望它很容易理解。

內容infile：

含量 script.awk

... 
    def notyet(): 
     wait for it 

    def ok_start(x): 
     stuff 
     stuff 
     def dontgettrickednow(): 
     keep going 
    #stuff 
     more stuff 

    def ok_stop_now(): 

：

BEGIN { 
     ## 'f' variable is the function to search, set a regexp with it. 
     f_regex = "^" f "[^a-zA-Z0-9]" 

     ## When set, print line. Otherwise omit line. 
     ## It is set when found the function searched. 
     ## It is unset when found any character different from '#' with less 
     ## spaces before it. 
     in_func = 0 
} 

## Found function. 
$1 == "def" && $2 ~ f_regex { 

     ## Get position of first 'd' in the line. 
     i = index($0, "d") 

     ## Sanity check. Never should success because the condition was 
     ## checked before. 
     if (i == 0) { 
       next 
     } 

     ## Get characters until matched index before, check that all of 
     ## them are spaces, and get its length. 
     indent = substr($0, 0, i - 1) 
     if (indent ~ /^[[:space:]]*$/) { 
       num_spaces = length(indent) 
     } 

     ## Set variable, print line and read next one. 
     in_func = 1 
     print 
     next 
} 

## When we are inside the function, line doesn't begin with '#' and 
## it's not a blank line (only spaces). 
in_func == 1 && $1 ~ /^[^#]/ && $0 ~ /[^[:space:]]/ { 

     ## Get how many characters there are until first non-space. The result 
     ## is the position of first non-blank, so substract one to get the number 
     ## of spaces. 
     spaces = match($0, /[^[:space:]]/) 
     spaces -= 1 

     ## If current indent is less or equal that the indent of function definition, then 
     ## end of function found, so end processing. 
     if (spaces <= num_spaces) { 
       in_func = 0 
     } 
} 

## Self-explanatory. 
in_func == 1 { 
     print 
} 

運行它喜歡：

awk -f script.awk -v f="ok_start" infile 

隨着下面的輸出：

def ok_start(x): 
     stuff 
     stuff 
     def dontgettrickednow(): 
     keep going 
    #stuff 
     more stuff 

Answer 2

爲什麼不只是讓python做到這一點？我認爲inspection模塊可以打印出一個功能的來源，所以你可以導入模塊，選擇功能並檢查它。不掛斷。幫你解決方案...

好的。原來，inspect.getsource功能交互方式定義的東西不起作用：

>>> def test(f): 
...  print 'arg:', f 
... 
>>> test(1) 
arg: 1 
>>> inspect.getsource(test) 
Traceback (most recent call last): 
    File "<stdin>", line 1, in <module> 
    File "C:\Python27\lib\inspect.py", line 699, in getsource 
    lines, lnum = getsourcelines(object) 
    File "C:\Python27\lib\inspect.py", line 688, in getsourcelines 
    lines, lnum = findsource(object) 
    File "C:\Python27\lib\inspect.py", line 529, in findsource 
    raise IOError('source code not available') 
IOError: source code not available 
>>> 

但對你的使用情況，它會工作：對於保存到磁盤模塊。就拿我test.py文件：

def test(f): 
    print 'arg:', f 

def other(f): 
    print 'other:', f 

而且比較這個交互式會話：

>>> import inspect 
>>> import test 
>>> inspect.getsource(test.test) 
"def test(f):\n print 'arg:', f\n" 
>>> inspect.getsource(test.other) 
"def other(f):\n print 'other:', f\n" 
>>> 

所以......你需要寫一個接受Python源文件名的簡單的Python腳本和一個函數/對象名稱作爲參數。然後它應該導入模塊並檢查功能並將其打印到STDOUT。