奇怪子行爲

Question

subprocess.call(["find", ".", "-exec", "sh", "-c", "echo testFirst", ";"]) 
subprocess.call(["find", ".", "-exec", "sh", "-c", "echo testSecond", ";"], shell=True) 
subprocess.call(["find . -exec sh -c 'echo testThird' \\;"], shell=True) 

subprocess.call(["find", ".", "-exec", "sh", "-c", "touch testFirst", ";"]) 
subprocess.call(["find", ".", "-exec", "sh", "-c", "touch testSecond", ";"], shell=True) 
subprocess.call(["find . -exec sh -c 'touch testThird' \\;"], shell=True) 

產出如下：

testFirst 
testFirst 
testFirst 
. 
./test.py 
./data 
testThird 
testThird 
testThird 
. 
./test.py 
./testFirst 
./data 

而且只有testFirst和testThird文件被創建。

行爲的解釋是什麼？

我會假設輸出爲testFirst,testSecond,testThird以及正在創建的三個文件。

Answer 1

https://stackoverflow.com/a/10661488/1663462

當你通過殼=真，POPEN期待一個字符串參數，而不是 列表。

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