将列表中的 URL 解析为子进程调用

Question

问题是在调用命令中获取调用以识别o.geturl（）命令。

from subprocess import call
import time
from urlparse import urlparse
def ListOfTests():
    myFile = open("testUrlToImport.txt", "r")
    myLines = list(myFile)
    myFile.close()
    for urls in myLines:
        urlsAsAString = str(urls)
        o =urlparse(urlsAsAString)
        call(["phantomjs", "yslow.js", "--info", "basic", "--format", "plain", "o.geturl()"])

任何指示都值得赞赏。谢谢！

Answer 1

您应该从o.geturl()中删除引号。这应该有效：

call(["phantomjs", "yslow.js", "--info", "basic", "--format", "plain", o.geturl()])