目录

时间:2015-09-21 20:28:12

标签: python pyqt subprocess

我试图运行子进程。我在一个目录上运行一个python文件(转换目录中的每个文件。)转换器工作,并已将其实现为gui(PYQT4)。这是我到目前为止所得到的:

def selectFile(self):



    self.listWidget.clear() # In case there are any existing elements in the list
    directory = QtGui.QFileDialog.getExistingDirectory(self,
                                                       "Pick a folder")


    if directory:
        for file_name in os.listdir(directory):
            if file_name.endswith(".csv"):
                self.listWidget.addItem(file_name)
                print (file_name)




def convertfile(self, directory):

    subprocess.call(['python', 'Createxmlfromcsv.py', directory], shell=True)

我得到的错误是..

Traceback (most recent call last):
  File "/Users/eeamesX/PycharmProjects/Workmain/windows.py", line 162, in convertfile
    subprocess.call(['python', 'Createxmlfromcsv.py', directory], shell=True)
  File "/opt/local/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/subprocess.py", line 522, in call
    return Popen(*popenargs, **kwargs).wait()
  File "/opt/local/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/subprocess.py", line 710, in __init__
    errread, errwrite)
  File "/opt/local/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/subprocess.py", line 1335, in _execute_child
    raise child_exception
TypeError: execv() arg 2 must contain only strings

对于初学者的任何帮助表示赞赏:)

2 个答案:

答案 0 :(得分:2)

从评论到问题,行:

    self.convertButton.clicked.connect(self.convertfile)
单击按钮时,

会向False方法发送convertfile,这就是您看到错误的原因。

您需要向convertfile添加一些代码,以获取list-widget中所选项目的目录路径。类似的东西:

    item = self.listWidget.currentItem()
    if item is not None:
        directory = unicode(item.text())
        subprocess.call(['python', 'Createxmlfromcsv.py', directory])

但请注意,您没有在list-widget中存储完整的目录路径,因此子进程调用可能会失败。您应该将项目添加到list-widget中,如下所示:

    self.listWidget.addItem(os.path.join(directory, file_name))

答案 1 :(得分:-1)

在" subprocess.call([' python',' Createxmlfromcsv.py',目录],shell = True)",'目录'变量不是String。