Question

我希望用户打开文件，文件名传递给使用该文件保存一些数据的函数。我为此使用PyQt5 getOpenFileName：

fileName, _ = QFileDialog.getOpenFileName(None, "Open Data File")

但是，我收到此错误消息：“进程完成，退出代码-1073740791（0xC0000409）”。我正在使用python 3.7.1和pyqt 5.9.2。我的问题是如何解决此错误。

编辑：这是示例代码：

import pandas as pd
import csv
from PyQt5.QtWidgets import QFileDialog
import os
def saveSubjectList(FileNamePath,subjectList):
    columns = list(subjectList[0].__dict__.keys())
    vals = []
    for iSub in range(len(subjectList)):
        valsSub = subjectList[iSub].__dict__.values()
        vals.append(valsSub)

    subject_pd = pd.DataFrame(vals, columns=columns)
    subject_pd.to_csv(FileNamePath)

def loadSubjectList(FileNamePath):
    csvFile = open(FileNamePath, 'r')
    csvFileArray = []
    for row in csv.reader(csvFile):
        csvFileArray.append(row[1:])

    keys = csvFileArray[0]
    subList = []

    for sub in csvFileArray[1:]:
        subject1 = Subject(sub[0])
        for val, key in zip(sub, keys):
            setattr(subject1, key, val)
        subList.append(subject1)
    for item in range(len(subList)):
        print(subList[item].__dict__)

class Subject:
    def __init__(self, Id):
        self.Id = Id
        self.Name = ""
        self.FrameNumberList = []
        self.Color = [0, 1, 0]


if __name__ == '__main__':

    subjectList = []
    for i in range(2):
        subject1 = Subject(i)
        subject1.Name = "Subject" + str(i)
        subject1.FrameNumberList = [i, 10, 23, 23]
        subjectList.append(subject1)
    FileNamePath = os.getcwd()
    fileName = 'New'
    fileNameSuffix = 'csv'
    csvPathAndName = os.path.join(FileNamePath, fileName + '.' + 
     fileNameSuffix)
    saveSubjectList(csvPathAndName, subjectList)
    loadSubjectList(csvPathAndName)

这是没有getOpenFileName的初始代码，可以正常工作。然后，我将行从FileNamePath = os.getcwd删除到csvPathAndName = ...，并将其替换为：

csvPathAndName, _ = QFileDialog.getOpenFileName(None, "Open Data File")

因为我希望用户能够在计算机中的任何位置创建或选择文件，但随后出现错误。

使用PyQt5 getOpenFileName打开文件时遇到问题

0 个答案: