2017-01-20 97 views
1

我在PyQt4中構建一個程序,它需要從多個文本文件中提取數據。我有一個按鈕,將選擇文件:它的代碼是在PyQt中讀取文件

qtCreatorFile = 'parser.ui' 

Ui_MainWindow, QtBaseClass = uic.loadUiType(qtCreatorFile) 

class MyApp(QtGui.QMainWindow, Ui_MainWindow): 
    def __init__(self): 
     QtGui.QMainWindow.__init__(self) 
     Ui_MainWindow.__init__(self) 
     self.setupUi(self) 

     self.file_selector.clicked.connect(self.File_Selector) 

     self.log 

    def File_Selector(self): 
     files_list = [] 
     filenames = str(QFileDialog.getOpenFileNames(self, "Select File", "", "*.txt")) 
     self.log.insertPlainText('Loading files ' + '\n') 
     self.log.insertPlainText(filenames + '\n') 

if __name__ == "__main__": 
     app = QtGui.QApplication(sys.argv) 
     window = MyApp() 
     window.show() 
     sys.exit(app.exec_()) 

當按鈕被按下時,我可以選擇我需要的文本文件,但我不能讀?當我問它打印的文件名在日誌中它給了我<PyQt4.QtCore.QStringList object at 0x0000000002BD0BA8>

我也試過:

text = open(filenames).read() 
    self.log.insertPlainText(text) 

但是,讓IOError: [Errno 22] invalid mode ('r') or filename: '<PyQt4.QtCore.QStringList object at 0x0000000002F00BA8>讓我怎麼做QStringList object可讀?

回答

1

QtGui.QFileDialog.getOpenFileNames(...)返回一個字符串列表,所以你不能打開它並加載它,你必須一個接一個地做。

def File_Selector(self): 
    filenames = QtGui.QFileDialog.getOpenFileNames(self, "Select File", "", "*.txt") 
    self.log.insertPlainText('Loading files ' + '\n') 
    self.log.insertPlainText(str(filenames) + '\n') 
    for filename in filenames: 
     text = open(filename).read() 
     self.log.insertPlainText(text) 
+0

啊好吧。所以如果我想加載多個文件,我必須得到文件路徑,然後將其傳遞到程序的另一部分,或選擇目錄? – Charlietrypsin