如何在填充整个单元格时左对齐Python tkinter网格列

Question

我正在尝试编写一个python类来以表格格式显示数据.我确信那里已经有类做同样的事情,但是,我正在使用这个练习来教自己Python和tkinter.在大多数情况下,我让这个类按照我想要的方式工作,但是当我向左对齐时,我无法获得标题和数据单元来填充整个单元格.这是我的班级当前为表格生成的内容:

我继续将单元格上的粘性更改为(W,E)而不是W,以显示我希望表格看起来如何,除了每个单元格左对齐.以下是我拍摄的内容:

基于研究,我已经做了,似乎我需要使用weight的属性grid_columnconfigure和grid_rowconfigure,但各方面我一直在使用他们试过,我不能,得到它的工作.

这是我的类的代码(我正在使用Python 3.4):

from tkinter import *
from tkinter import ttk
from tkinter import font

class TableData:

    def __init__(self,parent,attributes,columns,data):
        self.parent = parent
        self.tableName = StringVar()
        self.tableName.set(attributes['tableName'])
        self.columns = columns
        self.columnCount = 0
        self.borderColor = attributes['borderColor']
        self.titleBG = attributes['titleBG']
        self.titleFG = attributes['titleFG']
        self.titleFontSize = attributes['titleFontSize']
        self.headerBG = attributes['headerBG']
        self.headerFG = attributes['headerFG']
        self.headerFontSize = attributes['headerFontSize']
        self.dataRowColor1 = attributes['dataRowColor1']
        self.dataRowColor2 = attributes['dataRowColor2']
        self.dataRowFontSize = attributes['dataRowFontSize']
        self.dataRowFG = attributes['dataRowFG']
        self.data = data
        self.tableDataFrame = ttk.Frame(self.parent)
        self.tableDataFrame.grid(row=0,column=0)
        self.initUI()

    def countColumns(self):
        cnt = 0
        for i in self.columns:
            cnt += 1

        self.columnCount = cnt

    def buildTableTitle(self):
        tableTitleFont = font.Font(size=self.titleFontSize)
        Label(self.tableDataFrame,textvariable=self.tableName,bg=self.titleBG,fg=self.titleFG,font=tableTitleFont, highlightbackground=self.borderColor,highlightthickness=2).grid(row=0,columnspan=self.columnCount,sticky=(W,E), ipady=3)

    def buildHeaderRow(self):
        colCount = 0
        tableHeaderFont = font.Font(size=self.headerFontSize)
        for col in self.columns:
            Label(self.tableDataFrame,text=col,font=tableHeaderFont,bg=self.headerBG,fg=self.headerFG,highlightbackground=self.borderColor,highlightthickness=1).grid(row=1,column=colCount,sticky=W, ipady=2, ipadx=5)
            colCount += 1

    def buildDataRow(self):
        tableDataFont = font.Font(size=self.dataRowFontSize)
        rowCount = 2
        for row in self.data:
            if rowCount % 2 == 0:
                rowColor = self.dataRowColor2
            else:
                 rowColor = self.dataRowColor1
            colCount = 0
            for col in row:
                Label(self.tableDataFrame,text=col,bg=rowColor,fg=self.dataRowFG,font=tableDataFont,highlightbackground=self.borderColor,highlightthickness=1).grid(row=rowCount,column=colCount,sticky=W,ipady=1, ipadx=5)
                colCount += 1
            rowCount += 1

    def initUI(self):
        self.countColumns()
        self.buildTableTitle()
        self.buildHeaderRow()
        self.buildDataRow()

这是一个引用TableData类的测试文件:

from tkinter import *
from tkinter import ttk
from tableData import TableData
import sqlite3

root = Tk()
root.geometry('1000x400')

mainframe = ttk.Frame(root).grid(row=0,column=0)

attributes = {}
attributes['tableName'] = 'Title'
attributes['borderColor'] = 'black'
attributes['titleBG'] = '#1975D1'
attributes['titleFG'] = 'white'
attributes['titleFontSize'] = 16
attributes['headerBG'] = 'white'
attributes['headerFG'] = 'black'
attributes['headerFontSize'] = 12
attributes['dataRowColor1'] = 'white'
attributes['dataRowColor2'] = 'grey'
attributes['dataRowFontSize'] = 10
attributes['dataRowFG'] = 'black'

columns = ['Col 1', 'Column 2', 'Column 3','Column    4']

results = [('1','Key','Desc','Attribute'),('2','Key Column','Description Column','AttributeColumn')]

table = TableData(mainframe,attributes,columns,results)

root.mainloop()

提前感谢您的任何见解.如果有任何其他信息有用,请告诉我.

Answer 1

对于任何几何网格,添加选项sticky="W",例如,

self.tableDataFrame.grid(sticky="W", row=0, column=0)

Answer 2

如果希望标签中的文本左对齐,请使用锚点选项.它需要一个表示指南针上一个点的字符串(例如:"w"="west",表示文本固定在左侧):

for col in row:
    Label(..., anchor="w").grid(...)