Question

我正在尝试组织标题，以便它们与列对齐。下面是代码（创建表），但它不像我想要的那样工作。：

class Example(tk.Frame):
    def __init__(self, parent):
        tk.Frame.__init__(self, parent)               
        names = ["header1", "header2", "header3", "header4"]
        self.label1 = tk.Label(self, text=("        ".join(names)))
        self.table = SimpleTableInput(self, 3, 4)
        self.submit = tk.Button(self, text="Submit", command=self.on_submit)
        self.label1.pack(side="top")
        self.table.pack(side="top", fill="both", expand=True)
        self.submit.pack(side="bottom")

我很感激任何建议！感谢

整个脚本在这里：

import tkinter as tk

class SimpleTableInput(tk.Frame):
    def __init__(self, parent, rows, columns):
        tk.Frame.__init__(self, parent)

        self._entry = {}
        self.rows = rows
        self.columns = columns

        # register a command to use for validation
        vcmd = (self.register(self._validate), "%P")
        # create the table of widgets
        for row in range(self.rows):
            for column in range(self.columns):
                index = (row, column)
                e = tk.Entry(self, validate="key", validatecommand=vcmd)
                e.grid(row=row, column=column, stick="nsew")
                self._entry[index] = e
        # adjust column weights so they all expand equally
        for column in range(self.columns):
            self.grid_columnconfigure(column, weight=1)
        # designate a final, empty row to fill up any extra space
        self.grid_rowconfigure(rows, weight=1)

    def get(self):
        '''Return a list of lists, containing the data in the table'''
        result = []
        for row in range(self.rows):
            current_row = []
            for column in range(self.columns):
                index = (row, column)
                current_row.append(self._entry[index].get())
            result.append(current_row)
        return result

    def _validate(self, P):
        '''Perform input validation. 

        Allow only an empty value, or a value that can be converted to a float
        '''
        if P.strip() == "":
            return True

        try:
            f = float(P)
        except ValueError:
            self.bell()
            return False
        return True

class Example(tk.Frame):
    def __init__(self, parent):
        tk.Frame.__init__(self, parent)               
        names = ["header1", "header2", "header3", "header4"]
        self.label1 = tk.Label(self, text=("        ".join(names)))
        self.table = SimpleTableInput(self, 3, 4)
        self.submit = tk.Button(self, text="Submit", command=self.on_submit)
        self.label1.pack(side="top")
        self.table.pack(side="top", fill="both", expand=True)
        self.submit.pack(side="bottom")

    def on_submit(self):
        print(self.table.get())

root = tk.Tk()
Example(root).pack(side="top", fill="both", expand=True)
root.mainloop()

Answer 1

我建议的方法是在顶部创建一个框架，并用SELECT a.value1, a.value2, b.value3, c.value4 FROM table1 a LEFT JOIN table2 b ON a.some_value=b.some_value LEFT JOIN table3 c ON a.some_value=c.some_value WHERE a.some_value = 'some_text';内部分隔的标签填充它。它允许对齐列并即使窗口展开也保持这种对齐：

names

如何在tkinter表python中对齐列标题

1 个答案: