Question

我使用'for'循环来创建由'Label'对象制作的表格。除了'divider'之外，每个循环都附加到它自己的二维数组（每列有十三个'标签'）;我想在每个对象的各个表中给出一个索引位置，这样我就可以引用这些点并改变'Label'的内容。

def stock_append(entry_grid_exp, entry_grid_req):
for Label in entry_grid_exp:
    Label.configure(text="test")

提交按钮调用函数'stock_append'并传递'entry_grid_exp'和'entry_grid_req'参数。我读了一篇文章说，将列表放入'for'循环并将列表中的每个项目归为'Label'将允许我调用'configure'函数，但它似乎不起作用。此外，这个循环（如果有效）将配置'entry_grid_exp'中的每个项目而不是一个特定的位置。

from tkinter import *
root = Tk()
Med_font = ("Helvatica", 11)
Small_font = ("Helvatica", 9)
main = Frame(root, bg="#dcdcdc", height=600, width=600)
main.grid()
s_name = StringVar()
s_category = StringVar()
s_description = StringVar()
s_quantity = StringVar()
s_risk = StringVar()
s_ID = StringVar()


def stock_append(entry_grid_exp, entry_grid_req, program):
    for item in entry_grid_exp:
        Label.configure(program, text="test")


def stock_define(entry_grid_exp, entry_grid_req, program):
    stock_frame = Frame(main, bg="#dcdcdc", height=100, width=1000)
    stock_frame.grid()

    name = Entry(stock_frame, width=15, textvariable=s_name)
    name.grid(row=1, column=1, padx=10)

    category = Entry(stock_frame, width=15, textvariable=s_category)
    category.grid(row=1, column=2, padx=10)

    description = Entry(stock_frame, width=15, textvariable=s_description)
    description.grid(row=1, column=3, padx=10)

    quantity = Entry(stock_frame, width=15, textvariable=s_quantity)
    quantity.grid(row=1, column=4, padx=10)

    risk = Entry(stock_frame, width=15, textvariable=s_risk)
    risk.grid(row=1, column=5, padx=10)

    stock_id = Entry(stock_frame, width=15, textvariable=s_ID)
    stock_id.grid(row=1, column=6, padx=10)
    submit = Button(stock_frame, width=15, text="submit", command=lambda: stock_append(entry_grid_exp, entry_grid_req,
                                                                                  program))
    submit.grid(row=1, column=7, padx=10)


def stock_lists():
    program = Frame(main, bg="#dcdcdc", height=600, width=600)
    program.grid()

    title = Label(program, text="MAIN", font=Med_font, bg="#dcdcdc")
    title.grid(row=2, column=1)

    entry_grid_exp = []
    entry_grid_req = []

    exp_list = Frame(program, bg="#dcdcdc", height=500, width=600)
    exp_list.grid(row=3, column=1)

    stock_add = Button(exp_list, bg="#bebebe", width=10, text="ADDSTCK", command=lambda: stock_define(entry_grid_exp,
                                                                                                  entry_grid_req,
                                                                                                  program))
    stock_add.grid(row=4, column=15)

    experiment_list = Label(exp_list, width=10, bg="#dcdcdc", fg="#000000", text="Experiment list.")
    experiment_list.grid(row=4, column=1)

    for i in range(1, 16):
        entry_column = []
        for x in range(0, 13):
            entry_column.append(Label(exp_list, width=15, bg="#FFFFFF", relief=SUNKEN))
            entry_column[x].grid(row=x+6, column=i, padx=0, pady=0)
        entry_grid_exp.append(entry_column)

    for i in range(1, 16):
        divider = Label(exp_list, width=10, bg="#dcdcdc", fg="#dcdcdc", text="divider")
        divider.grid(row=21, column=i)

    request_list = Label(exp_list, width=10, bg="#dcdcdc", fg="#000000", text="Requested list.")
    request_list.grid(row=20, column=1)

    for i in range(1, 16):
        entry_column = []
        for x in range(0, 6):
            entry_column.append(Label(exp_list, width=15, bg="#FFFFFF", relief=SUNKEN))
            entry_column[x].grid(row=x+21, column=i, padx=0, pady=0)
        entry_grid_req.append(entry_column)

stock_lists()
mainloop()

我正在尝试完成这个项目而且完全迷失了，并且非常感谢任何帮助，因为我不知道我在做什么。

更新示例

\DB::table('table_name')
 ->where(function($q){
       $q->where('column', 'value1')
         ->orWhere('column', 'value2');
  })
  ->where('column2', 'LIKE', '%value3%');

如何在2D数组中配置'tkinter'对象？

0 个答案: