Question

import tkinter as tk

LABEL_BG = "#ccc"  # Light gray.
ROWS, COLS = 10, 6  # Size of grid.
ROWS_DISP = 3  # Number of rows to display.
COLS_DISP = 4  # Number of columns to display.

class MyApp(tk.Tk):
    def __init__(self, title="Sample App", *args, **kwargs):
        tk.Tk.__init__(self, *args, **kwargs)

        self.title(title)
        self.configure(background="Gray")
        self.columnconfigure(0, weight=1)
        self.rowconfigure(0, weight=1)

        master_frame = tk.Frame(self, bg="Light Blue", bd=3, relief=tk.RIDGE)
        master_frame.grid(sticky=tk.NSEW)
        master_frame.columnconfigure(0, weight=1)

        label1 = tk.Label(master_frame, text="Frame1 Contents", bg=LABEL_BG)
        label1.grid(row=0, column=0, pady=5, sticky=tk.NW)

        frame1 = tk.Frame(master_frame, bg="Green", bd=2, relief=tk.GROOVE)
        frame1.grid(row=1, column=0, sticky=tk.NW)

        cb_var1 = tk.IntVar()
        checkbutton1 = tk.Checkbutton(frame1, text="StartCheckBox", variable=cb_var1)
        checkbutton1.grid(row=0, column=0, padx=2)

        label2 = tk.Label(master_frame, text="Frame2 Contents", bg=LABEL_BG)
        label2.grid(row=2, column=0, pady=5, sticky=tk.NW)

        # Create a frame for the canvas and scrollbar(s).
        frame2 = tk.Frame(master_frame)
        frame2.grid(row=3, column=0, sticky=tk.NW)

        # Add a canvas in that frame.
        canvas = tk.Canvas(frame2, bg="Yellow")
        canvas.grid(row=0, column=0)

        # Create a vertical scrollbar linked to the canvas.
        vsbar = tk.Scrollbar(frame2, orient=tk.VERTICAL, command=canvas.yview)
        vsbar.grid(row=0, column=1, sticky=tk.NS)
        canvas.configure(yscrollcommand=vsbar.set)

        # Create a horizontal scrollbar linked to the canvas.
        hsbar = tk.Scrollbar(frame2, orient=tk.HORIZONTAL, command=canvas.xview)
        hsbar.grid(row=1, column=0, sticky=tk.EW)
        canvas.configure(xscrollcommand=hsbar.set)

        # Create a frame on the canvas to contain the buttons.
        buttons_frame = tk.Frame(canvas, bg="Red", bd=2)

        # Add the buttons to the frame.
        for i in range(1, ROWS+1):
            for j in range(1, COLS+1):
                button = tk.Button(buttons_frame, padx=7, pady=7, relief=tk.RIDGE,
                                   text="[%d, %d]" % (i, j))
                button.grid(row=i, column=j, sticky='news')

        # Create canvas window to hold the buttons_frame.
        canvas.create_window((0,0), window=buttons_frame, anchor=tk.NW)

        buttons_frame.update_idletasks()  # Needed to make bbox info available.
        bbox = canvas.bbox(tk.ALL)  # Get bounding box of canvas with Buttons.
        #print('canvas.bbox(tk.ALL): {}'.format(bbox))

        # Define the scrollable region as entire canvas with only the desired
        # number of rows and columns displayed.
        w, h = bbox[2]-bbox[1], bbox[3]-bbox[1]
        dw, dh = int((w/COLS) * COLS_DISP), int((h/ROWS) * ROWS_DISP)
        canvas.configure(scrollregion=bbox, width=dw, height=dh)

        label3 = tk.Label(master_frame, text="Frame3 Contents", bg=LABEL_BG)
        label3.grid(row=4, column=0, pady=5, sticky=tk.NW)

        frame3 = tk.Frame(master_frame, bg="Blue", bd=2, relief=tk.GROOVE)
        frame3.grid(row=5, column=0, sticky=tk.NW)

        cb_var2 = tk.IntVar()
        checkbutton2 = tk.Checkbutton(frame3, text="EndCheckBox", variable=cb_var2)
        checkbutton2.grid(row=0, column=0, padx=2)


if __name__ == "__main__":
    app = MyApp("Scrollable Canvas")
    app.mainloop()

我想要做的是，如果用户输入一个名字，它将被放在位置0，如果用户输入另一个名字，它将放在位置1所以0 - bob，1 - jon。我想要它做的是如果bob位置0被删除，它将把jons位置从1改为0。

Answer 1

使用List，删除后无需移动所有元素，但如果强制使用array，则可以在一个简单的循环中移动所有元素：

public void deleteQueue() {       
    int position;
    PassengerQueue();
    System.out.println("Enter Queue Position which you want to delete a customer from: ");
    position = input.nextInt();

    System.out.println("");

    for (int i = position; i < qitems.length-1; i++) {
        qitems[i] = qitems[i+1];
    }
    qitems[qitems.length] = null;
}

Answer 2

当试图弄清楚如何在＆＃34;低级＆＃34;对象就像一个String数组，你总是可以问：

是否有更高级别的＆＃34;已经存在的Java类可以做同样的事情吗？

在这种情况下，答案是肯定的，有一些......但我假设您不允许直接使用它们。

但是，您仍然可以查看他们的源代码，看看他们是如何完成您尝试做的事情。

例如，在Android Studio中，您可以创建一个新的ArrayList对象...或者只需键入＆＃34; ArrayList，＆＃34;然后右键单击它...并选择＆＃34;转到声明。＆＃34;

这会将您带到您点击的课程的源代码。

这是ArrayList的Android源代码执行remove（）的方式：

public E remove(int index) {
    if (index >= size)
        throw new IndexOutOfBoundsException(outOfBoundsMsg(index));

    modCount++;
    E oldValue = (E) elementData[index];

    int numMoved = size - index - 1;
    if (numMoved > 0)
        System.arraycopy(elementData, index+1, elementData, index,
                         numMoved);
    elementData[--size] = null; // clear to let GC do its work

    return oldValue;
}

你可以看到这与哈米德的答案相似。

Answer 3

你可以循环遍历数组重新排列位置，或者只是使用一些已经为你执行此操作的类such as a linked list for example，它实现了remove（int index）方法。

Answer 4

把

qitems[position-1] = qitems[position]

在for循环内循环遍历删除后的每个索引。

删除名称时如何将位置向上移动一个

4 个答案: