小编foz*_*ios的帖子

首先,所有相关代码

main.py

import string
import app
group1=[ "spc", "bspc",",","."]#letters, space, backspace(spans mult layers)
# add in letters one at a time
for s in string.ascii_lowercase:
    group1.append(s)
group2=[0,1,2,3,4,5,6,7,8,9, "tab ","ent","lAR" ,"rAR" , "uAR", "dAR"]
group3= []
for s in string.punctuation:
    group3.append(s)#punc(spans mult layers)
group4=["copy","cut","paste","save","print","cmdW","quit","alf","sWDW"] #kb shortcut
masterGroup=[group1,group2,group3,group4]
myApp =app({"testFKey":[3,2,2]})

app.py

import tkinter as tk
import static_keys
import dynamic_keys
import key_labels
class app(tk.Frame):

    def __init__(inputDict,self, master=None,):
        tk.Frame.__init__(self, master)
        self.grid(sticky=tk.N+tk.S+tk.E+tk.W)
        self.createWidgets(self, inputDict)
    def createWidgets(self,inDict):
        top=self.winfo_toplevel()
        top.rowconfigure(0, weight=1)
        top.columnconfigure(0, weight=1)
        self.rowconfigure(0, weight=1)
        self.columnconfigure(0, weight=1) …

我在xib中有uiButtons.我为所有人设置了恢复标识.我需要打印这些恢复ID列表.为此,我在viewDidload中调用以下代码:

-(void)loadViewFromNIB:(NSString *)nibName owner:(id)owner
{


NSArray *objects = [[NSBundle mainBundle] loadNibNamed:nibName owner:owner options:nil];
NSArray *subviews = [[objects objectAtIndex:0]subviews];
for (id key in subviews) {
        [key addTarget:self
        action:@selector(touchB:)
        forControlEvents:UIControlEventTouchDown];
        [key addTarget:self
                   action:@selector(touchE:)
         forControlEvents:UIControlEventTouchUpInside];
        NSString *ident = self.restorationIdentifier;
        NSLog(@"%@",ident);


}

我得到这个输出:

2013-02-24 13:05:38.817 fozbKEY[3939:11603] (null)
2013-02-24 13:05:38.822 fozbKEY[3939:11603] (null)
2013-02-24 13:05:38.824 fozbKEY[3939:11603] (null)

这只是重复了一堆.我做错了什么？我如何解决它？谢谢!

我有python 3代码不能按预期工作:

def addFunc(x,y):
    print (x+y)

def subABC(x,y,z):
    print (x-y-z)

def doublePower(base,exp):
    print(2*base**exp)

def  RootFunc(inputDict):
    for k,v in inputDict.items():
        if v[0]==1:
            d[k] = addFunc(*v[1:])
        elif v[0] ==2:
            d[k] = subABC(*v[1:])
        elif  v[0]==3:
            d[k] = doublePower(*v[1:])


d={"s1_7":[1,5,2],"d1_6":[2,12,3,3],"e1_3200":[3,40,2],"s2_13":[1,6,7],"d2_30":[2,42,2,10]}
RootFunc(d)

#test to make sure key var assignment works
print(d)

我明白了:

{'d2_30': None, 's2_13': None, 's1_7': None, 'e1_3200': None, 'd1_6': None}

我期望:

{'d2_30': 30, 's2_13': 13, 's1_7': 7, 'e1_3200': 3200, 'd1_6': 6}

怎么了？

半相关:我知道字典是无序的,但有什么理由为什么python选择了这个命令？它是通过随机数发生器运行的吗？

虽然图片基本上显示了我想做的事情，但是：

1. 所有标签都应该是相同大小的完美正方形
2. 2 列中的标签应该完全对齐
3. 底部的标签应该完全在其他两个标签的下方和中间。
4. 实际标签应该是实心的，标签文本周围应该有一点空间（文本不应占据整个标签）。

如果可能，应使用网格。我会用python3