学习Swift,好奇有没有更干净的方式来写这段代码?
var apple = 1
var orange = 1
var kiwi = 1
if (orange > apple) && (orange > kiwi) {
print("The orange is the best")
} else if (apple > orange) && (apple > kiwi) {
print("The apple is the best")
} else if (kiwi > apple)
我在使用kiwi时遇到了麻烦,它是来自node.js的包装载器
我已经安装了kiwi,并从终端获得了以下反馈:
bingomanatee@UbunTwo:/var/node/kiwi$ sudo kiwi install ejs
install : ejs
version : <!DOCTYPE HTML PUBLIC "-//IETF//DTD HTML 2.0//EN"> <html><head> <title>404 Not Found</title> </head><body>
我有个泡泡。我为文本匹配、连续排序和排序匹配生成了一个小的开始。
我的示例脚本:
var testtext1 = "apple banana and kiwi pineapple juice";
var testtext2 = "need apple banana pineapple milkshake";
var testtext3 = "apple pineapple lower prices";
var testtext4 = "only apple banana kiwi pineapple lovers";
var te
假设我有这个列表:
apple
orange
banana
kiwi
strawberry
如何将第2行(orange)的值与第4行(kiwi)的值交换以获得以下结果:
apple
kiwi # before "orange"
banana
orange # before "kiwi"
strawberry
输入df1和df2
df1:
Subcategory_Desc Segment_Desc Flow Side Row_no
APPLE APPLE LOOSE Apple Kanzi Front Row 1
APPLE APPLE LOOSE Apple Jazz Front Row 1
CITRUS ORANGES LOOSE Orange Navel Front Row 1
PEAR
我需要找到与所有查询短语匹配的记录,但忽略它们的出现顺序。
例如,我的查询字符串是apple banana kiwi。下列值应为真。
I like apple, banana and kiwi
Banana, kiwi and apple are fruits
下列值应为假
He does not like kiwi
如何在Oracle 11中实现SQL?
让我们考虑这样一个列表,
fruits = ['green apples', 'oranges', 'pears', 'berries', 'kiwi', 'red apples', 'kiwi']
如果我想将包含字符串'apples‘的每个元素都更改为瓜子,我会这样做,
fruits = [fruit if 'apples' not in fruit else 'melons' for fruit in fruits]
这给了我们,
[