mydict = {"key1":"value1", "key2":"value2"} 在Django模板中查找字典值的常规方法是{{mydict.key1}},{{mydict.key2}}.如果键是循环变量怎么办?即: {% for item in list %} # list中元素,都有一个NAME属性 {{ mydict.item.NAME }} # 取出字典中相对应的NAME属性的key对应的value {% end
Implement strStr(). Returns the index of the first occurrence of needle in haystack, or -1 if needle is not part of haystack. 额,当然我用的就是暴力搜索来,没用到KMP之类的算法,有时间再来补上辣,代码如下: class Solution { public: int strStr(string haystack, string needle) { ; ; ; i <= h
[抄题]: Given two strings A and B, find the minimum number of times A has to be repeated such that B is a substring of it. If no such solution, return -1. For example, with A = "abcd" and B = "cdabcdab". Return 3, because by repeating A
效果如下: 试过很多种办法,思路都在怎么控制<el-table-column type="expand">里面的type上,比如使用v-show等等,但是发现,要不就是展开图标全部没有,要不就是全部有,研究好久,终于想出来这么个办法: 使用getRowClass针对每一行添加类, getRowClass(row, index) { let res = [] if (!row.children)//即改行没有子元素时,添加row-expand-cover类 res.push(