项目背景需求是: 已知: var a=[{name:'jenny',age:18},{name:'john',age:19},{name:'jack',age:20}] var b ='jenny' 返回: {name:'jenny',age:18} 解题思路:现将数组转成字符串,判断改数组中是否存在该字符串,存在的情况下遍历该数组,返回匹配的数据 代码如下:
SELECT * FROM rsl a, (SELECT CODE, max(time_key) time_key FROM rsl GROUP BY CODE ) b WHERE a. CODE = b. CODE AND a.time_key = b.time_key AND a. CODE IN ('HK.00700', 'HK.03888'); table :rsl 然后查询出根据每一种的code 中最新的一组数据
使用pop()这个函数可以从数组中删除末尾的元素,shift方法可以删除数组中第一个元素.这些都是js中自带的函数,如果不使用这些函数的话,自己写的代码效率会很低的. <html> <head> <title>数组的字符串表示</title> <script type="text/javascript"> function B(){ var names1=["zhangsan1","lisi1&q
declare @t table(name varchar(),qy varchar(),je int) insert into @t union all union all union all union all union all --select * from @t a where not exists --这是取表中的NAME相同的最大值 --( -- from @t where name=a.name and je>a.je --) --第一个答案: SELECT NAME,QY,JE
引自:http://bbs.tianya.cn/post-414-38497-1.shtml 方法一: var ary = new Array("111","22","33","111"); var s = ary.join(",")+","; for(var i=0;i<ary.length;i++) { if(s.replace(ary[i]+",",&qu