Question

我需要确定最近连续几天的车牌已记录在表格中。板在给定的一天中可能会多次出现在桌子上，但是至少每24小时进行一次库存检查（以避免24小时的“结帐”）。

一天不是报告的24小时。 12hr <=天<= 24hr 。即-如果车辆在8:00 am进入库存，并且一直呆到第二天8:00 pm（共36小时），则将记录为2天。

如果车辆离开存货的时间超过假定的25小时以上，则当前停留已结束，任何进一步的停留都是新的停留。

仅应报告任何一个LicensePlate的最新停留时间。

示例数据：

LicensePlate |       TimeStamp     | Location
---------------------------------------------
abc123       | 1969-01-01 08:42:23 | West Lot   ---
abc123       | 1969-01-01 12:34:29 | West Lot      ---
abc123       | 1969-01-01 18:20:10 | West Lot         ---
abc123       | 1969-01-02 02:19:02 | West Lot             --- 2 Days Closed
abc123       | 1969-01-02 08:38:05 | West Lot         ---
abc123       | 1969-01-02 12:13:39 | West Lot      ---
abc123       | 1969-01-02 20:43:34 | West Lot   ---
123abc       | 1969-01-02 09:31:04 | South Lot  ---
123abc       | 1969-01-02 15:31:04 | South Lot     --- 1 Day (more than 12hr) Closed
123abc       | 1969-01-02 22:31:04 | South Lot  ---
abc123       | 1969-01-04 22:48:37 | North Lot  ---
abc123       | 1969-01-05 17:57:32 | North Lot     --- 4 Days (reported)
abc123       | 1969-01-06 14:12:48 | North Lot     --- 
abc123       | 1969-01-07 13:05:18 | North Lot  ---
123abc       | 1969-01-03 12:22:32 | East Lot   ---
123abc       | 1969-01-03 19:16:25 | East Lot      ---
123abc       | 1969-01-04 06:22:13 | East Lot         ---
123abc       | 1969-01-04 13:22:56 | East Lot         --- 3 Days (reported)
123abc       | 1969-01-05 11:22:32 | East Lot         ---
123abc       | 1969-01-06 08:22:44 | East Lot      ---
123abc       | 1969-01-06 12:22:08 | East Lot   ---

生成的报告将包含以下数据

LicensePlate |     EnteredDate     |  Location   |  DaysIn
----------------------------------------------------------
abc123       | 1969-01-04 12:48:37 |  North Lot  |     4
123abc       | 1969-01-03 12:22:32 |  North Lot  |     3

实际表在架构上有所不同，只需要我应该如何完成此操作的指针即可。

Answer 1

找到了一种完全可以完成我需要做的事情的方法。

<!doctype html>
<html lang="en">
<head>
  <meta charset="utf-8">
  <title>jQuery.inArray demo</title>
  <style>
  div {
    color: blue;
  }
  span {
    color: red;
  }

  .em{
  font-color:white;
  }
  </style>
  <script src="jquery-1.10.2.js"></script>
  <link href="emoji.css" rel="stylesheet">

</head>
<body>
 <div class ="see"></div>
<code>HEY</code>
 <input class="input">
 <div class ="see"></div>
  <div class ="see2"></div>
    <div class ="see3"></div>



<div class="add"></div>



<script>
var emojiss = [ "--abc", "--woman","--eye"];
var emoji=["-angel","-alien","----1","---1","--100","--1234","--8ball","--a","--ab","--abc","--abcd","--accept","--admission_tickets","--adult","--aerial_tramway","--airplane","--airplane_arriving","--airplane_departure"]

var input=$(".input");
var input2=$(".input").val();



 $(input).change(function(){

$(".see").html( $(".input").val());


if(jQuery.inArray($(".input").val(), emoji) != -1) {

var see=$(".see").html().replace('--', '').replace('--', '');


$(".see2").html(see);

var classs =see+"";

alert($(".see").html());




<!-- $(".see3").html(real); -->


    $(this).addClass("em em"+$(".see").html()).removeClass("see");

} else {
    alert("is NOT in array");


} 

});






</script>

</body>
</html>

在表格中查找连续的日期（MYSQL）

1 个答案: