将逗号移到另一个位置：

ADD_MONTHS( (SELECT MAX(INSPECTION_DATE) AS MAX_DATE FROM DDS_MQC_QAP_VW) -26,))+1,-1)
                                                                             ^
                                                                            here
ADD_MONTHS( (SELECT MAX(INSPECTION_DATE) AS MAX_DATE FROM DDS_MQC_QAP_VW),-26))+1,-1)

您创建的日期无效

new Date(moment.utc().format( ''yyyy-MM-dd''));

试试

vm.minDate = new Date();
vm.maxDate = new Date();
vm.endBy = new Date();

于是我找到了解决办法。我只是在触发关闭下拉菜单之前将 maxHeight 设置为 offsetHeight。

我还需要使用 0 毫秒的 setTimeout 来触发关闭它，但在运行以下函数后。

export function retreatDropdown(e: HTMLElement) {
  e.style.maxHeight = `${e.offsetHeight}px`;
}

我可以为打开动画做一个类似的技巧，但在打开之前你不会知道正确的 maxHeight，所以我猜这更棘手（但不如关闭动画 IMO 重要）。>

编辑：关闭/打开下拉菜单的完整（TypeScript）解决方案

const defaultTransitionTime = 200;

export function toggleDropdown(
  e: HTMLElement,value: boolean = null,toggleAction: () => void = null // Function that toggles true/false for accordion open
) {
  if (value) {
    openDropdown(e,toggleAction);
  } else {
    retreatDropdown(e,toggleAction);
  }
}

export function openDropdown(
  e: HTMLElement,openAction: () => void = null
) {
  if (openAction) setTimeout(() => openAction(),0);
  e.style.display = "block";
  setTimeout(() => {
    const offsetHeight = e.offsetHeight;
    e.style.maxHeight = "0px";
    setTimeout(() => {
      e.style.maxHeight = `${offsetHeight}px`;
    },5);
  },5);
  setTimeout(() => {
    e.style.maxHeight = "none";
  },defaultTransitionTime);
}

export function retreatDropdown(
  e: HTMLElement,closeAction: () => void = null
) {
  e.style.maxHeight = `${e.offsetHeight}px`;
  if (closeAction) setTimeout(() => closeAction(),0);
}

下拉CSS类：

.dropdown {
  transition: opacity 200ms linear,max-height 200ms linear;
  will-change: opacity,max-height;
}

Angular ngStyle（应用于手风琴的打开/关闭）：

elementStyle(e: HTMLElement) {
    return this.isOpen(HTMLElement) // I use a map of booleans for open state here...
      ? { display: "block","max-height": "none" }
      : { "max-height": 0,opacity: 0 };
  }

Max Sum of Max-K-sub-sequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 10831 Accepted Submission(s): 3970

Problem Description
Given a circle sequence A[1],A[2],A[3]…A[n]. Circle sequence means the left neighbour of A[1] is A[n] , and the right neighbour of A[n] is A[1].
Now your job is to calculate the max sum of a Max-K-sub-sequence. Max-K-sub-sequence means a continuous non-empty sub-sequence which length not exceed K.

Input
The first line of the input contains an integer T(1<=T<=100) which means the number of test cases.
Then T lines follow, each line starts with two integers N , K(1<=N<=100000 , 1<=K<=N), then N integers followed(all the integers are between -1000 and 1000).

Output
For each test case, you should output a line contains three integers, the Max Sum in the sequence, the start position of the sub-sequence, the end position of the sub-sequence. If there are more than one result, output the minimum start position, if still more than one , output the minimum length of them.

Sample Input
4
6 3
6 -1 2 -6 5 -5
6 4
6 -1 2 -6 5 -5
6 3
-1 2 -6 5 -5 6
6 6
-1 -1 -1 -1 -1 -1

Sample Output
7 1 3
7 1 3
7 6 2
-1 1 1

问题链接：HDU3415 Max Sum of Max-K-sub-sequence
问题简述：给定长度为 n 的循环序列，计算长度不超过 k 的子序列其和为最大。
问题分析：给代码不解释。
程序说明：（略）
参考链接：（略）
题记：（略）

AC 的 C++ 语言程序如下：

/* HDU3415 Max Sum of Max-K-sub-sequence */

#include <bits/stdc++.h>

using namespace std;

const int INF = 0x7F7F7F7F;
const int N = 100000;
int a[N * 2 + 1], q[N * 2 + 1];

int main()
{
    int t, n, k;
    scanf("%d", &t);
    while(t--) {
        scanf("%d%d", &n, &k);
        a[0] = 0;
        for(int i = 1; i <= n; i++) {
            scanf("%d", &a[i]);
            a[i] += a[i - 1];       // 计算前缀和
        }
        a[n + 1] = 0;
        for(int i = n + 1; i < n + k; i++)
            a[i] = a[n] + a[i - n];     // a[n + i] = a[i] + a[n]，i=1...k

        int end = n + k - 1, head = 0, tail = 0, maxSum = -INF, l, r;
        for(int i = 1; i <= end ; i++) {
            while(head < tail && a[i - 1] < a[q[tail - 1]]) tail--;
            q[tail++] = i - 1;
            while(head < tail && i - q[head] > k) head++;
            if(maxSum < a[i] - a[q[head]]) {
                maxSum = a[i] - a[q[head]];
                l = q[head] + 1;
                r = i > n ? i % n : i;
            }
        }

        printf("%d %d %d\n", maxSum, l, r);
    }

    return 0;
}