force scheduled shopware taks lik feed

bin/console scheduled-task:run --no-wait;
php -d memory_limit=2G -d max_execution_time=0 bin/console messenger:consume async --time-limit=600 -vv;

程式碼審核

請你對目前這個 repository 進行「全面程式碼審查」。

重要限制:
1. 先不要直接修改任何程式碼。
2. 不要只做表面掃描,請先理解專案架構、執行流程、驗證機制、資料存取、部署方式後再提出問題。
3. 所有發現都必須指出:
   - 嚴重程度:Critical / High / Medium / Low
   - 問題類型:安全性 / 架構 / 錯誤處理 / 效能 / 可維護性 / 測試不足 / 部署風險
   - 影響範圍
   - 相關檔案與程式碼位置
   - 為什麼這是問題
   - 建議修正方式
   - 是否需要補測試
4. 不要提出沒有依據的猜測;如果無法確認,請標註「需人工確認」。
5. 優先找出會造成資安風險、生產環境錯誤、資料不一致、權限繞過、Token / Cookie / CORS / CSRF 問題、例外未處理、日誌外洩、弱掃風險的問題。
6. 請依照實際程式碼內容審查,不要只給一般建議。

請依照以下順序執行:

第一階段:專案盤點
- 說明此 repository 的專案組成、主要技術棧、入口點、重要設定

Lighthouse MCP

請使用 lighthouse MCP 對以下登入頁執行行動裝置 Performance 稽核:

<Url>

請執行下列工作:

1. 記錄 Performance 分數、FCP、LCP、TBT、CLS 和 Speed Index。
2. 找出造成 LCP 超過 2.5 秒的主要原因。
3. 檢查 ASP.NET Core 10、Razor View、Razor Layout、CSS、JavaScript、圖片、Web Font、靜態資源、中介軟體及第三方資源。
4. 檢查靜態檔案壓縮、快取標頭、Response Compression、資源載入順序,以及是否存在 Render-blocking resources。
5. 依改善效益、修改風險與實作成本排序改善項目。
6. 修改程式碼時不得改變登入流程、驗證邏輯、頁面功能及既有 UI 行為。
7. 修改後再次使用 lighthouse MCP 執行相同條件的測試。
8. 比較修改前後的 Lighthouse 結果。
9. LCP 目標為 2.5 秒以下。
10. 最後列出:

* 修改的檔案

ERROR LOGS

var/log/system.log
var/log/exception.log
var/log/debug.log

Listening 1 IELTS official

9:30 am
Halindale
central street
792
?
1,80
19:30
7.15
Commuter
?early morning
A 13
B 14
C 11 12
-
first-year
balance
foreign students
Relaxation
Motivation
research
The secret garden
-
walk
motivation
abstract ideas
-
darkness to lightness
-
-
human companionship
Negative
pleasure
fate
Active
success
B
B/C
A
A
B

審核新舊程式碼

請只審查 `sha_old..sha_new` 範圍內的變更,不要修改任何檔案。

請以以下 Git 範圍為審查依據:

```bash
git log --oneline sha_old..sha_new
git diff --stat sha_old..sha_new
git diff sha_old..sha_new
```

審查範圍只限於 `sha_old` 之後到 `sha_new` 為止的變更。

可以閱讀其他未變更檔案作為上下文,但不要把原本就存在的舊問題列為本次問題,除非該問題是由 `sha_old..sha_new` 範圍內的變更引入、擴大或暴露。

本專案是 Flutter 專案,請特別審查:

* Widget lifecycle,例如 `initState`、`dispose`、Controller / Listener / Worker 是否釋放
* GetX 狀態管理是否有狀態殘留、重複刷新、Controller 過肥問題
* async / await 是否有 race condition、重複請求、頁面離開後仍更新狀

3532. Path Existence Queries in a Graph I

You are given an integer n representing the number of nodes in a graph, labeled from 0 to n - 1. You are also given an integer array nums of length n sorted in non-decreasing order, and an integer maxDiff. An undirected edge exists between nodes i and j if the absolute difference between nums[i] and nums[j] is at most maxDiff (i.e., |nums[i] - nums[j]| <= maxDiff). You are also given a 2D integer array queries. For each queries[i] = [ui, vi], determine whether there exists a path between nodes ui and vi. Return a boolean array answer, where answer[i] is true if there exists a path between ui and vi in the ith query and false otherwise.
/**
 * @param {number} n
 * @param {number[]} nums
 * @param {number} maxDiff
 * @param {number[][]} queries
 * @return {boolean[]}
 */
var pathExistenceQueries = function(n, nums, maxDiff, queries) {
    const comp = new Array(n).fill(0);

    // Build connected components
    for (let i = 1; i < n; i++) {
        if (nums[i] - nums[i - 1] <= maxDiff) {
            comp[i] = comp[i - 1];
        } else {
            comp[i] = comp[i - 1] + 1;
        }
    }

    // Answer queries
    const ans

3756. Concatenate Non-Zero Digits and Multiply by Sum II

You are given a string s of length m consisting of digits. You are also given a 2D integer array queries, where queries[i] = [li, ri]. For each queries[i], extract the substring s[li..ri]. Then, perform the following: Form a new integer x by concatenating all the non-zero digits from the substring in their original order. If there are no non-zero digits, x = 0. Let sum be the sum of digits in x. The answer is x * sum. Return an array of integers answer where answer[i] is the answer to the ith query. Since the answers may be very large, return them modulo 109 + 7.
/**
 * @param {string} s
 * @param {number[][]} queries
 * @return {number[]}
 */
var sumAndMultiply = function(s, queries) {
    const MOD = 1000000007;
    const MODB = BigInt(MOD);

    // Safe modular multiplication using BigInt
    const mulmod = (a, b) => Number((BigInt(a) * BigInt(b)) % MODB);

    // ------------------------------------------------------------
    // STEP 1: Extract all NON-ZERO digits and their original positions
    // --------------------------------------------------

3754. Concatenate Non-Zero Digits and Multiply by Sum I

You are given an integer n. Form a new integer x by concatenating all the non-zero digits of n in their original order. If there are no non-zero digits, x = 0. Let sum be the sum of digits in x. Return an integer representing the value of x * sum.
/**
 * @param {number} n
 * @return {number}
 */
var sumAndMultiply = function(n) {
    // Step 1: build x by concatenating non-zero digits
    let s = String(n);
    let filtered = "";
    for (let ch of s) {
        if (ch !== '0') filtered += ch;
    }

    let x = filtered.length === 0 ? 0 : Number(filtered);

    // Step 2: compute digit sum of x
    let sum = 0;
    while (x > 0) {
        sum += x % 10;
        x = Math.floor(x / 10);
    }

    // Step 3: multiply original x by sum
    r

1288. Remove Covered Intervals

Given an array intervals where intervals[i] = [li, ri] represent the interval [li, ri), remove all intervals that are covered by another interval in the list. The interval [a, b) is covered by the interval [c, d) if and only if c <= a and b <= d. Return the number of remaining intervals.
/**
 * @param {number[][]} intervals
 * @return {number}
 */
var removeCoveredIntervals = function(intervals) {
    // Step 1: Sort intervals by:
    //   - start ascending
    //   - if starts are equal, end descending
    //
    // This ensures that any interval that *could* cover another
    // appears before the interval it might cover.
    intervals.sort((a, b) => {
        if (a[0] === b[0]) {
            // For same start, put the longer interval first
            return b[1] - a[1];
    

1301. Number of Paths with Max Score

You are given a square board of characters. You can move on the board starting at the bottom right square marked with the character 'S'. You need to reach the top left square marked with the character 'E'. The rest of the squares are labeled either with a numeric character 1, 2, ..., 9 or with an obstacle 'X'. In one move you can go up, left or up-left (diagonally) only if there is no obstacle there. Return a list of two integers: the first integer is the maximum sum of numeric characters you can collect, and the second is the number of such paths that you can take to get that maximum sum, taken modulo 10^9 + 7. In case there is no path, return [0, 0].
/**
 * @param {string[]} board
 * @return {number[]}
 */
var pathsWithMaxScore = function(board) {
    const n = board.length;
    const MOD = 1e9 + 7;

    // dpScore[i][j] = max score to reach (i,j)
    // dpCount[i][j] = number of ways to achieve that score
    const dpScore = Array.from({ length: n }, () => Array(n).fill(-Infinity));
    const dpCount = Array.from({ length: n }, () => Array(n).fill(0));

    // Start at S (bottom-right)
    dpScore[n-1][n-1] = 0;
    dpCount[n-1][n-1] = 1;

2492. Minimum Score of a Path Between Two Cities

You are given a positive integer n representing n cities numbered from 1 to n. You are also given a 2D array roads where roads[i] = [ai, bi, distancei] indicates that there is a bidirectional road between cities ai and bi with a distance equal to distancei. The cities graph is not necessarily connected. The score of a path between two cities is defined as the minimum distance of a road in this path. Return the minimum possible score of a path between cities 1 and n. Note: A path is a sequence of roads between two cities. It is allowed for a path to contain the same road multiple times, and you can visit cities 1 and n multiple times along the path. The test cases are generated such that there is at least one path between 1 and n.
/**
 * @param {number} n
 * @param {number[][]} roads
 * @return {number}
 */
var minScore = function(n, roads) {
    const graph = Array.from({ length: n + 1 }, () => []);
    for (const [a, b, w] of roads) {
        graph[a].push([b, w]);
        graph[b].push([a, w]);
    }

    const seen = new Array(n + 1).fill(false);
    const stack = [1];
    seen[1] = true;

    // Find all nodes reachable from 1
    while (stack.length) {
        const node = stack.pop();
        for (const [nei] of gr

new-nim.md

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

3620. Network Recovery Pathways

You are given a directed acyclic graph of n nodes numbered from 0 to n − 1. This is represented by a 2D array edges of length m, where edges[i] = [ui, vi, costi] indicates a one‑way communication from node ui to node vi with a recovery cost of costi. Some nodes may be offline. You are given a boolean array online where online[i] = true means node i is online. Nodes 0 and n − 1 are always online. A path from 0 to n − 1 is valid if: All intermediate nodes on the path are online. The total recovery cost of all edges on the path does not exceed k. For each valid path, define its score as the minimum edge‑cost along that path. Return the maximum path score (i.e., the largest minimum-edge cost) among all valid paths. If no valid path exists, return -1.
/**
 * @param {number[][]} edges
 * @param {boolean[]} online
 * @param {number} k
 * @return {number}
 */
var findMaxPathScore = function(edges, online, k) {
    const n = online.length;

    // Build adjacency list only for edges whose endpoints are online
    const adj = Array.from({ length: n }, () => []);
    let maxEdgeCost = 0;

    for (const [u, v, cost] of edges) {
        if (online[u] && online[v]) {
            adj[u].push([v, cost]);
            if (cost > maxEdgeCost) maxEdgeCost 

Check if DNS-Records are propagated successfully (using dig)

dig +short A sub.domain.end