您可以使用分層查詢和CONNECT_BY_ROOT。 從層次結構的根開始，向下展開： SELECT id, CONNECT_BY_ROOT(id) AS root_idFROM entryWHERE id IN (6, 3)START WITH parent_id IS NULLCONNECT BY PRIOR id = parent_id; 或者，從條目返回到根目錄： SELECT CONNECT_BY_ROOT(id) AS id, id AS root_idFROM entryWHERE parent_id IS NULLSTART WITH id IN (6, 3)CONNECT BY PRIOR parent_id = id; 其中，對于樣本數據： CREATE TABLE entry( id, parent_id ) ASSELECT 1, NULL FROM DUAL UNION ALLSELECT 2, 1 FROM DUAL UNION ALLSELECT 3, 2 FROM DUAL UNION ALLSELECT 4, NULL FROM DUAL UNION ALLSELECT 5, 4 FROM DUAL UNION ALLSELECT 6, 5 FROM DUAL UNION ALLSELECT 7, 6 FROM DUAL Both output: I

https://v3.cn.vuejs.org/api/s...

您可以這樣做導出依賴于承諾的函數： let getTokens = async () => {var token = null;var user_id = null;if (typeof window !== 'undefined') { token = (await getItem('secure_token')); }if (typeof window !== 'undefined') { user_id = (await getItem('user_id')); }return {token, user_id };}export const API_ROUTE = getTokens().then((data)=> { return axios.create({ baseURL: API_URL, withCredentials: true, crossdomain: true, headers: { "Key": data.token, "Id": data.user_id } }); }); 您可以使用此承諾訪問模塊中的值。 那你就可以了

我想你只需要使用別名，所以試試這個： SELECT ename AS "EMP NAME", sal * 12 + nvl(comm, 0) AS "ANNUAL EARNINGS"FROM emp;

使用條件連接&動態查詢可以解決這個問題，因為問題是與ScanCode表連接時出現重復項，如果沒有ScanCode，我們可以進行條件連接，如果我們不連接，這樣我們就不會得到重復項。如果提供了ScanCode，那么每個ScanCode將只有一個項目&在這種情況下，我們不會得到重復的項目。 另外，如果我們對top(@p_limit記錄使用動態限制，那么在查詢結束時需要選項（重新編譯），這樣它就不會使用具有先前限制的緩存執行計劃，并且性能會下降。 Declare @p_scanCode BIGINT = '12345', @p_limit INT = 500 DECLARE @sql nvarchar(max) = N' SELECT TOP(@p_limit) i.ItemCode, i.StandardDescription FROM Item i ' + CASE WHEN @p_scanCode IS NOT NULL THEN N' LEFT OUTER JOIN SCANCODE AS SC ON I.ItemCode = SC.FK_RetailItemCode AND GETDATE() BETWEEN SC.EffectiveDate AND SC.TerminationDate ' ELSE N'' END + 'WHERE 1 = 1' + CASE

這是一個帶有祖先的遞歸查詢。即： with RECURSIVE ancestors AS ( SELECT Id, Name, parentLuId FROM Lu WHERE exists(select * from item i where i.Luid = LU.id and i.Name = 'baloon') UNION SELECT d.Id, d.Name, d.parentLuId FROM Lu inner JOIN ancestors c ON c.parentLuId = d.Id)SELECT *FROM ancestors; 這里是DBFiddle演示

Use df.rolling: df = pd.DataFrame(data)df.loc['Returns'] = df.loc['Net Income']/df.loc['Total Asset'].rolling(2).mean()print(df) 2012 2013 2014 2015 2016Total Asset 1000000.0 2000000.0 3000000.00 4.000000e+06 3.000000e+06Net Income 100000.0 300000.0 350000.00 2.600000e+05 3.000000e+05Returns NaN 0.2 0.14 7.428571e-02 8.571429e-02 將您的df轉換為： df_t = pd.DataFrame(data).Tdf_t['Returns'] = df_t['Net Income']/df_t['Total Asset'].rolling(2).mean()print(df_t) Total Asset Net Income Returns2012 1000000 100000 NaN2013 2000000 300000 0.2000002014 3000000 350000 0.1400002015 4000000 260000 0.0742862016 300

答：不，等待沒有top-level超時。 此功能實際上在Deno中用于Web服務器，例如： import { serve } from "https://deno.land/std@0.103.0/http/server.ts";const server = serve({ port: 8080 });console.log(`HTTP webserver running. Access it at: http://localhost:8080/`);console.log("A");for await (const request of server) { let bodyContent = "Your user-agent is:\n\n"; bodyContent += request.headers.get("user-agent") || "Unknown"; request.respond({ status: 200, body: bodyContent });}console.log("B"); 在本例中，“A”在控制臺中打印，“B”直到Web服務器關閉（這不會自動發生）。

使用遞歸函數從數據庫結果生成多維數組 我用這個代碼解決了我的問題 public function buildTree(array $elements, $parentId = 0) { $branch = array(); foreach ($elements as $element) { if ($element['parent_id'] == $parentId) { $children = $this->buildTree($elements, $element['id']); $isArray = filter_var($element['isArray'], FILTER_VALIDATE_BOOLEAN); if ($isArray) { $element['isArray'] = $children; } $branch[] = $element; } } return $branch; }

如果需要更正左上邊框，則不能使用邊框。您可以使用：：before css來更正此問題。 body { margin: 0; padding: 0; } header { border-top: 3px solid #107dcc; height: 55px; width: 100%; background: #CED3DC; top: 0; } .menu_btn { height: 38px; line-height: 38px; text-align: center; color: #000; float: left; border-right: 1px solid #107dcc; font-size: 14px; width: 100px; position: relative; } .menu_btn::before { position: absolute; content: ""; width: 100%; height: 3px; top: 0; left: 0; background: #107dcc; display: none; } .menu_btn:hover:before { display: block; } <header> <div id=""> <a href="#"> <div class="menu_

您需要用“let”而不是const定義var，然后在“then”中賦值。 let value;f(x).then(val => value = val) 也許你有更好的方法來做這件事，但有了contexte，我會這樣做。

根據這些線，存儲庫會自動處理路徑上的權重；因此，加載一個pre-trained權重（無論是.h5檢查點還是.weights進行遷移學習，其余的按照培訓筆記本進行； model = Yolov4(weight_path='mytraining.weights', class_name_path=class_name_path)

當用戶在“簽出”頁面上填寫特定簽出字段時，請使用以下命令禁用特定付款方式： // Jquery script that send the Ajax requestadd_action( 'woocommerce_after_checkout_form', 'custom_checkout_js_script' );function custom_checkout_js_script() { $field_key = 'billing_company'; // Here set the targeted field WC()->session->__unset('field_'.$field_key); ?> <script type="text/javascript"> jQuery(function($){ if (typeof wc_checkout_params === 'undefined') return false; var field = '[name="<?php echo $field_key; ?>"]'; $( 'form.checkout' ).on('input change', field, function() { $.ajax({ type: 'POST', url: wc_checkout_params.a

將基/根uuid&名稱保留在遞歸循環中有助于聚合。 下面的解決方案使用EXISTS來知道遞歸何時結束。 WITH RECURSIVE RCTE_NODES AS ( SELECT uuid , name , uuid as root_uuid , name as root_name , 1 as lvl , ARRAY[]::uuid[] as children , true as has_next FROM category WHERE parent_uuid IS null UNION ALL SELECT cat.uuid , cat.name , cte.root_uuid , cte.root_name , cte.lvl+1 , cte.children || cat.uuid , (exists(select 1 from category cat2 where cat2.parent_uuid = cat.uuid)) FROM RCTE_NODES cte JOIN category cat ON cat.parent_uuid = cte.uuid)SELECT root_uuid as uuid, root_name as name, array_agg(children) as childrenFROM RCTE_NODESWHERE has_nex

可以在連接表后使用group by完成此操作： select s.name , avg(p.commissionRate) from Purchases p inner JOIN Salesman s on s.empId=p.empId group by s.nameorder by 2 desc