因為這兩個數(shù)組應(yīng)該組合在同一個索引上，所以我們可以使用一個簡單的for循環(huán)并使用該索引來獲得這兩個值； <?php$first = [ 'john', 'jane' ];$last = [ 'doe', 'joe' ];$res = [];for ($i = 0; $i < sizeof($first); $i++) { $res[] = "{$first[$i]} {$last[$i]}";}var_dump($res); Will produce: array(2) { [0]=> string(8) "john doe" [1]=> string(8) "jane joe"} 在線試用！

基本思路：分組→按組合計→處理成結(jié)果視圖用 Lodashconst result = _(arr).groupBy(({ code, batch, store }) => [code, batch, store].join("|")) .values() .map(list => { const { code, batch, store } = list[0]; return { code, batch, store, number: _.sumBy(list, "number") }; }) .value();自己寫const result = Object.values( arr.reduce((r, { code, batch, store, number }) => { const key = `${code}|${batch}|${store}`; const item = r[key] ??= { code, batch, store, number: 0 }; item.number += number; return r; }, {}));

您需要將$fields存儲回$done，并對其進(jìn)行編碼。 不需要調(diào)用array_flip()來轉(zhuǎn)換回數(shù)組，請使用array_keys()。 $done = json_decode($res2["done_by"]); //$done output= {"done":[5415703999,1670937087,6887688688],"skip":[],"left":[]}$fields = array_flip($done->done);unset($fields['1670937087']);$done->done = array_keys($fields);$json = json_encode($done);print($json);

將字符串與<=>運算符或strcmp函數(shù)進(jìn)行比較： usort($a, function($a, $b) { return $a['0'] <=> $b['0'];});usort($a, function($a, $b) { return strcmp($a['0'], $b['0']);}); Fiddle here.

您可以使用foreach對結(jié)果進(jìn)行分組，然后根據(jù)需要顯示打印組。 <?php$categories = [];foreach ($data as $value) { $categoryname = $value['name']; $medication = $value['medication']; $arrmaincatname = explode('(', rtrim($categoryname, ')')); if ($categoryname == "Saba") { $categories[$arrmaincatname[0]][] = $medication; }}foreach ($categories as $categoryName => $category) {?> In <?php echo $categoryName; ?> category -- [<?php echo implode(', ', $category) ?>] <?php}

您只需獲取0到3之間的隨機(jī)數(shù)（可以使用下劃線的u2;.random），并將其用作索引，從組件數(shù)組中獲取元素。 var components = [component1, component2, component3, component4]var dots = []for (var i = 0; i < 17; ++i) dots.push(components[_.random(3)]);

strip()不會就地修改字符串（字符串在Python中是不可變的）。您需要將結(jié)果賦回變量。 title = title.strip()

你只需要使用foreach循環(huán)。這是解決辦法。 <?php$arr = [ ["id" => 1, "check" => 1, "option" => "mk"], ["id" => 2, "check" => 3, "option" => "zz"], ["id" => 3, "check" => 5, "option" => "mk"]];$newArr = [];foreach($arr as $a) { // check if array element is already there based on option key in $newArr. In that case just add check value. if(isset($newArr[$a['option']])) { $newArr[$a['option']]['check'] += $a['check']; continue; } $newArr[$a['option']] = $a;}$newArr = array_values($newArr);print_r($newArr); 這是輸出。 Array( [0] => Array ( [id] => 1 [check] => 6 [option] => mk )

只要用這個更新你的delelteItems函數(shù) deleteItems = itemToRemove => { const { List } = this.state; const newList = List.filter(item => item !== itemToRemove); this.setState({ List: newList }); }; 并像這樣從列表組件傳遞項引用 <button onClick={() => props.deleteItems(item)}>Done</button> 你也可以檢查這個https://stackblitz.com/edit/react-xugsaz?file=src%2FListItems.js

一種解決方案是使用array_slice函數(shù)將這兩部分分開，并將它們與array_merge結(jié)合起來： $data = [1, 2, 3, 4];$pos = 2;$ordered = array_merge( array_slice($data, $pos), array_slice($data, 0, $pos));// [3, 4, 1, 2]

您可以維護(hù)一個tracker對象來檢查同名對象是否已經(jīng)被訪問過。如果已訪問，則將對象推入duplicate數(shù)組，否則推入unique數(shù)組。試試這個- const data=[{createdBy:"ABC",createdDate:"2020",lastModifiedDate:"2021",name:"John Doe",section:{createdBy:"ABC",status:"ACTIVE"},sectionName:"Mechanical"},{createdBy:"ABC",createdDate:"2020",lastModifiedDate:"2021",name:"John Doe",section:{createdBy:"A2C",status:"ACTIVE"},sectionName:"Mechanical"},{createdBy:"A1C",createdDate:"2020",lastModifiedDate:"2021",name:"John Doe",section:{createdBy:"ABC",status:"ACTIVE"},sectionName:"Mechanical"}];const track = {};const unique = [];const duplicate = [];for (const item of data)

我在@Nigel Ren的幫助下找到了答案，他們提到了答案https://stackoverflow.com/a/3289602/1213708. <?phpfunction array_to_xml(array $arr, SimpleXMLElement $xml){ foreach ($arr as $k => $v) { is_array($v) ? $this->array_to_xml($v, $xml->addChild($k)) : $xml->addChild($v, $k); // this is the important part where the $key/$value is flipped and fixes the issue I was having } return $xml;}$test_array = array ( 'akey' => 'auth_key', 'cmd' => 'command', 'aid' => 'account_id', 'scode' => 'short_code', 'key' => 'keyword', 'words' => 'message', 'contacts' => [ '1' => 'contact_number', '2'

你不應(yīng)該對你的數(shù)據(jù)進(jìn)行多次傳遞。只需將name值用作臨時鍵，然后僅在其createAt值大于存儲值時保留重復(fù)行的數(shù)據(jù)。Re-index完成循環(huán)后，請單擊數(shù)組。 Code: (Demo) $result = [];foreach ($arr as $row) { if (!isset($result[$row['name']]) || (int)$row['createdAt'] > (int)$result[$row['name']]['createdAt']) { $result[$row['name']] = $row; }}var_export(array_values($result)); Output: array ( 0 => array ( 'id' => '6230061c0e88d709ca0d7bbc', 'name' => 'Mobile SamSung', 'slug' => 'mobile-samsung', 'createdAt' => '1648006346', ), 1 => array ( 'id' => '5d1eff529a426778d4b92383', 'name' => 'Mobile Iphone', 'slug' => 'mobile-iphone', 'createdAt'

從簡化這個過程開始，制作一個簡單的預(yù)定時間數(shù)組，這樣就可以在一個循環(huán)中使用in_array() $schedules = array();$schedules[0]['start_hour'] = "08:00:00";$schedules[1]['start_hour'] = "08:15:00";$schedules[2]['start_hour'] = "08:30:00";#Make simple array of booked times$booked = [];foreach ( $schedules as $sch ): $booked[] = strtotime($sch['start_hour']);endforeach;for ($i = strtotime("08:00:00"); $i <= strtotime("09:00:00"); $i = $i+15*60): if( in_array($i, $booked)): echo date("H:i", $i)." SCHEDULED! \n"; else: echo date("H:i", $i)." NOT SCHEDULED :( \n"; endif;endfor; RESULT 08:00 SCHEDULED! 08:15 SCHEDULED! 08:30 SCHEDULED! 08:45

您可以使用array_column來提取新數(shù)組中'holi_date'的所有值，其中只包含'holi_date'的值 然后使用srtotime將所有日期轉(zhuǎn)換為有效的時間戳。用array_map可以很容易地完成。 最后一步是使用in_array。這是一個例子。 $holidayArray = [ 0 => ["holi_date" => "2021-03-30"], 1 => ["holi_date" => "2021-06-25"], 2 => ["holi_date" => "2021-06-18"], 3 => ["holi_date" => "2021-07-05"], 4 => ["holi_date" => "2021-06-21"], 5 => ["holi_date" => "2022-01-20"]];if (in_array(strtotime('2021-07-05'), array_map('strtotime', array_column($holidayArray, 'holi_date')))) { echo 'yes';} else { echo 'no';}