我嘗試重構(gòu)我的代碼，但我的代碼表現(xiàn)不佳。目標(biāo)是獲得具有 unique 的數(shù)組列表"answer_id"，并獲得一些“分?jǐn)?shù)”、“投票”等。在我現(xiàn)有的元素中，我首先檢查 $most_voted 是否為 null，如果是，我分配第一個(gè)元素，然后我將在第二個(gè)元素中插入一個(gè)foreach新元素，或者我更新一個(gè)現(xiàn)有元素。但是從我的第二個(gè)開始foreach，我就過(guò)得很糟糕。關(guān)于這個(gè)邏輯有什么建議嗎？$answers = $history->toArray(); //here I have an array of arrays$most_voted = [];foreach ($answers as $key => $answer) { if (!empty($most_voted) ) { // in this if I create new arrays or I update existing ones foreach ($most_voted as $k => $v) { //If value already exists, increase Votes/Score/Percentage if (intval($most_voted[$k]['answer_id']) === intval($answer['lkp_answer_id'])) { $most_voted[$k]['Votes'] = $most_voted[$k]['Votes'] + 1; $most_voted[$k]['Score'] = $most_voted[$k]['Score'] + $answer['score']; $most_voted[$k]['Percentage'] = substr((($most_voted[$k]['Votes'] * 100) / $votes), 0, 5); $most_voted[$k]['Weight'] = substr((($most_voted[$k]['Score'] * 100) / $total_scoring), 0, 5); //Else add new array element } else { $name = LkpAnswer::where('id', '=', $answer['lkp_answer_id'])->pluck('name'); (isset($name[0]) && $name[0] !== null) ? $name = $name[0] : ''; if(! empty($answer['lkp_answer_id'])){ $most_voted[$key] = [ 'answer_id' => $answer['lkp_answer_id'], 'Name' => $name, 'Votes' => 1, 'Score' => $answer['score'], 'Percentage' => substr(((1 * 100) / $votes), 0, 5), 'Weight' => substr((($answer['score'] * 100) / $total_scoring), 0, 5), ]; } } } }
1 回答
尚方寶劍之說(shuō)
TA貢獻(xiàn)1788條經(jīng)驗(yàn) 獲得超4個(gè)贊
優(yōu)化的第一條規(guī)則（在我看來(lái)），
切勿在循環(huán)內(nèi)運(yùn)行查詢。
在循環(huán)開始之前：
// Fetch all data at once and use select() whenever possible
$LkpAnswers = LkpAnswer::whereIn('id', collect($answers)->pluck('lkp_answer_id')->toArray())->select('name')->get();
在你的其他條件下（在第二個(gè)循環(huán)內(nèi)）做這樣的事情。
// Before, this query was inside 2 for loops (n^2). For every n, there's 1 query.
// This right here is a Collection of LkpAnswer::model. Collection is very useful for optimization
$name = $LkpAnswers->where('id', $answer['lkp_answer_id'])->pluck('name')->toArray();
使用它并更改您的第二個(gè) else 條件。僅通過(guò)這樣做，您將減少近 50% 的運(yùn)行時(shí)間。另請(qǐng)查看有關(guān)如何在 laravel 中進(jìn)行優(yōu)化的答案。它討論了緩存、索引、select()和類型轉(zhuǎn)換。
讓我在下面的評(píng)論中發(fā)布。
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