Php 拉雷维尔：用雄辩的语言获得最常用的词语

我有一个表后，它有一个列的内容。内容列是文本。我想通过雄辩的方式获得今天内容中使用最多的单词，假设table name=post，field name=content，试试这个：

将获得表中第一个最常见的内容，并且：

$mostUsed = Post::groupBy('content')
    ->orderBy('content', 'desc')
    ->count('content');

---

将获得从最常见到最罕见的寄存器顺序。顺便说一下，这只是从MySQL到Eloquent的一个改编，根据

我想你要找的是这个

注意：由于您没有提供任何表结构，我不知道如何筛选今天的帖子。我希望有一个名为date的专栏

数一数今天用过的所有单词

// array of all the content strings.
$contents = Post::where('date', date('Y-m-d'))->pluck('content');

// final result will be stored here as a key value pair.
// used count against each word.
$word_count = [];

foreach($contents as $content) {

    // array of each word in the content separated by 'space'.
    $words = explode(' ', $content);

    foreach($words as $word) {

        // if the word has already used +1 the count, else set the count as 1.
        $count = array_key_exists($word, $word_count) ? ($word_count[$word] + 1) : 1;

        // set new word count to the array.
        array_set($word_count, $word, $count);
    }
}

$most_used_word = array_search(max($word_count), $word_count);

---

到目前为止，您尝试了什么？是laravel 5或更高版本吗？内容栏可能与文章正文类似，您想获取所有内容栏中的所有文本，并从中获取最常用的文本吗？谢谢，如果我只想获取htags？什么是htags？

// array of all the content strings.
$contents = Post::where('date', date('Y-m-d'))->pluck('content');

// final result will be stored here as a key value pair.
// used count against each word.
$word_count = [];

foreach($contents as $content) {

    // array of each word in the content separated by 'space'.
    $words = explode(' ', $content);

    foreach($words as $word) {

        // if the word has already used +1 the count, else set the count as 1.
        $count = array_key_exists($word, $word_count) ? ($word_count[$word] + 1) : 1;

        // set new word count to the array.
        array_set($word_count, $word, $count);
    }
}

$most_used_word = array_search(max($word_count), $word_count);