Php 计算数据库行中有多少特定单词

我有两个单词列表和一个数据库，里面有上千篇新闻文章

我想计算一下数据库中每篇文章中有多少来自列表$baddwords和$goodwords的单词。接下来，我想将每行的两个结果（$baddwords和$goodwords）保存在baddwords和goodwords列中。我将使用cronjob运行此脚本

我当前的表结构最后两行是空的

我想要的表格结构 最后两列中的$badwords和$goodwords数

我当前的PHP代码

<?php
//the wordlists
$badwords = "depressive horrible";
$goodwords = "great";

//connection to the database
$servername = "localhost";
$username = "user";
$password = "pass";
$dbname = "db";

$conn = new mysqli($servername, $username, $password, $dbname);

// here is my sql query

$sql = " UPDATE news
set badwords = (SELECT count (*) from news
where newscontent LIKE '.%$badwords%.')";    

//close the connection
$conn->close();
?>

---

如果我正确理解了您的问题，您希望检查您的数据库中是否存在某个单词列表。在这种情况下，您正在寻找这样的查询（根据您使用的db类，在查询中也使用转义，例如（））：

如果要显示每个单词列表在数据库中存在的次数，则需要：

SELECT    COUNT(*) AS `count`
         ,`newscontent` 
FROM      `news`
GROUP BY  `newscontent`

如果您想显示给定字数的字符串数量，这就是您要查找的内容：

<?php
  $sql = new mysqli($host, $user, $password, $database);
  $query = $sql->query('select * from `news`');
  $summary = [];

  while($record = $query->fetch_object()) { 
    $summary[count(explode(' ', $record->newscontent))]++;
  }

  echo '<pre>';
  print_r($summary);
  echo '</pre>';

你好，里昂，欢迎来到StackOverflow。请仔细阅读“.I'msorry Leon，但它仍然不清楚：即，您想在mysql或php中执行计数？“基于带单词的字符串”是什么意思？请添加更多细节并提供具体示例：表结构、字段内容示例、字符串示例、所需结果示例等…非常感谢@Ben！我重新措辞了我的问题，事实上，我在寻找一个稍微不同的解决方案。你的方法的组合应该可以做到这一点。。但是，到目前为止，我还没有找到解决方案。
SELECT    COUNT(*) AS `count`
         ,`newscontent` 
FROM      `news`
WHERE     `newscontent` = '" . $wordlist . "'

SELECT    COUNT(*) AS `count`
         ,`newscontent` 
FROM      `news`
GROUP BY  `newscontent`

<?php
  $sql = new mysqli($host, $user, $password, $database);
  $query = $sql->query('select * from `news`');
  $summary = [];

  while($record = $query->fetch_object()) { 
    $summary[count(explode(' ', $record->newscontent))]++;
  }

  echo '<pre>';
  print_r($summary);
  echo '</pre>';

<?php
  // your db connection ...

  // array with good and bad words
  $good = [
    'awesome',
    'neat',
    'fantastic',
    'great',
    // and so on
  ];

  $bad = [
    'horrible',
    'worst',
    'bad',
    'terrific',
    // and so on
  ];

  // if you keep using your string approach you can set $good and $bad with $good = explode(' ', $goodwords); and $bad = explode(' ', $badwords);

  // fetch the record you need
  $query = $sql->query('select * from `news` where `ID` = 1'); // insert parameter for your ID here instead of just 1
  $newsitem = $query->fetch_object();

  // set up good and bad word counters
  $totalGood = 0;
  $totalBad = 0;

  // check how many times each word is mentioned in newscontent
  foreach($good as $word) { 
    // add spaces arround the word to make sure the full word is matched, not a part
    $totalGood += substr_count($newsitem->newscontent, ' ' . $word . ' ');
  }

  // check how many times each word is mentioned in newscontent
  foreach($bad as $word) { 
    // add spaces arround the word to make sure the full word is matched, not a part
    $totalBad += substr_count($newsitem->newscontent, ' ' . $word . ' ');
  }  

  // update the record
  $sql->query("
    update `news` 
    set `badwords` = " . $totalBad . ",
        `goodword` = " . $totalGood . "
    where `ID` = " . $newsitem->ID);