只顯示來自內容的20個字符？顯示段落？

Question

我真的有單獨的問題，我需要幫助。我只想顯示來自'內容'的20個字符。

<?php 
    $output = ''; 
    if(isset($_GET['q']) && $_GET['q'] !== ' ') { 
     $searchq = $_GET['q']; 

     $q = mysqli_query($db, "SELECT * FROM article WHERE title LIKE '%$searchq%' OR content LIKE '%$searchq%'") or die(mysqli_error()); 
     $c = mysqli_num_rows($q); 
     if($c == 0) { 
      $output = 'No search results for <strong>"' . $searchq . '"</strong>'; 
     } else { 
      while($row = mysqli_fetch_array($q)) { 
       $id = $row['id']; 
       $title = $row ['title']; 
       $content = $row ['content']; 
      $output .= '<a href="article.php?id=' .$id. '"> 

       <h3>'.$title.'</h3></a>'.$content.''; 
      } 
     } 
    } else { 
      header("location: ./"); 
    } 
    print("$output"); 
    mysqli_close($db); 
?> 

Answer 1

我會回答你的第一個問題：

插入此行後：

$content = $row ['content']; 
if(strlen($content)>20) $content=substr ($content,0,19);