Question

我有一个单独的问题，我需要帮助。我只想显示来自＆＃39;内容＆＃39;。

的20个字符

    <?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

我将回答你的第一个问题：

在此之后插入此行：

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

仅显示内容中的20个字符？显示段落？

1 个答案: