Question

我有一个包含echo命令的php脚本，但它没有正确显示在我的网页上。这是代码：

<?php
$content_sql = "SELECT * FROM mag_books WHERE id = '$id'";
$content_res = mysqli_query($con, $content_sql);
while($content = mysqli_fetch_assoc($content_res)){

    $content_intro = nl2br($content["intro"]);

    $display_content = "

        <div class=\"pageSection text\">
            $content_article
        </div>
        <div class=\"pageSection text\">
            $folder = 'files/books/'.$post_year.'/'.$post_id.'/';
            $filetype = '*.{jpg}*';
            $files = glob($folder.$filetype, GLOB_BRACE);
            foreach ($files as $file)
            {
                echo '

                        <div class=\"galleryCell\">
                            <img class=\"galleryPhoto\" src=\"files/books/'.$file.'\" />
                        </div>

                ';
            }
        </div>

    ";

};
?>

在我的网站上显示：

= 'files/books/'.2016.'/'.1463391024.'/'; = '*.{jpg}*'; = glob(., GLOB_BRACE); foreach ( as ) { echo ' '; }

如何逃避代码以显示应该回显的内容？

Answer 1

你没有将你的PHP代码分配给变量。你可能会像这样使用

 $display_content ='';
 $display_content .= "<div class=\"pageSection text\">".$content_article."</div><div class=\"pageSection text\">";
                $folder = 'files/books/'.$post_year.'/'.$post_id.'/';
                $filetype = '*.{jpg}*';
                $files = glob($folder.$filetype, GLOB_BRACE);
                foreach ($files as $file)
                {


                    $display_content .="<div class=\"galleryCell\"><img class=\"galleryPhoto\" src=\"files/books/".$file."\" /></div>";
                }
               $display_content .= "</div>";
    echo $display_content;

Answer 2

试试这样：

$display_content = '';

while($content = mysqli_fetch_assoc($content_res)){

    $content_intro = nl2br($content["intro"]);

    $display_content .= '
        <div class="pageSection text">
            ' . $content_article . '
        </div>
        <div class="pageSection text">';

            $folder = 'files/books/'.$post_year.'/'.$post_id.'/';
            $filetype = '*.{jpg}*';
            $files = glob($folder . $filetype, GLOB_BRACE);
            foreach ($files as $file) {
                $display_content .= '
            <div class="galleryCell">
               <img class="galleryPhoto" src="files/books/'.$file.'" />
            </div>';
            }
        $display_content .= '
        </div>';
}

echo $display_content;

PHP脚本中的PHP Echo

2 个答案: