0

我的图像链接每个都假设以项目符号显示。问题是它只为所有图像链接显示一个项目符号,而不是为每个图像链接显示一个项目符号。我在我的 php/html 中做错了什么?

<table id="tableqanda" cellpadding="0" cellspacing="0">
    <thead>
    <tr>
        <th width="11%" class="image">Image</th>
    </tr>
    </thead>
    </table>
    <div id="tableqanda_onthefly_container">
    <table id="tableqanda_onthefly" cellpadding="0" cellspacing="0">
    <tbody>
    <?php

                function CreateLink($filename, $type){
 if($type == 'image'){
  return '<a href="previewimage.php?filename='.$filename.'" title="Click to view in New window" target="_blank">'.htmlspecialchars($filename).'</a>';
 }
}

          foreach ($arrQuestionId as $key=>$question) {
        echo '<tr class="tableqandarow">'.PHP_EOL;
        //start:procedure
        $img_result = '';
        if(empty($arrImageFile[$key])){
          $img_result = '&nbsp;';
        }else{
          $img_result .=  '<ul class="qandaul"><li>';
           if(is_array( $arrImageFile[$key] )){
            foreach($arrImageFile[$key] as $filename){
             $img_result.= CreateLink($filename, "image");
            }
           }else{
            $img_result.= CreateLink($arrImageFile[$key], "image");
           }
           $img_result.= '</li></ul>';
        }
        //end:procedure

        echo '<td width="11%" class="imagetd">'.$img_result.'</td>' . PHP_EOL;
        echo '</tr>'.PHP_EOL;

}

?>
    </tbody>
    </table>
    </div>

屏幕截图显示了它现在的样子:

在此处输入图像描述

4

1 回答 1

2

您可能希望每个链接都在<li>标签中:

代替:

$img_result.= CreateLink($filename, "image");

和:

$img_result.= '<li>' . CreateLink($filename, "image") . '</li>;
// the other places where you use that function as well...

并确保列表仅以<ul>标签开头和结尾(不是附加的<li>

于 2013-01-27T01:49:10.120 回答