我的图像链接每个都假设显示在项目符号中。问题是它只为所有图像链接显示一个项目符号点,而不是为每个图像链接显示一个项目符号点。我在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 = ' ';
}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>
屏幕截图显示了它的样子:
答案 0 :(得分:2)
您可能希望每个链接都在<li>
标记中:
替换:
$img_result.= CreateLink($filename, "image");
使用:
$img_result.= '<li>' . CreateLink($filename, "image") . '</li>;
// the other places where you use that function as well...
并确保列表仅以<ul>
标记开头和结尾(不是附加<li>
)