这可能非常简单,但我不习惯这种编码方式 - 如何更改此选择菜单:
$control .= '<select name="'. $this->_hash_value($hash, $xml_obj->value) .'" id="'. $this->_hash_value($hash, $xml_obj->value) .'" data-native-menu="false">';
foreach ($nameArr as $folder => $imageArr) {
foreach ($imageArr as $image) {
if (substr($folder, 1, strlen($folder)).$image == $xml_obj->value) {
$control .= '<option value="'. substr($folder, 1, strlen($folder)).$image .'" selected="selected">'. $image .'</option>';
} else {
$control .= '<option value="'. substr($folder, 1, strlen($folder)).$image .'">'. $image .'</option>';
}
}
}
$control .= '</select>';
进入单个列表项(具有当前所选项的值),该列表项会导致基本列表包含用户可以选择的所有项目?
答案 0 :(得分:1)
你应该可以这样做:
$control .= '<ul id="'. $this->_hash_value($hash, $xml_obj->value) .'" data-role="listview" data-theme="g">';
foreach($nameArr as $folder => $imageArr)
{
foreach ($imageArr as $image) {
$control .= '<li><a href="'. substr($folder, 1, strlen($folder)).$image .'">'. $image .'</a></li>';
}
}
$control .= "</ul>";
答案 1 :(得分:0)
清单1:
$control .= '<ul data-role="listview" data-inset="true">';
$control .= '<li>';
$control .= '<a href="'. site_url() .'/mobilegallery/gallery/'.$x[0]->attributes()->indexI.'">Bilder</a>';
$control .= '</li>';
$control .= '</ul>';
列表2,在新页面上:
$html .= '<ul data-role="listview" class="ui-listview" data-inset="true">';
for($i = 0, $c = $xml->COM->MOVIE->count(); $i < $c; $i++ ){
$html .= '<li>';
$html .= '<a>
<img src="https://[url]/'.rawurlencode($this->_decode_path($xml->COM->MOVIE[$i]->attributes()->dbIcoFilename)).'" id="imgThumb" alt="'.$xml->COM->MOVIE[$i]->attributes()->nameS.'" />
<h1>'.$xml->COM->MOVIE[$i]->attributes()->nameS.'</h1>
</a>';
$html .= ' </li>';
}
$html .= '</ul>';