【发布时间】:2022-01-21 03:09:34
【问题描述】:
我有简单的 html5 cgi 生成的文件网页,但我需要在 google 中有带有搜索选项的右键单击上下文菜单。
每个文件 div 看起来像这样:
if [[ $file =~ \.jpg$ || $file =~ \.JPG$ || $file =~ \.jpeg$ || $file =~ \.png$ || $file =~ \.PNG$ || $file =~ \.webp$ || $file =~ \.bmp$ ]]; then
{
echo '<div class=''"'lightbox-file'"'' data-src=''"'$file'"'' data-sub-html=''"'$file'"''>'
echo ' <div class="filemanager-photo">'
echo ' <a href=''"'$file'"''>'
echo ' <img src=''"'./.tmp/thumb - $file.jpg'"'' />'
echo ' <div class='"'filemanager-photo-name'"'>'$filename'</div>'
echo ' </a>'
echo ' </div>'
echo '</div>'
} >> "$url_local_path"/.tmp/list_photo
上下文菜单脚本:
# Context menu
echo ' <script src="/js/context-menu.js"></script> '
echo ' <script> '
echo ' var items = [ '
echo " { name: 'Pobierz', fn: function(target) { console.log('Pobierz', target); }}, "
echo " { name: 'Udostepnij', fn: function(target) { window.open('//facebook.com/sharer/sharer.php?u=' + window.location.href); }}, "
echo ' {}, '
echo " { name: 'Google', fn: function(target) { window.open('https://www.google.com/search?q=' + filemanager-photo-name??????????????????); }}, "
echo ' {}, '
echo " { name: 'Usuń', fn: function(target) { console.log('Usuń', target); }}, "
echo ' ]; '
echo " var cm1 = new ContextMenu('.filemanager-photo-name', items); "
#echo " var cm2 = new ContextMenu('.minimal', items, { className: 'ContextMenu--theme-custom', minimalStyling: true }); "
echo " cm1.on('shown', () => console.log('Context menu shown')); "
echo ' </script> '
当我右键单击页面上的照片链接时,我应该放什么来获得谷歌正确的关键字,即 $filemanager-photo-name 变量?
echo " { name: 'Google', fn: function(target) { window.open('https://www.google.com/search?q=' + filemanager-photo-name??????????????????); }}, "
【问题讨论】:
-
当我点击 div 到 java 变量时,如何将 div 与名称一起传递?我认为首先我需要将 div 名称变量传递给 contextmenu 列表,然后我可以在菜单选项中使用该变量