从 div 检索数据属性

Question

例如，我知道如何使用 .data('id'( 获取数据属性。但是我不知道如何获取"数据kgvid_video_vars"中包含的"attachment_id"值？

这就是我试图让它进入的内容，以便我可以在其他地方使用它。

 alert(jQuery(this).find('.kgvid_videodiv').data('id'));

data-kgvid_video_vars='{"id":"kgvid_16","attachment_id":116,"player_type":"Video.js","width":"640","height":"360","fullwidth":"true","countable":true,"count_views":"start_complete","start":"","autoplay":"false","pauseothervideos":"true","set_volume":"1","mute":"false","meta":true,"endofvideooverlay":"","resize":"true","auto_res":"automatic","pixel_ratio":"true","right_click":"on","playback_rate":"false","nativecontrolsfortouch":"false","locale":"en","enable_resolutions_plugin":false}'

Answer 1

要从 html 组件获取属性的值，请执行以下操作：

kgvid_video_vars= jQuery(this).find('.kgvid_videodiv').data('kgvid_video_vars');

要在 javascript 中获取 json 变量的值，请执行以下操作：

kgvid_video_vars= '{"id":"kgvid_16","attachment_id":116}';
jsonData = JSON.parse(kgvid_video_vars);
alert(jsonData.attachment_id);

参见 JSON.parse((

为了满足回答您的问题

kgvid_video_vars= jQuery(this).find('.kgvid_videodiv').data('kgvid_video_vars');
jsonData = JSON.parse(kgvid_video_vars);
alert(jsonData.attachment_id);

Answer 2

var data-kgvid_video_vars = JSON.parse(jQuery(this).find('.kgvid_videodiv').data('data-kgvid_video_vars'));
alert(data-kgvid_video_vars["attachment_id"])