从div

Question

我知道如何使用.data（＆＃39; id＆＃39;）获取数据属性，例如.. 但我无法弄清楚如何获得＆＃34; attachment_id＆＃34;值包含在＆＃34; data-kgvid_video_vars＆＃34; ？

这就是我想要进入的地方所以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

从html组件获取属性的值：

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

在javascript中获取json变量的值：

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

alert(jsonData.attachment_id);

请参阅JSON.parse()

完成回答你的问题

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"])