问题描述
我试图移植一个旧项目,它使用了Jquery-Layout.当我尝试启动应用程序时,它就到了线...
I am trying to port a legacy project and it used Jquery-Layout. When I try to start the application it gets to the line...
sC.selector = $N.selector.split(".slice")[0];
还有酒吧...
现在我对该库了解不多,但是如果我在故障点断点,我会看到$N instanceof jQuery
返回true,并且我在集合中看到一个DOM元素.是否可能没有选择器?
Now I don't know a lot about this library but if I breakpoint at the point of failure I see $N instanceof jQuery
returns true and I see a DOM element in the collection. Is it possible for it not to have a selector?
推荐答案
对于jquery.layout 1.4.4,只需注释第1831行:sC.selector = $ N.selector.split(.slice")[0];一切正常(在jQuery v3.3.1上测试)
For jquery.layout 1.4.4 just comment row 1831: sC.selector = $N.selector.split(".slice")[0]; and all works normally (tested on jQuery v3.3.1)
这篇关于jQuery-Layout引发错误,因为$ N.selector未定义的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!