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选择使用ExtJS不包含某种身份证件的内容
原标题:Selecting elements that do not contain a certain ID using ExtJS

页: 1 我如何利用ExtJS 2.3.0这样做? 我尝试如下:

Ext.query("select,input:not([id*=foo][id*=bar])", "SomeForm");`

......但它没有工作。 不包括有<代码>foo的身份证。 她们似乎从事以下工作:

Ext.query("select,input:not([id*=foo])", "SomeForm")`

我不敢肯定如何增加第二个国际发展指标。 任何想法?

最佳回答

不知道如何将这种选择结合起来,但我认为,如果你利用过滤功能,你可以筛选出第一个问题:

Ext.DomQuery.filter(Ext.query( input:not([id*=foo]) , SomeForm ), input:not([id*=bar]) );

我刚刚就这一职能做过更多的通知,这一职能可以发挥作用:

Ext.query( input:not([id*=foo]):not([id*=bar]) );

但是,这给我带来了你提到的同一错误:

Ext.query( input:not([id*=foo]):not([id*=bar]) , SomeForm );

因此,在我看来,当你通过一个具有多个属性选择者的具体根源时,似乎喜欢“ExtJS”查询功能。

问题回答

Ext.query ("select,input:not([id*=foo]):not([id*=bar]),“SomeForm”

Ext.query("select,input:not([id*=foo]):not([id*=bar])", "SomeForm")




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