選擇多個標籤內的ID與ID

Question

在jQuery中，我怎樣才能建立一個標籤h1，h2，h3和p內選擇#con？

像$('#con h1, #con h2, #con h3, #con p')，但沒有重複#con

Answer 1

你可以做任何以下的：

$("#con h1, #con h2, #con h3, #con p") // your original 

$("h1, h2, h3, p", $("#con"))   // pass jQuery object in as context 

$("h1, h2, h3, p", "#con")    // pass selector in as context 

$("#con").find("h1, h2, h3, p")   // do what jQuery ultimately does 
             // in the end when passing context 
             // as jQuery or as string selector 

I've created a nice little jsFiddle that demos all of this.

Answer 2

$('h1, h2, h3, ...', $('#con'))會工作。這將上下文設置爲#con並在上下文（#con）內搜索h1，h2等。