blocks|key|255456|text|这要好得多：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|255457|$('code').contents().unwrap().wrap('<pre/>');|code-block|syntax|javascript|255458|尽管无可否认，Felix+Kling's+solution近似于twice+as+fast|offset|length|255459|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/7093417/using-jquery-to-replace-one-tag-with-another/7093472#7093472|1|http://jsperf.com/substituting-one-tag-for-another-with-jquery^0|0|0|7|M|0|W|D|1|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|W|8|@]|9|@$I|X|J|Y|1|Z]|$I|10|J|11|1|12]]|A|$]]|$1|K|3|-4|5|6|7|13|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|R]]|S|$5|N|O|P|A|$Q|T]]]]

This is much nicer:

<pre><code>$('code').contents().unwrap().wrap('&lt;pre/&gt;');
</code></pre>

Though admittedly <a href="https://stackoverflow.com/questions/7093417/using-jquery-to-replace-one-tag-with-another/7093472#7093472">Felix Kling's solution</a> is approximately <a href="http://jsperf.com/substituting-one-tag-for-another-with-jquery" rel="noreferrer">twice as fast</a>:

blocks|key|2944329|text|您将始终获得第一个code的内容，这是正确的，因为$('code').html()将始终引用第一个元素，无论您在哪里使用它。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2944330|相反，您可以使用.each遍历所有元素并分别更改每个元素：|2944331|$('code').each(function()+{
++++$(this).replaceWith(+"<pre>"+%2B+$(this).html()+%2B+"</pre>"+);
++++//+this+function+is+executed+for+all+'code'+elements,+and
++++//+'this'+refers+to+one+element+from+the+set+of+all+'code'
++++//+elements+each+time+it+is+called.
});|code-block|syntax|javascript|2944332|entityMap|0|LINK|mutability|MUTABLE|url|http://api.jquery.com/each/^0|9|4|P|G|0|8|5|8|5|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]|$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|Z|8|@$9|10|A|11|B|C]]|D|@$9|12|A|13|1|14]]|E|$]]|$1|H|3|I|5|J|7|15|8|@]|D|@]|E|$K|L]]|$1|M|3|-4|5|6|7|16|8|@]|D|@]|E|$]]]|N|$O|$5|P|Q|R|E|$S|T]]]]

It's correct that you'll always obtain the first <code>code</code>'s contents, because <code>$('code').html()</code> will always refer to the first element, wherever you use it.

Instead, you could use <a href="http://api.jquery.com/each/"><code>.each</code></a> to iterate over all elements and change each one individually:

<pre><code>$('code').each(function() {
 $(this).replaceWith( "&lt;pre&gt;" + $(this).html() + "&lt;/pre&gt;" );
 // this function is executed for all 'code' elements, and
 // 'this' refers to one element from the set of all 'code'
 // elements each time it is called.
});
</code></pre>

blocks|key|2944334|text|试试这个：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2944335|$('code').each(function(){

++++$(this).replaceWith(+"<pre>"+%2B+$(this).html()+%2B+"</pre>"+);

});|code-block|syntax|javascript|2944336|http://jsfiddle.net/mTGhV/|offset|length|2944337|entityMap|0|LINK|mutability|MUTABLE|url^0|0|0|0|Q|0|0^^$0|@$1|2|3|4|5|6|7|R|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|S|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|T|8|@]|9|@$I|U|J|V|1|W]]|A|$]]|$1|K|3|-4|5|6|7|X|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|H]]]]

Try this:

<pre><code>$('code').each(function(){

 $(this).replaceWith( "&lt;pre&gt;" + $(this).html() + "&lt;/pre&gt;" );

});
</code></pre>

<a href="http://jsfiddle.net/mTGhV/" rel="noreferrer">http://jsfiddle.net/mTGhV/</a>

blocks|key|255428|text|这个怎么样？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|255429|$('code').each(function+()+{
++++$(this).replaceWith(+"<pre>"+%2B+$(this).html()+%2B+"</pre>"+);
});|code-block|syntax|javascript|255430|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

How about this?

<pre><code>$('code').each(function () {
 $(this).replaceWith( "&lt;pre&gt;" + $(this).html() + "&lt;/pre&gt;" );
});
</code></pre>

blocks|key|2944510|text|基于Felix的答案。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2944511|$('code').replaceWith(function()+{
++++var+replacement+=+$('<pre>').html($(this).html());
++++for+(var+i+=+0;+i+<+this.attributes.length;+i%2B%2B)+{
++++++++replacement.attr(this.attributes[i].name,+this.attributes[i].value);
++++}
++++return+replacement;
});|code-block|syntax|javascript|2944512|这将在替换pre标记中重现code标记的属性。|offset|length|style|CODE|2944513|编辑:这将替换那些在其他code标签的innerHTML中的code标签。|2944514|function+replace(thisWith,+that)+{
++++$(thisWith).replaceWith(function()+{
++++++++var+replacement+=+$('<'+%2B+that+%2B+'>').html($(this).html());
++++++++for+(var+i+=+0;+i+<+this.attributes.length;+i%2B%2B)+{
++++++++++++replacement.attr(this.attributes[i].name,+this.attributes[i].value);
++++++++}
++++++++return+replacement;
++++});
++++if+($(thisWith).length>0)+{
++++++++replace(thisWith,+that);
++++}
}

replace('code','pre');|2944515|entityMap^0|0|0|5|3|D|4|0|C|4|J|9|U|4|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@$I|V|J|W|K|L]|$I|X|J|Y|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|Z|8|@$I|10|J|11|K|L]|$I|12|J|13|K|L]|$I|14|J|15|K|L]]|9|@]|A|$]]|$1|O|3|P|5|D|7|16|8|@]|9|@]|A|$E|F]]|$1|Q|3|-4|5|6|7|17|8|@]|9|@]|A|$]]]|R|$]]

Building up on Felix's answer.

<pre><code>$('code').replaceWith(function() {
 var replacement = $('&lt;pre&gt;').html($(this).html());
 for (var i = 0; i &lt; this.attributes.length; i++) {
 replacement.attr(this.attributes[i].name, this.attributes[i].value);
 }
 return replacement;
});
</code></pre>

This will reproduce the attributes of the <code>code</code> tags in the replacement <code>pre</code> tags.

Edit: This will replace even those <code>code</code> tags that are inside the <code>innerHTML</code> of other <code>code</code> tags.

<pre><code>function replace(thisWith, that) {
 $(thisWith).replaceWith(function() {
 var replacement = $('&lt;' + that + '&gt;').html($(this).html());
 for (var i = 0; i &lt; this.attributes.length; i++) {
 replacement.attr(this.attributes[i].name, this.attributes[i].value);
 }
 return replacement;
 });
 if ($(thisWith).length&gt;0) {
 replace(thisWith, that);
 }
}

replace('code','pre');
</code></pre>

blocks|key|260599|text|从jQuery+1.4.2开始：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|260600|$('code').replaceWith(function(i,html)+{
++return+$('<pre+/>').html(html);
});​|code-block|syntax|javascript|260601|然后可以选择新图元：|260602|$('pre').css('color','red');|260603|来源：http://api.jquery.com/replaceWith/#comment-45493689|offset|length|260604|jsFiddle：http://jsfiddle.net/k2swf/16/|260605|entityMap|0|LINK|mutability|MUTABLE|url|http://api.jquery.com/replaceWith/#comment-45493689|1|http://jsfiddle.net/k2swf/16/^0|0|0|0|0|3|1F|0|0|9|T|1|0^^$0|@$1|2|3|4|5|6|7|10|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|12|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|13|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|14|8|@]|9|@$M|15|N|16|1|17]]|A|$]]|$1|O|3|P|5|6|7|18|8|@]|9|@$M|19|N|1A|1|1B]]|A|$]]|$1|Q|3|-4|5|6|7|1C|8|@]|9|@]|A|$]]]|R|$S|$5|T|U|V|A|$W|X]]|Y|$5|T|U|V|A|$W|Z]]]]

As of jQuery 1.4.2:

<pre><code>$('code').replaceWith(function(i,html) {
 return $('&lt;pre /&gt;').html(html);
});​
</code></pre>

You can then select the new elements:

<pre><code>$('pre').css('color','red');
</code></pre>

Source: <a href="http://api.jquery.com/replaceWith/#comment-45493689" rel="nofollow">http://api.jquery.com/replaceWith/#comment-45493689</a>

jsFiddle: <a href="http://jsfiddle.net/k2swf/16/" rel="nofollow">http://jsfiddle.net/k2swf/16/</a>

blocks|key|260649|text|如果您使用的是vanilla+JavaScript，您将：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|260650|260651|创建新元素|unordered-list-item|260652|将旧元素的子元素移动到新元素|260653|在旧元素之前插入新元素|260654|删除旧元素|260655|260656|以下是此过程的jQuery等价物：|260657|$("code").each(function+()+{
++++$("<pre></pre>").append(this.childNodes).insertBefore(this);
++++$(this).remove();
});|code-block|syntax|javascript|260658|下面是jsperf：|260659|http://jsperf.com/substituting-one-tag-for-another-with-jquery/7|offset|length|260660|PS:所有使用destructive.或.html()的解决方案都是.innerHTML|style|BOLD|CODE|260661|entityMap|0|LINK|mutability|MUTABLE|url^0|0|0|0|0|0|0|0|0|0|0|0|1S|0|0|7|B|K|7|Y|A|0^^$0|@$1|2|3|4|5|6|7|1B|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|1C|8|@]|9|@]|A|$]]|$1|C|3|D|5|E|7|1D|8|@]|9|@]|A|$]]|$1|F|3|G|5|E|7|1E|8|@]|9|@]|A|$]]|$1|H|3|I|5|E|7|1F|8|@]|9|@]|A|$]]|$1|J|3|K|5|E|7|1G|8|@]|9|@]|A|$]]|$1|L|3|-4|5|6|7|1H|8|@]|9|@]|A|$]]|$1|M|3|N|5|6|7|1I|8|@]|9|@]|A|$]]|$1|O|3|P|5|Q|7|1J|8|@]|9|@]|A|$R|S]]|$1|T|3|U|5|6|7|1K|8|@]|9|@]|A|$]]|$1|V|3|W|5|6|7|1L|8|@]|9|@$X|1M|Y|1N|1|1O]]|A|$]]|$1|Z|3|10|5|6|7|1P|8|@$X|1Q|Y|1R|11|12]|$X|1S|Y|1T|11|13]|$X|1U|Y|1V|11|13]]|9|@]|A|$]]|$1|14|3|-4|5|6|7|1W|8|@]|9|@]|A|$]]]|15|$16|$5|17|18|19|A|$1A|W]]]]

If you were using vanilla JavaScript you would:

<ul>
<li>Create the new element</li>
<li>Move the children of old element into the new element</li>
<li>Insert the new element before the old one</li>
<li>Remove the old element</li>
</ul>

Here is jQuery equivalent of this process:

<pre><code>$("code").each(function () {
 $("&lt;pre&gt;&lt;/pre&gt;").append(this.childNodes).insertBefore(this);
 $(this).remove();
});
</code></pre>

Here is the jsperf URL: 
<a href="http://jsperf.com/substituting-one-tag-for-another-with-jquery/7" rel="nofollow">http://jsperf.com/substituting-one-tag-for-another-with-jquery/7</a>

PS: All solutions that use <code>.html()</code> or <code>.innerHTML</code> are destructive.

blocks|key|2944663|text|另一种简单的方法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2944664|$('code').wrapInner('<pre+/>').contents();|code-block|syntax|javascript|2944665|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Another short &amp; easy way:

<pre><code>$('code').wrapInner('&lt;pre /&gt;').contents();
</code></pre>

blocks|key|259568|text|这里给出的所有答案都假定(如问题示例所示)标签中没有属性。如果对以下内容运行接受的答案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|259569|<code+class='cls'>A</code>|code-block|syntax|javascript|259570|如果将替换为|259571|<pre>A</pre>|259572|如果你也想保留这些属性--这就是替换一个标签的意思……？这就是解决方案：|259573|$("code").each(+function(){
++++var+content+=+$(+"<pre>"+);

++++$.each(+this.attributes,+function(){+
+++++++++content.attr(+this.name,+this.value+);
++++}+);

++++$(+this+).replaceWith(+content+);
}+);|259574|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

All answers given here assume (as the question example indicates) that there are no attributes in the tag. If the accepted answer is ran on this:

<pre><code>&lt;code class='cls'&gt;A&lt;/code&gt;
</code></pre>

if will be replaced with 

<pre><code>&lt;pre&gt;A&lt;/pre&gt;
</code></pre>

What if you want to keep the attributes as well - which is what replacing a tag would mean... ? This is the solution:

<pre><code>$("code").each( function(){
 var content = $( "&lt;pre&gt;" );

 $.each( this.attributes, function(){ 
 content.attr( this.name, this.value );
 } );

 $( this ).replaceWith( content );
} );
</code></pre>

blocks|key|2944482|text|$('code').each(function(){
++++$(this).replaceWith(+"<pre>"+%2B+$(this).html()++%2B+"</pre>"+);
++++});|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|2944483|最好的和干净的方法。|unstyled|2944484|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>$('code').each(function(){
 $(this).replaceWith( "&lt;pre&gt;" + $(this).html() + "&lt;/pre&gt;" );
 });
</code></pre>

Best and clean way.

blocks|key|257584|text|您可以使用jQuery的html函数。下面是一个示例，它将代码标记替换为pre标记，同时保留了代码的所有属性。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|257585|++++$('code').each(function()+{
++++++++var+temp=$(this).html();
++++++++temp=temp.replace("code","pre");
++++++++$(this).html(temp);
++++});|code-block|syntax|javascript|257586|这可以与需要交换的任何一组html元素标签一起工作，同时保留先前标签的所有属性。|257587|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

You could use jQuery's html function. Below is a sample the replaces a code tag with a pre tag while retaining all of the attributes of the code.

<pre><code> $('code').each(function() {
 var temp=$(this).html();
 temp=temp.replace("code","pre");
 $(this).html(temp);
 });
</code></pre>

This could work with any set of html element tags that needed to be swapped while retaining all the attributes of the previous tag.

blocks|key|2944613|text|制作了jquery插件，还维护了属性。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2944614|$.fn.renameTag+=+function(replaceWithTag){
++this.each(function(){
++++++++var+outerHtml+=+this.outerHTML;
++++++++var+tagName+=+$(this).prop("tagName");
++++++++var+regexStart+=+new+RegExp("%5E<"%2BtagName,"i");
++++++++var+regexEnd+=+new+RegExp("</"%2BtagName%2B">$","i")
++++++++outerHtml+=+outerHtml.replace(regexStart,"<"%2BreplaceWithTag)
++++++++outerHtml+=+outerHtml.replace(regexEnd,"</"%2BreplaceWithTag%2B">");
++++++++$(this).replaceWith(outerHtml);
++++});
++++return+this;
}|code-block|syntax|javascript|2944615|用法：|2944616|$('code').renameTag('pre')|2944617|entityMap^0|3|6|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|U|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|W|8|@]|D|@]|E|$]]]|P|$]]

Made <code>jquery</code> plugin, maintaining attributes also.

<pre><code>$.fn.renameTag = function(replaceWithTag){
 this.each(function(){
 var outerHtml = this.outerHTML;
 var tagName = $(this).prop("tagName");
 var regexStart = new RegExp("^&lt;"+tagName,"i");
 var regexEnd = new RegExp("&lt;/"+tagName+"&gt;$","i")
 outerHtml = outerHtml.replace(regexStart,"&lt;"+replaceWithTag)
 outerHtml = outerHtml.replace(regexEnd,"&lt;/"+replaceWithTag+"&gt;");
 $(this).replaceWith(outerHtml);
 });
 return this;
}
</code></pre>

Usage:

<pre><code>$('code').renameTag('pre')
</code></pre>

<h2>Goal:</h2>

Using jQuery, I'm trying to replace all the occurrences of: 

<pre><code>&lt;code&gt; ... &lt;/code&gt;
</code></pre>

with:

<pre><code>&lt;pre&gt; ... &lt;/pre&gt;
</code></pre>

<h3>My solution:</h3>

I got as far as the following, 

<pre><code>$('code').replaceWith( "&lt;pre&gt;" + $('code').html() + "&lt;/pre&gt;" );
</code></pre>

<h3>The problem with my solution:</h3>

but the issues is that it's replacing everything between the (second, third, fourth, etc)"code" tags with the content between the first "code" tags. 

e.g. 

<pre><code>&lt;code&gt; A &lt;/code&gt;
&lt;code&gt; B &lt;/code&gt;
&lt;code&gt; C &lt;/code&gt;
</code></pre>

becomes

<pre><code>&lt;pre&gt; A &lt;/pre&gt;
&lt;pre&gt; A &lt;/pre&gt;
&lt;pre&gt; A &lt;/pre&gt;
</code></pre>

I think I need to use "this" and some sort of function but I'm afraid I'm still learning and don't really understand how to piece a solution together.

Using jQuery to replace one tag with another

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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目标：使用jQuery，我尝试替换所有出现的：<code> ... </code>通过以下方式：<pre> ... </pre>我的解决方案是：我得到了如下结论：$('code').replaceWith( "<pre>" + $('code').html() + "</pre>" );我的解决方案的问题是：但问题是，它用first“代码”标记之间的内容替换了(第二、第三、第四等)“代码”标记之间

问使用jQuery将一个标记替换为另一个标记
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问使用jQuery将一个标记替换为另一个标记EN

回答 12

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问使用jQuery将一个标记替换为另一个标记
EN