blocks|key|4182113|text|请尝试使用以下代码。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4182114|<script>
var+str1+=+'helloworld';
var+str2+=+'hello';
var+percent+=+0;
var+newstr+=+str1.search(str2);
if(newstr+!=+-1){

++++percent+=+(str2.length+*+100)/str1.length;
+++alert(percent%2B'%25');

}else{
++alert(percent%2B'%25');
}

</script>|code-block|syntax|javascript|4182115|如果我理解你的问题，我相信这会对你有帮助。|4182116|谢谢。|4182117|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|P|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Please try with below code. 

<pre><code>&lt;script&gt;
var str1 = 'helloworld';
var str2 = 'hello';
var percent = 0;
var newstr = str1.search(str2);
if(newstr != -1){

 percent = (str2.length * 100)/str1.length;
 alert(percent+'%');

}else{
 alert(percent+'%');
}

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

if i understood your question, i'm sure this will help you.

Thanks.

blocks|key|4182133|text|我不是一个数学家，也不能向你展示这样一个神奇的公式，但是这个例子会让你从递归开始--请仔细看看代码注释：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4182134|/*+same_words+array+should+hold+the+following:
+++["john",+"bbc",+"hot",+"red,",+"xyz,",+"apples,",+"and",+"likes"]

+++NOTES:
++++-+The+more+words+you+add,+the+less+the+percentage+will+be...+that's+how+percent+works.
++++-+milk+and+MILK+are+the+same+word,+but:++MILK+and+MILK,+aren't+the+same+because+of+the+comma+(same+goes+for+periods).
*/

var++jSmith+=++'JOHN+smith+LIKES+bananas,+HOT+chocolate,+AND+APPLES,+nonsense,+RED,+321,+XYZ,+BBC+or+npr',
jSmithArray+=++jSmith.trim().toLowerCase().split('+');+//Makes+an+array+of+words+for+jSmith

var++jDoe+=++'JOHN+doe+LIKES+oranges,+milk,+AND+APPLES,+XYZ,+123,+RED,+green,+HOT+and...+cnn+sucks,+BBC+rocks',
jDoeArray+=++jDoe.trim().toLowerCase().split('+');+//Makes+an+array+of+words+for+jDoe

var+bothJohns+=+jSmithArray.concat(jDoeArray);+//console.log(bothJohns);

var+c+=+0;+//counter
var+same_words+=+[];+//collection+container


//collects+all+the+words+that+occur+in+both+sentences
function+collectSimilarWords(word)+{
++same_words.push(word);
}


//The+w+parameter+holds+the+word+"john"+for+the+1st+time.
//The+fn+parameter+holds+always+the+collectSimilarWords()+function.
function+recur(w,+fn)+{
++for+(var+p+in+jSmithArray)+{+//Every+property+of+the+jSmithArray+Array+holds+a+word+string.
++++if+(w)+{
+++++++if+(w+==+jSmithArray[p])+{+//If+the+word+we+passed+in+as+a+parameter+is+the+same+word+as+one+of+the+jSmithArray+elements...
+++++++++++fn(+jSmithArray[p]+);+//...then+pass+this+word+to+the+collectSimilarWords()+function.
+++++++}
+++++++c+%2B=+1;+//Increase+c+so+we+can+move+to...
+++++++recur(jDoeArray[c],+collectSimilarWords);+//...the+next+word+recursively.
++++}
++}
}

//Call+recur()+and+pass+in+the+first+word+of+the+jDoeArray+as+the+1st+param,+and+a+function+as+the+2nd+param.
recur(jDoeArray[c],+collectSimilarWords);


function+calcWhatever(samewords)+{+//Feel+free+to+calculate+however+you+want.
+++var+percent+=+(+(samewords.length+*+100)+/+bothJohns.length+).toFixed(2);
+++var+msg+=+"Out+of+Smith\'s+plus+Doe's+words+(total:+"+%2B+bothJohns.length+%2B+"),+about+"+%2B+percent+%2B+"%25+(or+"+%2B+same_words.length+%2B+")+of+them+are+found+in+both+of+their+sentences!";
+++return+msg;
}

//Hopefuly+setTimeout()+will+log+the+same_words+array+when+the+recursive+function+has+finished.
window.setTimeout(function()+{
++++console.log(same_words);
++++console.log(++calcWhatever(same_words)++);
},+1000);|code-block|syntax|javascript|4182135|除了for+in循环之外，您还可以像这样轻松地使用for循环：|offset|length|style|CODE|4182136|for+(var+p+=+0;+p<jSmithArray.length;+p%2B%2B)+{+/*+...+*/+}|4182137|entityMap^0|0|0|2|6|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]]|$1|G|3|H|5|6|7|S|8|@$I|T|J|U|K|L]]|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|$]]

I am not a mathematician and can't show you a magic formula for something like this, however this example will get you started with recursion - please look at the code comments carefully:

<pre><code>/* same_words array should hold the following:
 ["john", "bbc", "hot", "red,", "xyz,", "apples,", "and", "likes"]

 NOTES:
 - The more words you add, the less the percentage will be... that's how percent works.
 - milk and MILK are the same word, but: MILK and MILK, aren't the same because of the comma (same goes for periods).
*/

var jSmith = 'JOHN smith LIKES bananas, HOT chocolate, AND APPLES, nonsense, RED, 321, XYZ, BBC or npr',
jSmithArray = jSmith.trim().toLowerCase().split(' '); //Makes an array of words for jSmith

var jDoe = 'JOHN doe LIKES oranges, milk, AND APPLES, XYZ, 123, RED, green, HOT and... cnn sucks, BBC rocks',
jDoeArray = jDoe.trim().toLowerCase().split(' '); //Makes an array of words for jDoe

var bothJohns = jSmithArray.concat(jDoeArray); //console.log(bothJohns);

var c = 0; //counter
var same_words = []; //collection container


//collects all the words that occur in both sentences
function collectSimilarWords(word) {
 same_words.push(word);
}


//The w parameter holds the word "john" for the 1st time.
//The fn parameter holds always the collectSimilarWords() function.
function recur(w, fn) {
 for (var p in jSmithArray) { //Every property of the jSmithArray Array holds a word string.
 if (w) {
 if (w == jSmithArray[p]) { //If the word we passed in as a parameter is the same word as one of the jSmithArray elements...
 fn( jSmithArray[p] ); //...then pass this word to the collectSimilarWords() function.
 }
 c += 1; //Increase c so we can move to...
 recur(jDoeArray[c], collectSimilarWords); //...the next word recursively.
 }
 }
}

//Call recur() and pass in the first word of the jDoeArray as the 1st param, and a function as the 2nd param.
recur(jDoeArray[c], collectSimilarWords);


function calcWhatever(samewords) { //Feel free to calculate however you want.
 var percent = ( (samewords.length * 100) / bothJohns.length ).toFixed(2);
 var msg = "Out of Smith\'s plus Doe's words (total: " + bothJohns.length + "), about " + percent + "% (or " + same_words.length + ") of them are found in both of their sentences!";
 return msg;
}

//Hopefuly setTimeout() will log the same_words array when the recursive function has finished.
window.setTimeout(function() {
 console.log(same_words);
 console.log( calcWhatever(same_words) );
}, 1000);
</code></pre>

Instead of a <code>for in</code> loop, you could as easily use a for loop like so:

<pre><code>for (var p = 0; p&lt;jSmithArray.length; p++) { /* ... */ }
</code></pre>

I have to compare two strings, in that how much percentage str1 is contains str2.

<pre><code>str1="Hello wolrd"
str2="hello"
</code></pre>

means <code>distance(str1, str2)</code> should return 50%.. like that.

Compute string distance/difference in JavaScript?

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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我必须比较两个字符串，因为str1包含str2的百分比是多少。str1="Hello wolrd"str2="hello"意味着distance(str1, str2)应该返回50%..就像这样。

问是否以JavaScript为单位计算字符串距离/差？
EN

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问是否以JavaScript为单位计算字符串距离/差？EN