blocks|key|1085933|text|在为NSObject的子类重载-description时，我也遇到了同样的问题。在这种情况下，我可以使用回车符(\r)而不是换行符(\n)来创建换行符。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1085934|[NSString+stringWithFormat:@"%25@\r%25@",+mystring1,mystring2];|offset|length|style|CODE|1085935|entityMap^0|0|0|1N|0^^$0|@$1|2|3|4|5|6|7|J|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|K|8|@$D|L|E|M|F|G]]|9|@]|A|$]]|$1|H|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|I|$]]

I just ran into the same issue when overloading -description for a subclass of NSObject. In this situation I was able to use carriage return (\r) instead of newline (\n) to create a line break.

<code>[NSString stringWithFormat:@"%@\r%@", mystring1,mystring2];</code>

blocks|key|1301188|text|如果字符串以UIView+(例如，UILabel)的形式存在，则还需要将行数设置为0|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1301189|myView.numberOfLines=0;|code-block|syntax|javascript|1301190|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|$]]

If your string is going in a UIView (e.g a UILabel), you also need to set the number of lines to 0

<pre><code>myView.numberOfLines=0;
</code></pre>

blocks|key|1301418|text|我发现当我从.plist文件中读入字符串时，出现的"\n"被解析为"\\n"。对我来说，解决方案是用"\n"替换出现的"\\n"。例如，假设从我的.plist文件中读入一个名为myString的NSString实例，我必须调用...|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1301419|myString+=+[myString+stringByReplacingOccurrencesOfString:@"\\n"+withString:@"\n"];|code-block|syntax|javascript|1301420|..。在将其分配给我的UILabel实例之前...|1301421|myLabel.text+=+myString;|1301422|entityMap^0|6|6|P|4|X|5|1E|4|1N|5|21|6|2G|8|2P|8|0|0|B|7|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]|$9|T|A|U|B|C]|$9|V|A|W|B|C]|$9|X|A|Y|B|C]|$9|Z|A|10|B|C]|$9|11|A|12|B|C]|$9|13|A|14|B|C]|$9|15|A|16|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|17|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|18|8|@$9|19|A|1A|B|C]]|D|@]|E|$]]|$1|M|3|N|5|H|7|1B|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|1C|8|@]|D|@]|E|$]]]|P|$]]

I found that when I was reading strings in from a <code>.plist</code> file, occurrences of <code>"\n"</code> were parsed as <code>"\\n"</code>. The solution for me was to replace occurrences of <code>"\\n"</code> with <code>"\n"</code>. For example, given an instance of <code>NSString</code> named <code>myString</code> read in from my <code>.plist</code> file, I had to call...

<pre><code>myString = [myString stringByReplacingOccurrencesOfString:@"\\n" withString:@"\n"];
</code></pre>

... before assigning it to my <code>UILabel</code> instance...

<pre><code>myLabel.text = myString;
</code></pre>

blocks|key|1301255|text|NSString+*str1+=+@"Share+Role+Play+Photo+via+Facebook,+or+Twitter+for+free+coins+per+photo.";
NSString+*str2+=+@"Like+Role+Play+on+facebook+for+50+free+coins.";
NSString+*str3+=+@"Check+out+'What's+Hot'+on+other+ways+to+receive+free+coins";


NSString+*msg+=+[NSString+stringWithFormat:@"%25@\n%25@\n%25@",+str1,+str2,+str3];|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1301256|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>NSString *str1 = @"Share Role Play Photo via Facebook, or Twitter for free coins per photo.";
NSString *str2 = @"Like Role Play on facebook for 50 free coins.";
NSString *str3 = @"Check out 'What's Hot' on other ways to receive free coins";


NSString *msg = [NSString stringWithFormat:@"%@\n%@\n%@", str1, str2, str3];
</code></pre>

blocks|key|1086150|text|试试这个(+stringWithFormat必须以小写字母开头)|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1086151|+[NSString+stringWithFormat:@"%25@\n%25@",string1,string2];|code-block|syntax|javascript|1086152|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|$]]

try this ( stringWithFormat has to start with lowercase)

<pre><code> [NSString stringWithFormat:@"%@\n%@",string1,string2];
</code></pre>

blocks|key|1086045|text|\n\r看起来很适合我。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1086046|我正在使用带有IOS+6.0的Xcode+4.6作为目标。已在iPhone+4S上测试。你自己试试吧。|1086047|冯球|1086048|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|H|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|I|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|J|8|@]|9|@]|A|$]]|$1|F|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|G|$]]

\n\r seems working for me.

I am using Xcode 4.6 with IOS 6.0 as target. Tested on iPhone 4S. Try it by yourself.

Feng Chiu

blocks|key|1086263|text|NSString的换行符是\r|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1086264|正确使用[NSString+StringWithFormat:@"%25@\r%25@",string1,string2];|offset|length|style|CODE|1086265|\r+->回车|1086266|entityMap^0|0|4|1I|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|M|8|@$D|N|E|O|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|P|8|@]|9|@]|A|$]]|$1|J|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|K|$]]

Line breaks character for NSString is \r 

correct way to use <code>[NSString StringWithFormat:@"%@\r%@",string1,string2];</code>

\r ----> carriage return

blocks|key|1301507|text|\n是首选的换行方式。\r也可以工作。\n\r或\r\n过于夸张，可能会在以后给您带来问题。Cocoa，具有特定的段落和换行符，用于特定用途NSParagraphSeparatorCharacter、NSLineSeparatorCharacter。Here是我以上所有内容的来源。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1301508|entityMap|0|LINK|mutability|MUTABLE|url|https://developer.apple.com/library/mac/documentation/Cocoa/Conceptual/Strings/Articles/stringsParagraphBreaks.html^0|0|2|B|2|J|4|O|4|1Y|T|2S|O|3H|4|0|0^^$0|@$1|2|3|4|5|6|7|N|8|@$9|O|A|P|B|C]|$9|Q|A|R|B|C]|$9|S|A|T|B|C]|$9|U|A|V|B|C]|$9|W|A|X|B|C]|$9|Y|A|Z|B|C]]|D|@$9|10|A|11|1|12]]|E|$]]|$1|F|3|-4|5|6|7|13|8|@]|D|@]|E|$]]]|G|$H|$5|I|J|K|E|$L|M]]]]

<code>\n</code> is the preferred way to break a line. <code>\r</code> will work too. <code>\n\r</code> or <code>\r\n</code> are overkill and may cause you issues later. Cocoa, has specific paragraph and line break characters for specific uses <code>NSParagraphSeparatorCharacter</code>, <code>NSLineSeparatorCharacter</code>. <a href="https://developer.apple.com/library/mac/documentation/Cocoa/Conceptual/Strings/Articles/stringsParagraphBreaks.html" rel="nofollow">Here</a> is my source for all the above.

blocks|key|1086513|text|如果\n或\r不起作用，并且您正在使用uiwebview并尝试使用<+br+>标记来插入新行来加载html。不要只在NSString+stringWithFormat中使用<+br+>标记。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1086514|取而代之的是通过追加来使用相同的方法。即通过使用stringByAppendingString|1086515|+yourNSString+=+[yourNSString+stringByAppendingString:@" "];|code-block|syntax|javascript|1086516|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|L|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|M|8|@]|9|@]|A|$G|H]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

In case \n or \r is not working and if you are working with uiwebview and trying to load html using &lt; br > tag to insert new line. Don't just use &lt; br > tag in NSString stringWithFormat.

Instead use the same by appending. i.e by using stringByAppendingString

<pre><code> yourNSString = [yourNSString stringByAppendingString:@"&lt;br&gt;"];
</code></pre>

blocks|key|1301590|text|在Swift+3中，它要简单得多|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1301591|let+stringA+=+"Terms+and+Conditions"
let+stringB+=+"Please+read+the+instructions"

yourlabel.text+=++"\(stringA)\n\(stringB)"|code-block|syntax|javascript|1301592|或者如果您使用的是textView|1301593|yourtextView.text+=++"\(stringA)\n\(stringB)"|1301594|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]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

In Swift 3, its much simpler

<pre><code>let stringA = "Terms and Conditions"
let stringB = "Please read the instructions"

yourlabel.text = "\(stringA)\n\(stringB)"
</code></pre>

or if you are using a textView

<pre><code>yourtextView.text = "\(stringA)\n\(stringB)"
</code></pre>

Does anyone know how to use line breaks in NSString? I need to do something like this -

<pre><code>[NSString stringWithFormat:@"%@,\n%@", mystring1,mystring2];
</code></pre>

NSString with \n or line break

有人知道如何在NSString中使用换行符吗？我需要做这样的事情-[NSString stringWithFormat:@"%@,\n%@", mystring1,mystring2];

问带\n或换行符的NSString
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