In tableview, index is used to process English characters, Chinese characters (narrow characters and wide characters), and tableview is used to process English characters.

In tableview, index is used to process English characters, Chinese characters (narrow characters and wide characters), and tableview is used to process English characters.
[Problem description]

I encountered a tableview index problem when developing the iOS Address Book Project.

The test student found A bug: adding A section named "wide character A" cannot be merged into index A. Instead, an index named "A" is added,

Medium: An exception occurs at the end of the Right index. ABX is not merged into a normal index, but appears between Z and # Of a normal index.

[Problem Analysis]

Think about it, it should be because the encoding of the width Character A and A is inconsistent.

The previous code is like this. Check that the second line uses a regular expression to determine whether the first letter starts with a Latin letter and then mark this person as starting with a certain character. The Code is as follows:

NSPredicate * HanziPred = [NSPredicate attributes: @ "self matches % @", @ "[\ u4e00-\ u9fa5]"]; NSPredicate * EnglishPred = [NSPredicate attributes: @ "self matches % @", @ "^ [A-Z]. * $ "]; NSPredicate * numberPred = [NSPredicate predicateWithFormat: @" self matches % @ ", @" ^ [0-9]. * $ "]; if ([EnglishPred evaluateWithObject: first]) {// The Latin letter self. firstLetter = self. latinString. length> 0? [Self. latinString substringToIndex: 1]: @ "#";}

Assign the first value to firstLetter using a regular expression. Regular Expressions are really powerful and can be compatible with wide characters and narrow characters. In this way, both A and A conform to the regular expression, but they are actually two different characters.

[Solution]

My solution is to create a static global dictionary in advance and initialize it. When using regular expressions, first determine whether the dictionary can match to wide characters, if it can be forcibly replaced with a narrow character.

Static NSDictionary * widthLetter;-(id) init {self = [super init]; self. firstLetter = @ "#"; widthLetter = @ {@ "A": @ "A", @ "B": @ "B", @ "C ": @ "C", @ "D": @ "D", @ "E": @ "E", @ "F": @ "F", @ "G ": @ "G", @ "H": @ "H", @ "I": @ "I", @ "J": @ "J", @ "K ": @ "K", @ "L": @ "L", @ "M": @ "M", @ "N": @ "N", @ "O ": @ "O", @ "P": @ "P", @ "Q": @ "Q", @ "R": @ "R", @ "S ": @ "S", @ "T": @ "T", @ "U": @ "U", @ "V": @ "V", @ "W ": @ "W", @ "X": @ "X", @ "Y": @ "Y", @ "Z": @ "Z"}; return s Elf;} if ([EnglishPred evaluateWithObject: first]) {// Latin letters // compatible with wide characters if (widthLetter [first]) {self. firstLetter = self. latinString. length> 0? WidthLetter [first]: @ "#";} else {self. firstLetter = self. latinString. length> 0? First :@"#";}}

Result]

Both A and A can enter the index of,