你好!这个问题很长,所以一定要先坐下:)
在我的代码中,我从用户那里得到了一个字符串,然后我分析了那个字符串。通过分析,我的意思是遍历每个字符并将其与用户在 NSPredicateEditor 中设置的谓词进行数学运算。谓词以这种方式以编程方式设置:
NSArray* keyPaths = [NSArray arrayWithObjects:
[NSExpression expressionForKeyPath: @"Character"],
[NSExpression expressionForKeyPath: @"Character Before"],
[NSExpression expressionForKeyPath: @"Character After"],
nil];
// -----------------------------------------
NSArray* constants = [NSArray arrayWithObjects:
[NSExpression expressionForConstantValue: @"Any letter"],
[NSExpression expressionForConstantValue: @"Letter (uppercase)"],
[NSExpression expressionForConstantValue: @"Letter (lowercase)"],
[NSExpression expressionForConstantValue: @"Number"],
[NSExpression expressionForConstantValue: @"Alphanumerical"],
[NSExpression expressionForConstantValue: @"Ponctuation Mark"],
nil];
// -----------------------------------------
NSArray* compoundTypes = [NSArray arrayWithObjects:
[NSNumber numberWithInteger: NSNotPredicateType],
[NSNumber numberWithInteger: NSAndPredicateType],
[NSNumber numberWithInteger: NSOrPredicateType],
nil];
// -----------------------------------------
NSArray* operatorsA = [NSArray arrayWithObjects:
[NSNumber numberWithInteger: NSEqualToPredicateOperatorType],
[NSNumber numberWithInteger: NSNotEqualToPredicateOperatorType],
nil];
NSArray* operatorsB = [NSArray arrayWithObjects:
[NSNumber numberWithInteger: NSInPredicateOperatorType],
nil];
// -----------------------------------------
NSPredicateEditorRowTemplate* template1 = [[NSPredicateEditorRowTemplate alloc] initWithLeftExpressions: keyPaths
rightExpressions: constants
modifier: NSDirectPredicateModifier
operators: operatorsA
options: 0];
NSPredicateEditorRowTemplate* template2 = [[NSPredicateEditorRowTemplate alloc] initWithLeftExpressions: keyPaths
rightExpressionAttributeType: NSStringAttributeType
modifier: NSDirectPredicateModifier
operators: operatorsB
options: 0];
NSPredicateEditorRowTemplate* compound = [[NSPredicateEditorRowTemplate alloc] initWithCompoundTypes: compoundTypes];
// -----------------------------------------
NSArray* rowTemplates = [NSArray arrayWithObjects: template1, template2, compound, nil];
[myPredicateEditor setRowTemplates: rowTemplates];
所以你可以看到我有三个键路径和一些可以与之比较的常量。在分析字符串时,我基本上想这样做,在伪代码中:
originalString = [string from NSTextView]
for (char in originalString)
bChar = [character before char]
aChar = [character after char]
predicate = [predicate from myPredicateEditor]
// using predicate - problem!
result = [evaluate predicate with:
bChar somehow 'linked' to keypath 'Character Before'
char 'linked' to 'Character'
aChar 'linked' to 'Character After' // These values change, of course
and also:
constant "All letters" as "abc...WXYZ"
constant "Numbers" as "0123456789"
etc for all other constants set up // These don't
]
if (result) then do something with char, bChar and aChar
你可以看到我的问题基本上出在哪里:
由于空间的原因,'Character Before/After' 不能是键路径,但我想保持这种方式,因为它对用户来说更漂亮(想象一下有一些作为 'characterBefore' 的东西......)
诸如“数字”之类的常量实际上代表了诸如“0123456789”之类的字符串,我也无法向用户显示
我能够找到解决此问题的方法,但现在它不适用于每个角色,而且效率也很低(至少在我看来)。我所做的是从谓词中获取谓词格式,替换我必须替换的内容,然后评估新格式。现在来看一些解释这一点的真实代码:
#define kPredicateSubstitutionsDict [NSDictionary dictionaryWithObjectsAndKeys: \
@"IN 'abcdefghijklmnopqrstuvwxyz'", @"== \"Letter (lowercase)\"", \
@"IN 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'", @"== \"Letter (uppercase)\"", \
@"IN 'abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ'", @"== \"Any letter\"", \
@"IN '1234567890'", @"== \"Number\"", \
@"IN 'abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ1234567890'", @"== \"Alphanumerical\"", \
@"IN ',.;:!?'", @"== \"Ponctuation Mark\"", \
\
@"MATCHES '[^a-z]'" , @"!= \"Letter (lowercase)\"", \
@"MATCHES '[^A-Z]'" , @"!= \"Letter (uppercase)\"", \
@"MATCHES '[^a-zA-Z]'" , @"!= \"Any letter\"", \
@"MATCHES '[^0-9]'" , @"!= \"Number\"", \
@"MATCHES '[^a-zA-Z0-9]'" , @"!= \"Alphanumerical\"", \
@"MATCHES '[^,\.;:!\?]'" , @"!= \"Ponctuation Mark\"", \
\
nil]
// NSPredicate* predicate is setup in another place
// NSString* originalString is also setup in another place
NSString* predFormat = [predicate predicateFormat];
for (NSString* key in [kPredicateSubstitutionsDict allKeys]) {
prefFormat = [predFormat stringByReplacingOccurrencesOfString: key withString: [kPredicateSubstitutionsDict objectForKey: key]];
}
for (NSInteger i = 0; i < [originalString length]; i++) {
NSString* charString = [originalString substringWithRange: NSMakeRange(i, 1)];
NSString* bfString;
NSString* afString;
if (i == 0) {
bfString = @"";
}
else {
bfString = [originalString substringWithRange: NSMakeRange(i - 1, 1)];
}
if (i == [originalString length] - 1) {
afString = @"";
}
else {
afString = [originalString substringWithRange: NSMakeRange(i + 1, 1)];
}
predFormat = [predFormat stringByReplacingOccurrencesOfString: @"Character Before" withString: [NSString stringWithFormat: @"\"%@\"", bfString]];
predFormat = [predFormat stringByReplacingOccurrencesOfString: @"Character After" withString: [NSString stringWithFormat: @"\"%@\"", afString]];
predFormat = [predFormat stringByReplacingOccurrencesOfString: @"Character" withString: [NSString stringWithFormat: @"\"%@\"", charString]];
NSPredicate* newPred = [NSPredicate predicateWithFormat: predFormat];
if ([newPred evaluateWithObject: self]) { // self just so I give it something (nothing is actually gotten from self)
// if predicate evaluates to true, then do something exciting!
}
}
所以,给你,这是我正在做的简化版本。如果您看到任何拼写错误,很可能它们不在我的代码中,因为我已经对此进行了相当多的编辑,所以它会更简单。
总结一下:
- 我需要评估用户针对许多字符所做的谓词,对其进行大量修改,同时尽可能提高效率
我发现我的方法存在的问题是:
- 我觉得一点都不干净
- 我不能保证它适用于所有情况(当其中一个字符是换行符/输入字符时,谓词会引发错误,说它无法理解格式)
这就是所有的人!感谢您到目前为止的阅读。解决这个烂摊子时,愿你的上帝与你同在!
编辑:为了澄清一些事情,我要补充一点,这个问题的触发因素是我不能在设置谓词编辑器的一开始就定义一个带有名称的常量(显示给用户) 和一个表示该常量并以谓词格式插入的值。键路径也是如此:如果我可以有一个显示名称,然后一个值就是谓词的那些 var 字符串($VAR 或其他任何东西),那么所有问题都将得到解决。如果这是可能的,请告诉我如何。如果不可能,那么请关注我描述的其他问题。