Ios 替换主字符串中的多个不同子字符串

我有一个URL字符串，需要用它替换值

例如：

http:\\www.domain.com\id=[USER\u id]&record=[CALL\u KEY]&someother=[OTHER\u STUFF]

我需要在字符串中循环查找[]，然后用其中的内容替换它

现在，我必须对每个键进行硬编码，并将其替换为值

- (NSString*) injectValuesIntoURL:(NSString*) url{


    NSString * injected = @"";

    //[CALL KEY]
    injected = [url stringByReplacingOccurrencesOfString:@"[CALL_KEY]" withString:[[NSUserDefaults standardUserDefaults] objectForKey:@"kCallKey"]];

    //[USER_ID]
    injected = [injected stringByReplacingOccurrencesOfString:@"[USER_ID]" withString:[[NSUserDefaults standardUserDefaults] objectForKey:@"kUsername"]];

        //[OTHER_STUFF]
        injected = [injected stringByReplacingOccurrencesOfString:@"[OTHER_STUFF]" withString:[[NSUserDefaults standardUserDefaults] objectForKey:@"kOtherStuff"]];


    return injected;

}

我有很多不同的值可以插入，我不想硬编码[CALL KEY]，而是想动态读取其中的内容并注入值。因此，初始URL可能是

http:\\www.domain.com\id=[kUsername]&record=[kCallKey]&someother=[kOtherStuff]

所以我可以循环遍历字符串，找到[]并用其中的任何内容替换它

---

如何搜索字符串并找到[字符，然后将该字符串复制到]，然后继续搜索下一个字符串[如此等等？

我通过这种方式解决了一个类似的问题，我使用了一个字典，它将关键字存储为关键字，替换字符串作为值

NSDictionary * stringsReplacedByDictionary = NSDictionary *dictionary = @{
@"kUserName" : Username,
@"kCallKey" : CallKey,
@"kOtherStuff" : otherStuff 
};

// Enumerate the Dictionary containg all Strings to be replaced.
for (NSString *keys in stringsReplacedByDictionary) {

// Here I builded the search string in your case would be [kUsername]
NSString *keysWithParenthesis = [NSString stringWithFormat:@"[$%@]", keys];
NSString *value = [stringsReplacedByDictionary objectForKey:keys];


 //Check if there is a string to replace!
 if ([stringToScan containsString:keysWithParenthesis]) {
 NSRange replacingStringRange = [[textStorage string] rangeOfString:keysWithParenthesis];
 [textStorage replaceCharactersInRange:replacingStringRange withString:value];
        }

---

我通过这种方式解决了一个类似的问题，我使用了一个字典，它将关键字存储为关键字，将替换字符串存储为值

NSDictionary * stringsReplacedByDictionary = NSDictionary *dictionary = @{
@"kUserName" : Username,
@"kCallKey" : CallKey,
@"kOtherStuff" : otherStuff 
};

// Enumerate the Dictionary containg all Strings to be replaced.
for (NSString *keys in stringsReplacedByDictionary) {

// Here I builded the search string in your case would be [kUsername]
NSString *keysWithParenthesis = [NSString stringWithFormat:@"[$%@]", keys];
NSString *value = [stringsReplacedByDictionary objectForKey:keys];


 //Check if there is a string to replace!
 if ([stringToScan containsString:keysWithParenthesis]) {
 NSRange replacingStringRange = [[textStorage string] rangeOfString:keysWithParenthesis];
 [textStorage replaceCharactersInRange:replacingStringRange withString:value];
        }

试试这个

    NSString *url = @"http:\\www.domain.com\\id=[USER_ID]&record=[CALL_KEY]&someother=[OTHER_STUFF]";
    NSArray *values = @[@"one", @"two", @"three"];
    NSString *pattern = @"\\[([^\\]\\[]+)]";

    NSRegularExpression *regex = [[NSRegularExpression alloc] initWithPattern:pattern options:0 error:nil];
    NSInteger count = [regex matchesInString:url options:0 range:NSMakeRange(0, url.length)].count;

    if (count == 0 || count != values.count) {
        // handle error ..
    }

    NSTextCheckingResult *result = [regex firstMatchInString:url options:0 range:NSMakeRange(0, url.length)];

    for (int i = 0; i < values.count; i++) {
        url = [url stringByReplacingCharactersInRange:result.range withString:values[i]];
        result = [regex firstMatchInString:url options:0 range:NSMakeRange(0, url.length)];

        if (result.range.location == NSNotFound) { // optional
            break;
        }
    }

    NSLog(@"%@", url);

NSString*url=@“http:\\www.domain.com\\id=[USER\u id]&record=[CALL\u KEY]&someother=[OTHER\u STUFF]”；
NSArray*值=@[@“一”、“二”、“三”]；
NSString*模式=@“\\[（[^\\]\\[]+）]”；
NSRegularExpression*regex=[[NSRegularExpression alloc]initWithPattern:pattern选项：0错误：nil]；
NSInteger count=[regex matchesisnstring:url选项：0范围：NSMakeRange（0，url.length）].count；
if（count==0 | | count！=values.count）{
//处理错误。。
}
NSTextCheckingResult*result=[regex firstMatchInString:url选项：0范围：NSMakeRange（0，url.length）]；
对于（int i=0；i

试试这个

    NSString *url = @"http:\\www.domain.com\\id=[USER_ID]&record=[CALL_KEY]&someother=[OTHER_STUFF]";
    NSArray *values = @[@"one", @"two", @"three"];
    NSString *pattern = @"\\[([^\\]\\[]+)]";

    NSRegularExpression *regex = [[NSRegularExpression alloc] initWithPattern:pattern options:0 error:nil];
    NSInteger count = [regex matchesInString:url options:0 range:NSMakeRange(0, url.length)].count;

    if (count == 0 || count != values.count) {
        // handle error ..
    }

    NSTextCheckingResult *result = [regex firstMatchInString:url options:0 range:NSMakeRange(0, url.length)];

    for (int i = 0; i < values.count; i++) {
        url = [url stringByReplacingCharactersInRange:result.range withString:values[i]];
        result = [regex firstMatchInString:url options:0 range:NSMakeRange(0, url.length)];

        if (result.range.location == NSNotFound) { // optional
            break;
        }
    }

    NSLog(@"%@", url);

NSString*url=@“http:\\www.domain.com\\id=[USER\u id]&record=[CALL\u KEY]&someother=[OTHER\u STUFF]”；
NSArray*值=@[@“一”、“二”、“三”]；
NSString*模式=@“\\[（[^\\]\\[]+）]”；
NSRegularExpression*regex=[[NSRegularExpression alloc]initWithPattern:pattern选项：0错误：nil]；
NSInteger count=[regex matchesisnstring:url选项：0范围：NSMakeRange（0，url.length）].count；
if（count==0 | | count！=values.count）{
//处理错误。。
}
NSTextCheckingResult*result=[regex firstMatchInString:url选项：0范围：NSMakeRange（0，url.length）]；
对于（int i=0；i

我会使用正则表达式；要想知道，请看这里：请看。您案例中的正则表达式是

@“\\[（.+？）]”“

或

@”\[（[^\\\]\[]+）]”“

。我会使用正则表达式；要想知道，请看这里：请看。您案例中的正则表达式是

@“\\\[（.+？）]”“

或

\\[（[^\\\\]\[]）”

。