如何在不破坏URL的情况下转义此字符串中的* s和％s？

Question

我在这里有这个动作：

- (IBAction)searchButton:(id)sender {
NSString *textString = self.symbolSearchField.text;
NSURL *sourceURL = [[NSURL alloc]  initWithString: [NSString stringWithFormat:@"http://query.yahooapis.com/v1/public/yql?q=select%%20*%%20from%%20yahoo.finance.quotes%%20where%%20symbol%%20in%%20%%28%%22%@%%22%%29&env=store://datatables.org/alltableswithkeys", textString]];
NSXMLParser *parser = [[NSXMLParser alloc] initWithContentsOfURL:sourceURL];
parser.delegate = self;
[parser parse];

我希望它从搜索字段中获取文本，使用该文本修改URL，并从URL解析XML。但是，当我转义* s和％s时，似乎URL变为“已损坏”，并且它不会解析。如果我保留URL而不转义...

    NSURL *sourceURL = [[NSURL alloc]  initWithString: [NSString stringWithFormat:@"http://query.yahooapis.com/v1/public/yql?q=select%20*%20from%20yahoo.finance.quotes%20where%20symbol%20in%20%28%22%@%22%29&env=store://datatables.org/alltableswithkeys", textString]];

...然后我收到“无效的转化说明符'*'”和“比数据参数更多'％'次转化”警告。

所以我的问题是，如何在不破坏URL的情况下转义符号？

修改

为清楚起见，这是我做出一些调整后的代码。

NSString *textString = self.symbolSearchField.text; NSString *query = [NSString stringWithFormat:@"select * from yahoo.finance.quotes where symbol in (\"%@\")", textString]; NSString *escapedStoreURL = [@"store://datatables.org/alltableswithkeys" stringByAddingPercentEncodingWithAllowedCharacters:[NSCharacterSet URLHostAllowedCharacterSet]]; NSString *escapedQuery = [query stringByAddingPercentEncodingWithAllowedCharacters:[NSCharacterSet URLHostAllowedCharacterSet]]; NSString *URLString = [NSString stringWithFormat:@"http://query.yahooapis.com/v1/public/yql?q=%@&env=%@", escapedQuery, escapedStoreURL]; NSURL *sourceURL = [NSURL URLWithString:URLString]; NSXMLParser *parser = [[NSXMLParser alloc] initWithContentsOfURL:sourceURL]; parser.delegate = self; [parser parse];

Answer 1

试试这个：

NSString *firstPart = @"http://query.yahooapis.com/v1/public/yql?q=select%20*%20from%20yahoo.finance.quotes%20where%20symbol%20in%20%28%22";
NSString *lastPart = @"%22%29&env=store://datatables.org/alltableswithkeys";
NSURL *sourceURL = [[NSURL alloc]  initWithString: [NSString stringWithFormat:@"%@%@%@", firstPart, textString, lastPart]];

修改

您可能还想考虑以编程方式转义该查询，而不是事先在那里进行所有转义。商店网址也需要进行网址编码。

NSString *query = [NSString stringWithFormat:@"select * from yahoo.finance.quotes where symbol in (\"%@\")", textString];
NSString *escapedStoreUrl = [@"store://datatables.org/alltableswithkeys" stringByAddingPercentEncodingWithAllowedCharacters:[NSCharacterSet URLHostAllowedCharacterSet]];
NSString *escapedQuery = [query stringByAddingPercentEncodingWithAllowedCharacters:[NSCharacterSet URLHostAllowedCharacterSet]];
NSString *urlString = [NSString stringWithFormat:@"http://query.yahooapis.com/v1/public/yql?q=%@&env=%@", escapedQuery, escapedStoreUrl];

编辑2

所以我回来了，再次看到这个问题，事实证明在yahoo.finance.quotes中的u和o与alltableswithkeys的两个Ls之间存在一些不可见的字符（具体来说，U+200c）中的两个。删除这些应该可以解决您的问题原来这是caused by StackOverflow。

Answer 2

为什么不直接使用NSMutableString，并逐步附加字符串？

NSString *textString = @"AAPL";
NSMutableString *query = [[NSMutableString alloc] init];
NSString *startString = @"http://query.yahooapis.com/v1/public/yql?q=select%20*%20from%20yahoo.finance.quotes%20where%20symbol%20in%20%28%22";
NSString *endString = @"%22%29&env=store://datatables.org/alltableswithkeys";

[query appendString:startString];
[query appendFormat:@"%@",textString];
[query appendString:endString];

NSURL *sourceURL = [[NSURL alloc] initWithString:query];

现在，变量sourceURL是您的网址，其符号为“已转义”。