Question

经过大约一周没有解决方案的研究后，我放弃了，我向你们寻求建议。

我需要相同的php脚本以相同的方式处理POST请求，无论它们的来源是简单的HTML表单还是iOS应用程序。没有涉及JSON。

问题：当POST发件人是iOS应用时，某些字符会丢失。在这一点上，我会说我不知道它为什么会这样。显然，Web浏览器会在POST请求的构建中正确地转义所有字符，而我的obj-c代码却没有。

在objective-c代码中，你可以看到我做过的5个试验，并附上了一些错误。

我真的希望你能帮忙

tester.php文件：

<?php
header('Content-Type: text/html; charset=utf-8');

if (isset($_POST['email']))
{
    echo "email: " . $_POST['email'];
    echo "\n";
    echo "password: " . $_POST['password'];
}
else
{
?>

<form method="post">
    <input type="text" id="email" name="email" value="切换到中文@gmail.com" />
    <br>
    <input type="text" id="password" name="password" value="&+=/切/().%&" />
    <br>
    <input type="submit" />

</form>

<?php
}

Objectice -c HTTP POST执行：

// Get values
NSString *email    = @"切换到中文@gmail.com";
NSString *password = @"&+=/切/().%&";

NSLog(@"Before: %@ and %@", email, password);

// Escape tentative 1: fails: % disappears
//email = [email stringByAddingPercentEncodingWithAllowedCharacters:[NSCharacterSet alphanumericCharacterSet]];
//password = [password stringByAddingPercentEncodingWithAllowedCharacters:[NSCharacterSet alphanumericCharacterSet]];

// Escape tentative 2: failed: % disappears, + becomes a space, who knows what else
// email = [email stringByReplacingOccurrencesOfString: @"&" withString: @"%26"];
// password = [password stringByReplacingOccurrencesOfString: @"&" withString: @"%26"];

// Escape tentative 3: % disappears
// NSString *charactersToEscape = @"!*'();:@&=+$,/?%#[]\" ";
// NSCharacterSet *allowedCharacters = [[NSCharacterSet characterSetWithCharactersInString:charactersToEscape] invertedSet];
// email = [email stringByAddingPercentEncodingWithAllowedCharacters:allowedCharacters];
// password = [password stringByAddingPercentEncodingWithAllowedCharacters:allowedCharacters];

// Escape tentative 4: % disappears
/*CFStringRef cf_email = (__bridge CFStringRef)(email);
CFStringRef cf_password = (__bridge CFStringRef)(password);

email = CFBridgingRelease(CFURLCreateStringByAddingPercentEscapes(kCFAllocatorDefault,
                                                                  cf_email,
                                                                  NULL,
                                                                  CFSTR("%:/?#[]@!$&'()*+,;="),
                                                                  kCFStringEncodingUTF8));
password = CFBridgingRelease(CFURLCreateStringByAddingPercentEscapes(kCFAllocatorDefault,
                                                                  cf_password,
                                                                  NULL,
                                                                  CFSTR("%:/?#[]@!$&'()*+,;="),
                                                                  kCFStringEncodingUTF8));*/

// Escape tentative 5: password becomes "(null)"
/*email = [email stringByReplacingOccurrencesOfString: @"%" withString: @"%25"];
email = [email stringByReplacingOccurrencesOfString: @"&" withString: @"%26"];
email = [email stringByReplacingOccurrencesOfString: @"+" withString: @"%2b"];


password = [password stringByReplacingOccurrencesOfString: @"%" withString: @"%25"];
password = [password stringByReplacingOccurrencesOfString: @"&" withString: @"%26"];
password = [password stringByReplacingOccurrencesOfString: @"+" withString: @"%2b"];*/



// Execute HTTP request
NSString *postData = [NSString stringWithFormat:@"email=%@&password=%@", email, password];

NSURLSessionConfiguration *sessionConfiguration = [NSURLSessionConfiguration defaultSessionConfiguration];
NSURLSession *session = [NSURLSession sessionWithConfiguration:sessionConfiguration];

NSURL *url = [NSURL URLWithString:@"http://localhost:8080/tester.php"];

NSMutableURLRequest *request = [NSMutableURLRequest requestWithURL:url];

request.HTTPBody = [postData dataUsingEncoding:NSUTF8StringEncoding];
request.HTTPMethod = @"POST";

NSURLSessionDataTask *postDataTask = [session dataTaskWithRequest:request    completionHandler:^(NSData *data, NSURLResponse *response, NSError *error) {
    // Print response
    NSLog([[NSString alloc] initWithData:data encoding:NSUTF8StringEncoding]);
}];
[postDataTask resume];

如果我使用HTML表单（正是我在表单中输入的内容）的结果：

email: 切换到中文@gmail.com password: &+=/切/().%&

我从目标c代码中得到的结果（一些字符消失），最后一次测试之一：

2016-09-29 21:28:59.519 test[44752:2438179] Before: 切换到中文@gmail.com (15) and &+=/切/().%& (11)
2016-09-29 21:28:59.705 test[44752:2438721] email: 切换到中文@gmail.com password: &+=/切/().&

Answer 1

这就是我的方式。我尝试的一切都很好用。首先，您需要将该函数添加到NSString：

@implementation NSString (NSStringWebStructure)
-(NSString*)stringToWebStructure
{
    NSString* webString = [self stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding];

    webString = [webString stringByReplacingOccurrencesOfString:@"&" withString:@"%26"];
    webString = [webString stringByReplacingOccurrencesOfString:@"?" withString:@"%3F"];

    return webString;
}
@end

现在您需要请求函数：

-(NSString*)launchURL:(NSURL*)url withPostString:(NSString*)post
{
    NSData *postData = [post dataUsingEncoding:NSASCIIStringEncoding allowLossyConversion:YES];
    NSString *postLength = [NSString stringWithFormat:@"%lu", (unsigned long)[postData length]];

    NSMutableURLRequest *request = [[NSMutableURLRequest alloc] init];
    [request setURL:url];
    [request setHTTPMethod:@"POST"];
    [request setValue:postLength forHTTPHeaderField:@"Content-Length"];
    [request setHTTPBody:postData];

    NSError *error = nil;
    NSHTTPURLResponse *response = nil;
    NSData *urlData = [NSURLConnection sendSynchronousRequest:request returningResponse:&response error:&error];

    if (response.statusCode >= 200 && response.statusCode < 300)
    {
        NSString *responseData = [[NSString alloc] initWithData:urlData encoding:NSUTF8StringEncoding];
        if (responseData.length > 0) return responseData;
    }
    else if (error) return [error description];

    return nil;
}
-(NSString*)launchURL:(NSURL*)url withPostValues:(NSDictionary*)postDict
{
    NSArray* postKeys = postDict.allKeys;
    NSMutableArray* postLines = [[NSMutableArray alloc] init];

    for (NSString* key in postKeys)
    {
        [postLines addObject:[NSString stringWithFormat:@"%@=%@",key,[postDict[key] stringToWebStructure]]];
    }

    return [self launchURL:url withPostString:[postLines componentsJoinedByString:@"&"]];
}

考虑到你的例子，你只需要这样做：

NSString *email    = @"切换到中文@gmail.com";
NSString *password = @"&+=/切/().%&";

NSURL *url = [NSURL URLWithString:@"http://localhost:8080/tester.php"];
NSDictionary* postValues = @{@"email":email, @"password": password};

NSString* response = [self launchURL:url withPostValues: postValues];
NSLog(@"%@",response);

stringByAddingPercentEscapesUsingEncoding:在macOS 10.11中已被弃用，但是如果你要支持旧的macOS版本，就像我一样，这是一个功能方法（从macOS 10.6到macOS 10.12测试）。

iOS - 目标C - POST字符串转义/编码问题

1 个答案: