Question

我有以下字典：

NSDictionary *param = @{@"schoolid":@"schooldb1",

                        @"token":del.tokenString,

                        @"mobile":del.phoneString

                        };

NSLog(@"param:%@",param);

我想将此参数（schoolid，token，mobile）发送到Web视图。但我不知道如何发送。我试图在互联网上搜索，但我没有为我的问题找到任何正确的解决方案。

我的主要网址是：

     NSString *url=@"https://MyURL.com/School/AppSingleTrack";

我将打电话给UIWebview，如下所示：

NSString *finalurl=[NSString stringWithFormat:@"https://MyURL.com/School/AppSingleTrack/?%@",param];
NSURL *nsurl=[NSURL URLWithString:finalurl];
NSURLRequest *nsrequest=[NSURLRequest requestWithURL:nsurl];
[_webview loadRequest:nsrequest];
[self.view addSubview:_webview];

Answer 1

试试这个，

NSDictionary *param = @{@"schoolid":@"schooldb1",

                        @"token":del.tokenString,

                        @"mobile":del.phoneString

                        };

NSLog(@"param:%@",param);

NSString *url=@"https://24x7tracker.com/School/AppSingleTrack";


NSString *finalurl=[NSString stringWithFormat:@"https://24x7tracker.com/School/AppSingleTrack/"];
NSURL *nsurl=[NSURL URLWithString:finalurl];
NSMutableURLRequest *nsrequest=[NSMutableURLRequest requestWithURL:nsurl];
NSData *data = [NSJSONSerialization dataWithJSONObject:param options:0 error:nil];

[nsrequest setHTTPBody:data];

[_webview loadRequest:nsrequest];
[self.view addSubview:_webview];

如果需要[nsrequest setHTTPMethod:@"GET"];或POST以及内容类型等，请设置请求的必要条件。

您应该使用AFNetworking，这样会更容易。

Answer 2

使用此代码，

    NSString *sUrl =  @"https://24x7tracker.com/School/AppSingleTrack";

    NSMutableURLRequest *res = [NSMutableURLRequest requestWithURL:[NSURL URLWithString:sUrl]];

    [res setHTTPMethod:@"POST"];

    NSDictionary *params; = [NSDictionary dictionaryWithObjectsAndKeys:
                                @"schooldb1",@"schoolid",
                                del.tokenString,@"token",
                                del.phoneString,@"mobile",
                                nil];

    NSMutableArray *pairArray = [[NSMutableArray alloc] initWithCapacity:0];
    for (NSString *key in params)
        [pairArray addObject:[NSString stringWithFormat:@"%@=%@", key, params[key]]];

    [res setHTTPBody:[[pairArray componentsJoinedByString:@"&"] dataUsingEncoding:NSUTF8StringEncoding]];


    [NSURLConnection sendAsynchronousRequest:res
                                   queue:[NSOperationQueue mainQueue]
                       completionHandler:^(NSURLResponse *response, NSData *data, NSError *error) {

                           NSLog(@"request URL : %@",res.URL);
                           NSLog(@"request Method : %@",res.HTTPMethod);
                           NSLog(@"parameters : %@",params);
                           NSLog(@"response : %@",response);

                           Resp *r = [ [Resp alloc] initWithDictionary:[NSJSONSerialization JSONObjectWithData: data options: NSJSONReadingMutableContainers error:nil]];

                                   // Success - Show Sucess message
                                    if ([r.sCode isEqualToString:@"success"]) {

                                         NSLog(@"response message : %@",r.sData); 
                                     }

    }];

使用Class Resp：

Resp.h

#import <Foundation/Foundation.h>

@interface Resp : NSObject

@property (nonatomic, copy) NSString *sCode;
@property (nonatomic, copy) NSString *sMessage;
@property (nonatomic, copy) NSString *sData;


- (id)initWithDictionary:(NSDictionary *)dictionary;

@end

Resp.m

#import "Resp.h"

@implementation Resp

@synthesize sCode             = _id;
@synthesize sMessage          = _title;
@synthesize sData             = _data;


- (id)initWithDictionary:(NSDictionary *)dictionary {

    self = [super init];

    if (self) {
        self.sCode           = [dictionary objectForKey:@"code"];
        self.sMessage        = [dictionary objectForKey:@"message"];
        self.sData           = [dictionary objectForKey:@"data"];
    }

    return self;
}

@end

然后终于得到了答复，希望其有用

Answer 3

首先，与您的同事交谈或查看URL的文档，确认API需要的参数格式以及请求方法，例如GET或POST。

其次，将params连接到正确的格式，不要忘记转义参数。

如果您的网址需要正常参数，请尝试以下操作：

NSDictionary *params = @{@"schoolid" : @"",
                         @"token" : @"",
                         @"mobile" : @""};
NSMutableArray *keyValues = [NSMutableArray array];
for (NSString *key in params) {
    [keyValues addObject:[NSString stringWithFormat:@"%@=%@&", key, params[key]]];
}
NSString *paramsString = [keyValues componentsJoinedByString:@"&"];
paramsString = [paramsString stringByAddingPercentEncodingWithAllowedCharacters:[NSCharacterSet URLQueryAllowedCharacterSet]];

// Don't add / at last unless the URL has, because / is another path
NSString *baseURL = @"https://24x7tracker.com/School/AppSingleTrack";

// If GET, you can use these two lines, or use below
// NSString *urlString=[NSString stringWithFormat:@"%@?%@", baseURL, paramsString];
// NSURLRequest *request=[NSURLRequest requestWithURL:[NSURL URLWithString:urlString]];


// if POST
NSMutableURLRequest *request = [NSMutableURLRequest requestWithURL:[NSURL URLWithString:baseURL]];
request.HTTPMethod = @"POST"; // Default is GET, you can send get request by default,
request.HTTPBody = [paramsString dataUsingEncoding:NSUTF8StringEncoding];

[webView loadRequest:request];

编辑：根据@ Shubhank的猜测，如果webview的请求是通过ajax，你应该确认javascript的功能，并在webview的委托webViewDidFinishLoad中尝试这些代码：

- (void)webViewDidFinishLoad:(UIWebView *)webView {
    [webView stringByEvaluatingJavaScriptFromString:[NSString stringWithFormat:@"javascriptFunction(%@, %@, %@)", param1, param2, param3]];
}

如何将URL中的多个参数传递给iOS中的Web View？

3 个答案: