我有以下代码从我的iOS应用程序登录到Web服务器。虽然显示“登录失败”警报，但应用程序正确登录。此外，如果输入了错误的用户名/密码，则应用程序崩溃。
我不确定错误处理是否错误或语法错误。
崩溃时控制台输出（输入的用户名/通道不正确），响应代码为500：
PostData: username=ghh&password=hhhh
2013-07-19 16:10:18.477 appName [11414:907] Response code: 500
这里是登录成功的输出（用户名/传递是正确的），响应码：200：
PostData: username=user-2&password=user-2
2013-07-19 16:15:28.757 appName [11430:907] Response code: 200
2013-07-19 16:15:28.758 appName [11430:907] Response ==> {"meta":[],"data":{"token":"JXVJdC05X0t9NnNqR1RBOzYkbFhUbk91KClRTH11fGk3MmVRcjZEaSVSO3hhemJ5WTQ2OFp5U2htZDtzc01tdHMkTCVGbi18JG4pMXlGITUpezZqSnJWR2dVMmQpWTh5c2h3RTVdVHgodUNdZFIpRTFjIWxKUXVJcnRnLXF0OUo="}}
2013-07-19 16:15:28.764 appName 11430:907] {
data = {
token = "JXVJdC05X0t9NnNqR1RBOzYkbFhUbk91KClRTH11fGk3MmVRcjZEaSVSO3hhemJ5WTQ2OFp5U2htZDtzc01tdHMkTCVGbi18JG4pMXlGITUpezZqSnJWR2dVMmQpWTh5c2h3RTVdVHgodUNdZFIpRTFjIWxKUXVJcnRnLXF0OUo=";
};
meta = (
);
这是登录方法：
- (IBAction)loginClicked:(id)sender {
@try {
if([[txtUsername text] isEqualToString:@""] || [[txtPassword text] isEqualToString:@""] ) {
[self alertStatus:@"Please enter both Username and Password" :@"Login Failed!"];
} else {
NSString *post =[[NSString alloc] initWithFormat:@"username=%@&password=%@",[txtUsername text],[txtPassword text]];
NSLog(@"PostData: %@",post);
NSURL *url=[NSURL URLWithString:@"http://*******.net/api/token"];
NSData *postData = [post dataUsingEncoding:NSASCIIStringEncoding allowLossyConversion:YES];
NSString *postLength = [NSString stringWithFormat:@"%d", [postData length]];
NSMutableURLRequest *request = [[NSMutableURLRequest alloc] init];
[request setURL:url];
[request setHTTPMethod:@"POST"];
[request setValue:postLength forHTTPHeaderField:@"Content-Length"];
[request setValue:@"application/json" forHTTPHeaderField:@"Accept"];
[request setValue:@"application/x-www-form-urlencoded" forHTTPHeaderField:@"Content-Type"];
[request setHTTPBody:postData];
[NSURLRequest setAllowsAnyHTTPSCertificate:YES forHost:[url host]];
NSError *error = [[NSError alloc] init];
NSHTTPURLResponse *response = nil;
NSData *urlData=[NSURLConnection sendSynchronousRequest:request returningResponse:&response error:&error];
NSLog(@"Response code: %d", [response statusCode]);
if ([response statusCode] >=200 && [response statusCode] <300)
{
NSString *responseData = [[NSString alloc]initWithData:urlData encoding:NSUTF8StringEncoding];
NSLog(@"Response ==> %@", responseData);
// extract token from json
NSError *error = nil;
NSDictionary *responseDict = [NSJSONSerialization JSONObjectWithData:urlData options:0 error:&error];
NSString *token = [[responseDict objectForKey:@"data"] objectForKey:@"token"];
// save token string to nsuserdefaults
NSString *valueToSave = token;
[[NSUserDefaults standardUserDefaults]
setObject:valueToSave forKey:@"token"];
SBJsonParser *jsonParser = [SBJsonParser new];
NSDictionary *jsonData = (NSDictionary *) [jsonParser objectWithString:responseData error:nil];
NSLog(@"%@",jsonData);
NSInteger success = [(NSNumber *) [jsonData objectForKey:@"success"] integerValue];
NSLog(@"%d",success);
if(success == 1)
{
NSLog(@"Login SUCCESS");
[self alertStatus:@"Logged in Successfully." :@"Login Success!"];
} else {
NSString *error_msg = (NSString *) [jsonData objectForKey:@"error_message"];
[self alertStatus:error_msg :@"Login Failed!"];
}
} else {
if (error) NSLog(@"Error: %@", error);
[self alertStatus:@"Connection Failed" :@"Login Failed!"];
}
}
}
@catch (NSException * e) {
NSLog(@"Exception: %@", e);
[self alertStatus:@"Login Failed." :@"Login Failed!"];
}
}
不确定这是否有帮助，但这是一些错误输出。第5行是它被捕获的地方 - “陷阱”：
CoreFoundation`CFHash:
0x32119530: push {r4, r7, lr}
0x32119532: mov r4, r0
0x32119534: add r7, sp, #4
0x32119536: cbnz r4, 0x32119544 ; CFHash + 20
**0x32119538: trap**
0x3211953a: blx 0x32217a7c ; symbol stub for: -[NSOrderedSet intersectsSet:]
0x3211953e: movs r1, #9
0x32119540: blx 0x32217b4c ; symbol stub for: __61-[NSOrderedSet sortedArrayFromRange:options:usingComparator:]_block_invoke_0
0x32119544: ldrd r2, r3, [r4]
0x32119548: cmp r2, #0
0x3211954a: ubfx r0, r3, #8, #10
0x3211954e: beq 0x32119572 ; CFHash + 66
0x32119550: movw r1, #61284
0x32119554: movt r1, #2114
0x32119558: add r1, pc
0x3211955a: ldr r1, [r1]
0x3211955c: cmp r2, r1
0x3211955e: beq 0x32119572 ; CFHash + 66
0x32119560: movw r1, #57172
0x32119564: movt r1, #2114
0x32119568: add r1, pc
0x3211956a: ldr.w r1, [r1, r0, lsl #2]
0x3211956e: cmp r2, r1
0x32119570: bne 0x32119588 ; CFHash + 88
0x32119572: movw r1, #53054
0x32119576: movt r1, #2114
0x3211957a: add r1, pc
0x3211957c: ldr.w r0, [r1, r0, lsl #2]
0x32119580: ldr r1, [r0, #24]
0x32119582: mov r0, r4
0x32119584: cbnz r1, 0x3211959e ; CFHash + 110
0x32119586: pop {r4, r7, pc}
0x32119588: movw r0, #2272
0x3211958c: movt r0, #2113
0x32119590: add r0, pc
0x32119592: ldr r1, [r0]
0x32119594: mov r0, r4
0x32119596: pop.w {r4, r7, lr}
0x3211959a: b.w 0x32215fbc ; objc_msgSend$shim
0x3211959e: pop.w {r4, r7, lr}
0x321195a2: bx r1
感谢您对此的任何帮助：）
答案 0 :(得分：2)
始终显示“连接失败”警报的原因是，根据您发布的JSON响应，返回的JSON字典中没有关键字“成功”的对象。这意味着objectForKey返回nil，并且在{nil上调用的integerValue返回0。
崩溃的原因是，当登录失败时，将没有令牌（我假设）。因此，token将为零并且将nil传递给-[NSUserDefaults setObject:forKey:]
引发无效的参数异常。
您需要做的是更改代码，如下所示。
// extract token from json
NSError *error = nil;
NSDictionary *responseDict = [NSJSONSerialization JSONObjectWithData:urlData options:0 error:&error];
NSString *token = [[responseDict objectForKey:@"data"] objectForKey:@"token"];
if (token) // make sure token is not nil
{
// save token string to nsuserdefaults
NSString *valueToSave = token;
[[NSUserDefaults standardUserDefaults] setObject:valueToSave forKey:@"token"];
}
SBJsonParser *jsonParser = [SBJsonParser new];
NSDictionary *jsonData = (NSDictionary *) [jsonParser objectWithString:responseData error:nil];
NSLog(@"%@",jsonData);
NSInteger success = token ? 1 : 0; // token was sent = login successful
NSLog(@"%d",success);