PHP返回的json，Obj-C解析的一个例子

来源：互联网发布：无线网无网络访问权限编辑：程序博客网时间：2024/06/07 21:57

PHP返回的json，Obj-C无法解析

PHP代码如下：

<?php

$arr = array(

'name' =>'abc',

'nick' => 'bbc'

);

$json_string = json_encode($arr);

echo $json_string;

只是简单的返回一串json，OC代码如下：

NSString *name = _username.text;

NSString *pass = _password.text;

NSURL *url = [NSURL URLWithString:@"http://127.0.0.1:8888/login.php"];

NSLog(@"%@",url);

NSMutableURLRequest *request = [[NSMutableURLRequest alloc]initWithURL:url cachePolicy:NSURLRequestUseProtocolCachePolicy timeoutInterval:10];

[request setHTTPMethod:@"POST"];

NSString *str = [NSString stringWithFormat:@"name=%@&pass=%@",name,pass];

NSData *data = [str dataUsingEncoding:NSUTF8StringEncoding];

[request setHTTPBody:data];

NSData *received = [NSURLConnection sendSynchronousRequest:request returningResponse:nil error:nil];

NSError *error;

NSDictionary *result = [NSJSerialization JSONObjectWithData:received options:NSJSONReadingAllowFragments error:&error];

NSString *str1 = [[NSString alloc]initWithData:received encoding:NSUTF8StringEncoding];

NSLog(@"%@",str1);

NSLog(@"%@",error);

打印出的日志是这样：

2014-03-04 10:17:17.179 word[50534:70b] http://127.0.0.1:8888/login.php

2014-03-04 10:17:17.469 word[50534:70b] string{"name":"abc","nick":"bbc"}

2014-03-04 10:17:17.470 word[50534:70b] Error Domain=NSCocoaErrorDomain Code=3840 "The operation couldn’t be completed. (Cocoa error 3840.)" (Invalid value around character 0.) UserInfo=xxxxxxxx {NSDebugDescription=Invalid value around character 0.}

有人遇到过吗？折腾了好几个小时，完全没有进展。

你的php返回内容应该是{"name":"abc","nick":"bbc"}, 而不是string{"name":"abc","nick":"bbc"}

由于多了一个string, 因此json解析失败了

0 0