Question

我是iPhone的新手，我正在尝试将NSMutable数组转换为json字符串然后使用请求将此字符串发送到php文件，然后使用响应NSLog再次打印它以确保它已成功发送。所以我在viewDidLoad

中编写了以下代码

NSString *phpUrl = @"http://dt-works.com/eman/bookMarks.php";

NSMutableArray *arrayKey = [[NSMutableArray alloc] initWithObjects:@"one", @"two", @"three", nil];
NSMutableArray *arrayValue1 = [[NSMutableArray alloc] initWithObjects:@"1", @"2", @"3", nil]; 
NSMutableArray *arrayValue2 = [[NSMutableArray alloc] initWithObjects:@"11", @"22", @"33", nil]; 

NSDictionary *theReqDictionary1 = [NSDictionary dictionaryWithObjects:arrayValue1 forKeys:arrayKey];
NSDictionary *theReqDictionary2 = [NSDictionary dictionaryWithObjects:arrayValue2 forKeys:arrayKey];

NSMutableArray *myArray = [NSMutableArray arrayWithObjects:theReqDictionary1,theReqDictionary2, nil];

NSString *jsonString = [myArray JSONRepresentation];

NSLog(@"%@", jsonString);



 NSDictionary *questions = nil;
NSURL *link = [NSURL URLWithString:[phpUrl stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding]];
ASIFormDataRequest *request = [ASIFormDataRequest requestWithURL:link];
[request setPostValue:jsonString forKey:@"jsonString"];
[request setTimeOutSeconds:120];
[request setDelegate:self];
NSError *error = [request2 error];
[request2 startSynchronous];

if (!error) { 
    NSData *response = [request responseData];
    NSString *json = [[NSString alloc] initWithData:response encoding:NSUTF8StringEncoding];
    questions = [json objectFromJSONString];
    NSLog(@"Data: %@", questions);

}

并且bookMarks.php中的代码是：

$handle = fopen('php://input','r');
$jsonInput = $_POST['jsonString'];
print_r($jsonInput);

但它给了我：

[{"one":"1","two":"2","three":"3"},{"one":"11","two":"22","three":"33"}]
Data: (null)

我希望数据为：[{“one”：“1”，“two”：“2”，“three”：“3”}，{“one”：“11”，“two” ： “22”， “三”： “33”}] 该怎么做??

Answer 1

尝试这一点，而不是使用密钥发布JSON字符串，您需要将其作为数据发布到您的脚本中以便读入。目前在您的脚本中，您甚至不使用$ handle ...

NSURL *link = [NSURL URLWithString:[phpUrl     stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding]];
ASIFormDataRequest *request = [ASIFormDataRequest requestWithURL:link];
[request appendPostData:[jsonString  dataUsingEncoding:NSUTF8StringEncoding]]; 
[request setTimeOutSeconds:120];
[request setDelegate:self];

你的PHP代码..

$handle = fopen('php://input','r');
$jsonInput = fgets($handle);
$decoded = json_decode($jsonInput,true);
print_r($decoded);

使用ASIFormDataRequest将json发送到php

1 个答案: