如何从iphone post请求获取json数据 [英] how to get json data from iphone post request

问题描述

我正在使用JSON框架开发一个iPhone应用程序，我调用一个PHP脚本来更新本地服务器上的MySQL数据库。使用这些代码：

I am developing an iPhone App using JSON framework, I am calling a PHP script to update MySQL database on local server. Using these code:

NSString *jsonString = [sendData JSONRepresentation];
NSMutableURLRequest *request = [[NSMutableURLRequest alloc] init];
NSString*post = [NSString stringWithFormat:@"&json=%@", jsonString];
NSData*postData = [post dataUsingEncoding:NSASCIIStringEncoding allowLossyConversion:NO];

[request setURL:[NSURL URLWithString:@"http://localhost:8888/update.php"]];
[request setHTTPMethod:@"POST"];
[request setValue:@"application/x-www-form-urlencoded" forHTTPHeaderField:@"content-type"];
[request setHTTPBody:postData];

我想问：

什么是PHP脚本从该请求中获取数据，以及如何解码JSON数据到PHP对象以更新到数据库？

What is the PHP script to get data from that request, and how I can decode JSON data to PHP object to update to database?

任何帮助将不胜感激。 p>

Any help would be appreciated.

推荐答案

您将在变量$ _POST ['json']中找到您的数据，您可以查看通过POST使用：

You will find your data in the variable $_POST['json'], you can take a look at what you have received via POST by using:

<?php print_r($_POST); ?>

一旦你确定了你的数据在哪里，你可以从JSON传递到PHP数据使用：

Once you have identified where is your data, you can pass from JSON to PHP data using:

<?php $phpObj = json_decode($_POST['json']); ?>

同样，您可以使用print_r来查看数据的结构：

Again, you can use print_r to look at the structure of your data:

<?php print_r($phpObj); ?>

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