从 Xcode 到 JSON 到 PHP 到 mySql 的数组/字典
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Array/Dict From Xcode to JSON to PHP to mySql
提问by James Jesst
This is driving me nuts. It seems so simple, and I am likely missing something obvious - and strongly suspect it's because my PHP/mysql skills are lacking, but I can't get it to work. I have looked hard elsewhere (and stolen) a number of code snippets from StackOverflow trying to solve the issue, but I'm still not convinced I've got it working.
这让我发疯。看起来很简单，我可能遗漏了一些明显的东西 - 并且强烈怀疑这是因为我缺乏 PHP/mysql 技能，但我无法让它工作。我已经在其他地方努力寻找（并偷走了）来自 StackOverflow 的许多代码片段试图解决这个问题，但我仍然不相信我已经让它工作了。
From Xcode I'm trying to encode an NSDictionary object into JSON (using JSON framework) so that I can dynamically store the array in mysql (ideally as a single flattened object - I know, I know) using PHP POST method.
在 Xcode 中，我试图将 NSDictionary 对象编码为 JSON（使用 JSON 框架），以便我可以使用 PHP POST 方法将数组动态存储在 mysql（理想情况下作为单个扁平对象 - 我知道，我知道）。
Code is below. I can create the json ok. I can connect ok, I can change variables that arent array variables & that do not need to be send via json, I can do just about anything. I can't seem to pass that json and store it in mysql.
代码如下。我可以创建json。我可以正常连接，我可以更改不是数组变量和不需要通过 json 发送的变量，我可以做任何事情。我似乎无法传递该 json 并将其存储在 mysql 中。
Yes- im a noob.
是的 - 我是个菜鸟。
Thx...
谢谢...
I've got this far:
我已经走了这么远：
in xcode
在 xcode 中
NSDictionary *loginDict = [NSDictionary dictionaryWithObjectsAndKeys:
@"aname", @"username",
@"hello", @"password",
nil];
NSString *jsonString = [loginDict JSONRepresentation];
NSMutableURLRequest *request = [[NSMutableURLRequest alloc] init];
NSString *post = [NSString stringWithFormat:@"json=%@", jsonString];
NSData *postData = [post dataUsingEncoding:NSASCIIStringEncoding allowLossyConversion:NO];
[request setURL:[NSURL URLWithString:@"http://domain.com/post_dict.php"]];
[request setHTTPMethod:@"POST"];
[request setHTTPBody:postData];
[[NSURLConnection alloc] initWithRequest:request delegate:self];
at post_dict.php
在 post_dict.php
<?php
$rawJsonData = $_POST['json'];
$decodedData = json_decode($rawJsonData); //do i even need to decode if i want to store a flattened json object in mysql?
//Connect To Database
$hostname='**BLACKEDOUT**.com';
$username='**BLACKEDOUT**';
$password='**BLACKEDOUT**';
$dbname='**BLACKEDOUT**';
$usertable='users';
//I want to update the Records field with the array
$recordsfield = 'Records';
mysql_connect($hostname,$username, $password) OR DIE ('Unable to connect to database! Please try again later.');
mysql_select_db($dbname);
$query = "UPDATE $usertable SET $recordsfield = '$decodedData' ";//do i encode? serialize? dunno
$result = mysql_query($query);
if(!$result)
{
mysql_close();
echo mysql_error();
return;
}
mysql_close();
?>
回答by Nakkeeran
If your are sending your JSON in POST method , It can be received in PHP with the below code
如果您以 POST 方法发送 JSON，则可以使用以下代码在 PHP 中接收
<?php $handle = fopen('php://input','r');
$jsonInput = fgets($handle);
// Decoding JSON into an Array
$decoded = json_decode($jsonInput,true);
?>
回答by Jonathan Amend
To answer your comment: //do i even need to decode if i want to store a flattened json object in mysql?
要回答您的评论： //do i even need to decode if i want to store a flattened json object in mysql?
No, you should not json_decode()the data; instead save the $rawJsonDatato MySQL. You should however escape it using mysql_real_escape_string(). Here's an example:
不，你不应该json_decode()数据；而是将其保存$rawJsonData到 MySQL。但是，您应该使用mysql_real_escape_string(). 下面是一个例子：
$rawJsonData = $_POST['json']; //i don't to decode if i want to store a flattened json object in mysql.
//Connect To Database $hostname='BLACKEDOUT.com'; $username='BLACKEDOUT'; $password='BLACKEDOUT'; $dbname='BLACKEDOUT'; $usertable='users'; //I want to update the Records field with the array $recordsfield = 'Records';
mysql_connect($hostname,$username, $password) OR DIE ('Unable to connect to database! Please try again later.'); mysql_select_db($dbname);
//Escape the JSON data for MySQL
$mysqlEncodedJsonData = mysql_real_escape_string($rawJsonData);
$query = "UPDATE $usertable SET $recordsfield = '$mysqlEncodedJsonData' ";//inserted variable should be mysql_real_escape_string()'d as it is above
$result = mysql_query($query);
if(!$result) { mysql_close(); echo mysql_error(); return; }
mysql_close();
