我目前正在开发一个iOS应用程序，该应用程序允许用户点击TableViewCells为歌曲投票。
我已经成功地从连接到mySQL表的PHP脚本中加载了带有解析的JSON对象（歌曲）的tableviewcells
我还成功地使用POST方法向数据库发布了一个查询。
然而，现在我正试图在查询中使用POST变量，该变量返回JSON对象供我解析。
以下是我使用POST方法更新数据库的脚本：
let queryID = self.partyID[0]
let queryGenre = self.recievedGenre
var loadIntoVotes = "queryID='(queryID)&queryGenre='(queryGenre)"
let URL: NSURL = NSURL(string: "http://myserverURL/loadSongsIntoTable.php")!
let request:NSMutableURLRequest = NSMutableURLRequest(URL:URL)
request.HTTPMethod = "POST"
request.HTTPBody = loadIntoVotes.dataUsingEncoding(NSUTF8StringEncoding);
NSURLConnection.sendAsynchronousRequest(request, queue: NSOperationQueue.mainQueue())
{
(response, data, error) in
}
下面是我解析返回JSON对象的脚本：
let url=NSURL(string:"http://myserverURL/grabID.php")
let allSongsData=NSData(contentsOfURL:url!)
var allSongs: AnyObject! = NSJSONSerialization.JSONObjectWithData(allSongsData!, options: NSJSONReadingOptions(0), error: nil)
if let json = allSongs as? Array<AnyObject> {
for index in 0...json.count-1 {
let songID : AnyObject? = json[index]
let collection = songID! as Dictionary<String, AnyObject>
let ID : AnyObject! = collection["ID"]
self.partyID.append(ID as String)
}
}
因此，从本质上讲，我如何在PHP脚本中使用POST变量，然后根据返回的结果解析JSON？
不久前我也这么做了。我的PHP脚本看起来像：
<?php
include 'database.php';
// get data
$data = $_POST['json'];
$query = json_decode($data)->{'query'};
// connecting to mysql
$connection = mysql_connect($host, $username, $password);
mysql_select_db($db_name, $connection);
// check connection
if (!$connection)
{
die("Couldn't connect: " . mysql_error());
}
$result = mysql_query($query, $connection);
// check results
if (!$result)
{
die("Error getting results: " . mysql_error());
}
if (mysql_num_rows($result))
{
while (($row = mysql_fetch_row($result)))
{
$array[] = $row;
}
mysql_close($connection);
echo json_encode($array);
}
else
{
mysql_close($connection);
}
?>
以及在objective-c++中：
std::vector<std::vector<std::string>> getQuery(std::string query)
{
// convert cstring to obj-cstring
NSString *getQuery = [NSString stringWithCString:query.c_str() encoding:NSUTF8StringEncoding];
// create dict for json
NSDictionary *jsonDict = [NSDictionary dictionaryWithObjectsAndKeys:getQuery, @"query", nil];
// initialize json
NSError *error = nil;
NSData *jsonData = [NSJSONSerialization dataWithJSONObject:jsonDict options:0 error:&error];
NSString *jsonString = [[NSString alloc] initWithData:jsonData encoding:NSUTF8StringEncoding];
if (error)
{
NSLog(@"%s: JSON encode error: %@", __FUNCTION__, error);
}
// start request
std::string num_rows_string = LOCALHOST;
num_rows_string += "getQuery.php";
NSURL *url = [NSURL URLWithString:[NSString stringWithCString:num_rows_string.c_str() encoding:NSUTF8StringEncoding]];
NSMutableURLRequest *request = [NSMutableURLRequest requestWithURL:url];
NSString *queryString = [NSString stringWithFormat:@"json=%@", [jsonString stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding]];
NSData *sendData = [queryString dataUsingEncoding:NSUTF8StringEncoding];
[request setHTTPMethod:@"POST"];
[request setHTTPBody:sendData];
// execute request
NSURLResponse *response = nil;
NSData *getResult = [NSURLConnection sendSynchronousRequest:request returningResponse:&response error:&error];
if (error)
{
NSLog(@"%s: NSURLConnection error: %@", __FUNCTION__, error);
}
// ----------DEBUGGING TESTS----------
NSString *something = [[NSString alloc] initWithData:getResult encoding:NSASCIIStringEncoding];
NSLog(@"GETRESULT: %@", something);
// store result in array
std::vector<std::vector<std::string>> data;
if (getResult.length != 0)
{
NSArray *storeResult = [NSJSONSerialization JSONObjectWithData:getResult options:0 error:&error];
if (error)
{
NSLog(@"%s: jSONObjectWithData error: %@", __FUNCTION__, error);
}
for (int i = 0; i < storeResult.count; i++)
{
NSArray *row = storeResult[i];
std::vector<std::string> temp;
for (int a = 0; a < row.count; a++)
{
temp.push_back([row[a] UTF8String]);
}
data.push_back(temp);
}
}
return data;
}
您可能想要将objective-c代码更改为swift。不应该太难。