UITableView 拉取每个用户的特定请求

Posted 2023-03-04

【中文标题】UITableView 拉取每个用户的特定请求

【英文标题】：UITableView Pull Specific Request Per User

【发布时间】：2013-12-22 20:00:34

【问题描述】：

我不确定如何尝试为每个应用用户提取特定请求，例如每次为每个应用用户显示单独的朋友列表。我在 NSURL 中尝试了 POST/GET 请求，但没有运气。如何在加载时指定用户参数以根据登录用户加载不同的表？目前，表格视图中没有返回任何内容。如果您需要更多信息，请告诉我。

NSURL *url = [NSURL URLWithString:@"https://***/friendsList.php"];

    NSMutableURLRequest *request = [NSMutableURLRequest requestWithURL:url cachePolicy:NSURLRequestUseProtocolCachePolicy timeoutInterval:60];

    NSURLConnection *conneection = [[NSURLConnection alloc]initWithRequest:request delegate:self];

    [request setHTTPMethod:@"GET"];

    NSString *postString = savedUser;

    [request setHTTPBody:[postString dataUsingEncoding:NSUTF8StringEncoding]];

    //NSURLRequest *request = [NSURLRequest requestWithURL:url];

    //[[NSURLConnection alloc] initWithRequest:request delegate:self];


-(void)connection:(NSURLConnection *)connection didReceiveResponse:(NSURLResponse *)response

    data = [[NSMutableData alloc]init];


- (void)connection:(NSURLConnection *)connection didReceiveData:(NSData *)theData

    [data appendData:theData];


- (void)connectionDidFinishLoading:(NSURLConnection *)connection

    [UIApplication sharedApplication].networkActivityIndicatorVisible = NO;

    self.friendsList = [NSJSONSerialization JSONObjectWithData: data options:kNilOptions error: nil];

    [friendsListTable reloadData];

PHP：

<?php

    header('Content-type: application/json');

    $username = 's***'; 
    $password = '***'; 
    $host = 'localhost'; 
    $dbname = '***'; 

    $link = mysql_connect('localhost', $username, $password);

if (!$link) 

    die('Could not connect: ' . mysql_error());

else


$appUser = $_POST['username'];

// make foo the current db
$db_selected = mysql_select_db($dbname, $link);
if (!$db_selected) 
    die ('Can\'t login to shipstudent : ' . mysql_error());


$get_news = mysql_query("Select username, first_name, last_name from Users u inner join Friends f on f.friendid = u.iduser where u.username = '$appUser'", $link);
$articles = array();

while($row = mysql_fetch_assoc($get_news))
   
    $articles [] = $row;


echo json_encode($articles);


?>

【参考方案1】：

我是这样做的：

NSString *userParam = [NSString stringWithFormat:@"https://***.php?username=%@",savedUser];

NSURL *url = [NSURL URLWithString:userParam];