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我正在从服务器获取 json 格式的数据,但我希望用户在调用 json 服务时发送 user_id,以便它应该显示我正在使用以下代码的所需结果

如何通过传递 user_id PHP 代码从 iphone 调用数据

我用过

   $user_id=$_POST['user_id'];
   $query = mysql_query("SELECT s.*, u.user_id FROM survey_master AS s
JOIN user_profile AS u on u.user_id = s.user_id where s.user_id=$user_id");

    $rows = array();
    while($row = mysql_fetch_assoc($query)) {
   $rows[] = $row;
   }
  echo json_encode($rows);

iPhone 代码

  NSArray *tempArray =[[DataManager staticVersion] startParsing:@"http://celeritas-solutions.com/emrapp/surveyDescription.php"];

for (int i = 0; i<[tempArray count]; i++) {

    id *item = [tempArray objectAtIndex:i];
    NSDictionary *dict = (NSDictionary *) item;
    ObjectData *theObject =[[ObjectData alloc] init];
    [theObject setSurvey_title:[dict objectForKey:@"survey_Title"]];
    [theObject setSurvey_Description:[dict objectForKey:@"survey_Description"]];    
    [theObject setDate_Created:[dict objectForKey:@"date_Created"]];

    [surveyList addObject:theObject];
    [theObject release];
    theObject=nil;
}
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1 回答 1

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您可以使用 GET 发送 user_id。

在 PHP;

$user_id=$_GET['user_id'];

在 iPhone;

  NSString *url = [NSString stringWithFormat:@"http://celeritas-solutions.com/emrapp/surveyDescription.php?user_id=%@", user_id];
  NSArray *tempArray =[[DataManager staticVersion] startParsing:url];
于 2012-07-25T09:03:21.137 回答