从iOS应用程序发送帖子到PHP脚本不工作...简单的解决方案就像

Question

我之前已经做了好几次了但是由于某些原因我无法通过这个帖子...我尝试了设置为_POST且没有的变量的php脚本......当它们未设置为发布时它工作精细.这是我的iOS代码:

NSDate *workingTill = timePicker.date;
NSDateFormatter *formatter = [[NSDateFormatter alloc] init];
[formatter setDateFormat:@"HH:mm"];
NSString *time = [formatter stringFromDate:workingTill];

NSString *post = [NSString stringWithFormat:@"shift=%@&username=%@", time, usernameString];

NSData *postData = [post dataUsingEncoding:NSUTF8StringEncoding allowLossyConversion:NO];
NSString *postLength = [NSString stringWithFormat:@"%d", [post length]];

NSURL *url = [NSURL URLWithString:@"http://wowow.php"];
NSMutableURLRequest *request = [NSMutableURLRequest requestWithURL:url cachePolicy:NSURLRequestUseProtocolCachePolicy timeoutInterval:60.0];
[request setHTTPMethod:@"POST"];
NSLog(@"%@", post);
[request setValue:postLength forHTTPHeaderField:@"Content-Length"];
[request setHTTPBody:postData];

[NSURLConnection connectionWithRequest:request delegate:nil];

[self.navigationController popToRootViewControllerAnimated:YES];

这里是php的一大块,POST变量不在正确的位置？

<?php
function objectsIntoArray($arrObjData, $arrSkipIndices = array())
{
    $arrData = array();

    // if input is object, convert into array
    if (is_object($arrObjData)) {
        $arrObjData = get_object_vars($arrObjData);
    }

    if (is_array($arrObjData)) {
        foreach ($arrObjData as $index => $value) {
            if (is_object($value) || is_array($value)) {
                $value = objectsIntoArray($value, $arrSkipIndices); // recursive call
            }
            if (in_array($index, $arrSkipIndices)) {
                continue;
            }
            $arrData[$index] = $value;
        }
    }
    return $arrData;
}

    $newShift = $_POST('shift');
    $bartenderUsername = $_POST('username');

    mysql_connect("host", "name", "pw") or die(mysql_error());  
    mysql_select_db("harring4") or die(mysql_error());

    $result = mysql_query("SELECT * FROM BartenderTable WHERE username='".$bartenderUsername."'") or die(mysql_error());  

    $row = mysql_fetch_array($result);
    $newfname = $row['fname'];

我想这对于更有经验的开发人员来说是一个相当简单的答案,感谢您的帮助!

Answer 1

$_POST是一个数组,而不是一个函数.您需要使用方括号来访问数组索引:

$newShift = $_POST['shift'];
$bartenderUsername = $_POST['username'];