Question

我收到警告：mysqli_query（）希望参数1为mysqli，在第13行的/home/jbellmy1/public_html/wbg450/getGames.php中给出null作为错误但我可以在我的登录页面上连接并且所有返回精细。我知道这意味着我错过了一个参数，但是当我包含我的设置页面时，我似乎无法理解为什么我的$ link为null，该页面的链接是我与mysqli的连接。任何帮助将不胜感激。

<?php  //WBGPOKER getGames.php

require_once("settings.inc.php");

$p_id = $_POST['p_id'];
if(!isset($p_id)){
$p_id = $_GET['p_id'];
}

function getPlayerName($g_id){
$query =  "SELECT p.`p_name` FROM `player` as p INNER JOIN `game_player` AS gp 
            WHERE gp.`g_id`='$g_id' AND gp.`p_id`=p.`p_id` ORDER BY gp.`p_id` ASC LIMIT 1";
$result = mysqli_query($link, $query);  


}else{
if ($row = mysqli_fetch_row($result)){
    return $row[0];
}else{
    die("myStatus=dbERROR");
}
}
} // end function

$query = "SELECT `g_id`,`g_gameInit` FROM `game` WHERE `g_gameInit` IS NOT NULL AND `g_gameStart` IS NULL ORDER BY `g_gameInit` ASC";
$result = mysqli_query($link, $query);  


if(mysqli_num_rows($result) > 0){
// Init vars
$status = "myStatus=OK";
$outputString = "&output="; 
//process each row (game row) and append into output string delimited by          "|" and "^" for fields and rows
$num = mysqli_num_rows($result);
for($i=0;$i < $num;$i++){
    $row = mysqli_fetch_array($result); 
    $g_id           = $row["g_id"];
    $player_name    = getPlayerName($g_id);
    $gameInitTime   = $row["g_gameInit"];
    if ($i < $num-1){
        $outputString .= $g_id . "|" . trim($player_name) . "|" . $gameInitTime . "^";
    }else{
        $outputString .= $g_id . "|" . trim($player_name) . "|" . $gameInitTime;
    }
} //end for

print trim($status) . trim($outputString) . "&dummy=dummy"; 
}else{
print "myStatus=NOTOK&dummy=dummy";

} // end if-else

mysqli_close($link);

?>

连接MYSQLI错误

0 个答案: