mysql多查询功能不正确

Question

As I read the php manual on this function, I though that I was executing it correctly. Unfortunately I'm getting this error:

Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given on line 94

with this code:

$upSQL = "SELECT * FROM rated_teams WHERE server='$server' AND name='$teamname' AND master='2' ORDER BY id ASC;";
$upSQL .="SELECT name, rating FROM rated_teams WHERE server='$server' AND master='1'";
//echo $upSQL. "<br />";
$upresult=mysqli_multi_query($con, $upSQL);
$i=1;
$j=1;
$myrating=0;
while($row = mysqli_fetch_array($upresult)) { //LINE 94
    if ($row['win'] == 1 && $i <= 3) {      
            echo $i++ . "first 3 wins <br />";
            $myrating+=10;
            $j++;

        } else {
            if ($row['name'] == $opposer && $row['master'] == 1) {
                echo $opposer . " " . $row['rating'];
                echo $j++. " j<br />";
            }

            }
}
echo $myrating;

So, the direct question is: Why is this code incorrect?

Answer 1

The mysqli_multi_query method returns a boolean, while the method mysqli_fetch_array is supposed to be passed a mysqli_result, as the error tells you. You need to instead use the mysqli_store_result method as seen here:

http://php.net/manual/en/mysqli.multi-query.php

Using this method, you can loop through the results, as so:

if (mysqli_multi_query($con, $upSQL)) {
do {
    if ($result = mysqli_store_result($con)) {
        while ($row = mysqli_fetch_row($result)) {
            $currentresult = $row[0];
        }
        mysqli_free_result($result);
    }
} while (mysqli_next_result($con));
}

The do-while makes sure that your first result is there before going to the others, and gets that result, and then gives you the next results in the while segment of the loop. Hope this was helpful.