如何从 PHP 多查询语句获取第二个结果集


How do I get second result set from PHP multiquery statement?

我只想从这个多查询语句中的第二个查询中获取结果。目前,我构造它的方式我从两个查询语句中得到结果:

$query = ("call calcfields2_new('$_SESSION[Userid]');");
$query .= "SELECT * FROM CalcFields WHERE Userid=$_SESSION[Userid]"; 
if (mysqli_multi_query($dbc, $query)) {
do {
    if ($result = mysqli_store_result($dbc)) {
        while ($row = mysqli_fetch_assoc($result)) {
    $array1[]=$row;
        }
        mysqli_free_result($result);
        }
    if (mysqli_more_results($dbc)) {
    }
} 
while (mysqli_next_result($dbc));
}
}
echo(json_encode($array1));     

这对你有用吗?

$query = ("call calcfields2_new('$_SESSION[Userid]');");
$query .= "SELECT * FROM CalcFields WHERE Userid=$_SESSION[Userid]";
preg_match_all("/(?:(.*?)'s.*?(?:;|$))/",$query,$first_word);
if(mysqli_multi_query($dbc,$query)){
    do{
        if($result=mysqli_store_result($dbc)){
            if(array_shift($first_word[1])=="SELECT"){
                while($row=mysqli_fetch_assoc($result)) {
                    $array1[]=$row;
                }
            }
            mysqli_free_result($result);
        }
    } while(mysqli_more_results($dbc) && mysqli_next_result($dbc));
}
// if($error_mess=mysqli_error($dbc)){echo "Error: $error_mess";}
echo(json_encode($array1));