Why this PHP error occurs: “Strict standards: mysqli::next_result(): There is no next result set.”?

问题

I have code, which is basically a copy of a php.net's code, but for some reason it does not work. Here is the code on php.net:

<?php
$mysqli = new mysqli("localhost", "my_user", "my_password", "world");

/* check connection */
if (mysqli_connect_errno()) {
    printf("Connect failed: %s\n", mysqli_connect_error());
    exit();
}

$query  = "SELECT CURRENT_USER();";
$query .= "SELECT Name FROM City ORDER BY ID LIMIT 20, 5";

/* execute multi query */
if ($mysqli->multi_query($query)) {
    do {
        /* store first result set */
        if ($result = $mysqli->store_result()) {
            while ($row = $result->fetch_row()) {
                printf("%s\n", $row[0]);
            }
            $result->free();
        }
        /* print divider */
        if ($mysqli->more_results()) {
            printf("-----------------\n");
        }
    } while ($mysqli->next_result());
}

/* close connection */
$mysqli->close();
?>

The first change I made was the connection:

$mysqli = new mysqli("localhost", "root", "", "fanfiction");

The second change I made was the queries:

$query = "SELECT first FROM tests;";
$query .= "SELECT second FROM tests;";
$query .= "SELECT third FROM tests;";
$query .= "SELECT fourth FROM tests";

EDIT: The full code with my changes

<?php
$mysqli = new mysqli("localhost", "root", "", "fanfiction");

/* check connection */
if (mysqli_connect_errno()) {
    printf("Connect failed: %s\n", mysqli_connect_error());
    exit();
}

$query = "SELECT first FROM tests;";
$query .= "SELECT second FROM tests;";
$query .= "SELECT third FROM tests;";
$query .= "SELECT fourth FROM tests";

/* execute multi query */
if ($mysqli->multi_query($query)) {
    do {
        /* store first result set */
        if ($result = $mysqli->store_result()) {
            while ($row = $result->fetch_row()) {
                printf("%s\n", $row[0]);
            }
            $result->free();
        }
        /* print divider */
        if ($mysqli->more_results()) {
            printf("-----------------\n");
        }
    } while ($mysqli->next_result());
}

/* close connection */
$mysqli->close();
?>

The error I get:

Strict standards: mysqli::next_result(): There is no next result set. Please, call mysqli_more_results()/mysqli::more_results() to check whether to call this function/method in address on line line number

I searched a solution over the net, and particularly here on StackOverflow, but I did not find helpful solutions. Most of the solutions I found were one of those two:

In this solution,@Hammerite says to change the loop from do-while to while. This suggest that php.net's code has a problem in its logic, and I find it very hard to believe. But more importantly, it just does not work for me.
In this solution, @mickmackusa suggests to add a condition in the while and change $mysqli->next_result() to $mysqli->next_result() && $mysqli->more_results(), but this solution do not work quite well. It does indeed removes the error but it omits the last result.

回答1:

Try it with

} while ($mysqli->more_results() && $mysqli->next_result());

sscce:

<?php
ini_set('display_errors', 'on');
error_reporting(E_ALL|E_STRICT);

$mysqli = new mysqli("localhost", "localonly", "localonly", "test");
/* check connection */
if (mysqli_connect_errno()) {
    printf("Connect failed: %s\n", mysqli_connect_error());
    exit();
}

$mysqli->query('CREATE TEMPORARY TABLE City (ID int auto_increment, `Name` varchar(32), primary key(ID))') or die($mysqli->error);

$stmt = $mysqli->prepare("INSERT INTO City (`Name`) VALUES (?)") or die($mysqli->error);
$stmt->bind_param('s', $city) or die($stmt->error);
foreach(range('A','Z') as $c) {
    $city = 'city'.$c;
    $stmt->execute() or die($stmt->error);
}

$query  = "SELECT CURRENT_USER();";
$query .= "SELECT Name FROM City ORDER BY ID LIMIT 20, 5";

/* execute multi query */
if (!$mysqli->multi_query($query)) {
    trigger_error('multi_query failed: '.$mysqli->error, E_USER_ERROR);
}
else {
    do {
        /* store first result set */
        if ($result = $mysqli->store_result()) {
            while ($row = $result->fetch_row()) {
                printf("'%s'\n", $row[0]);
            }
            $result->free();
        }
        /* print divider */
        if ($mysqli->more_results()) {
            printf("-----------------\n");
        }
    } while ($mysqli->more_results() && $mysqli->next_result());
}

prints

'localonly@localhost'
-----------------
'cityU'
'cityV'
'cityW'
'cityX'
'cityY'

without warnings/notices.

来源：https://stackoverflow.com/questions/33220075/why-this-php-error-occurs-strict-standards-mysqlinext-result-there-is-no

标签

php

mysqli

mysqli-multi-query