查询未使用直接SQL正常工作的php返回结果

Question

- So I have found that the cause of the problem is that I am running a previous query (in this case a multi_query) that appears to be out of sync with the next query. If I close and reopen the connection, it works fine, but there is probably a better way to go about solving this problem. Any pointers on a better way are appreciated.

Complete code:

if (filter_var($_POST['survey']['contact']['contactEmail'], FILTER_VALIDATE_EMAIL)) {
    $survey = filter_input_array(INPUT_POST, FILTER_SANITIZE_STRING);
    $survey = $survey['survey'];
    $email = $survey['contact']['contactEmail'];
    include 'databaseconnection.php';
    /*
     * This part works...
     */
    $query = "BEGIN;INSERT INTO interests(contactEmail) VALUES ('$email');INSERT INTO equipment(contactEmail) VALUES ('$email'); INSERT INTO classes(contactEmail) VALUES ('$email');INSERT INTO housing(contactEmail) VALUES ('$email');INSERT INTO newSkills(contactEmail) VALUES ('$email');INSERT INTO selfIdentity(contactEmail) VALUES ('$email');INSERT INTO areYouA(contactEmail) VALUES ('$email');INSERT INTO locationIdeas(contactEmail) VALUES ('$email');INSERT INTO involved(contactEmail) VALUES ('$email');INSERT INTO contact(contactEmail) VALUES ('$email');";
    foreach ($survey as $key => $val) {
        foreach ($val as $k => $v) {
            $query.= "UPDATE $key SET $key.$k = '$v' WHERE $key.contactEmail='$email';";
        }
    }

    $query.="COMMIT;";
    if (!($mysqli->multi_query($query))) {
        echo "Error: (" . $mysqli->errno . ") " . $mysqli->error;
    }
    /*
     * This subsequent query does not work unless I close the connection and restart it, giving an " Error: (2014) Commands out of sync; you can't run this command now "
     */
/*******************************************/
    $mysqli->close();
    include 'databaseconnection.php';
/*******************************************/

    $query = "SELECT distinct * FROM interests, equipment, classes, housing, newSkills, selfIdentity, areYouA, locationIdeas, involved, contact WHERE interests.contactEmail='$email' AND equipment.contactEmail='$email' AND classes.contactEmail='$email' AND housing.contactEmail='$email' AND newSkills.contactEmail='$email' AND selfIdentity.contactEmail='$email' AND areYouA.contactEmail='$email' AND locationIdeas.contactEmail='$email' AND involved.contactEmail='$email' AND contact.contactEmail='$email';";
    if (!($result = $mysqli->query($query))) {
        echo "Error: (" . $mysqli->errno . ") " . $mysqli->error;
    }
    $statistics = $result->fetch_array(MYSQLI_ASSOC);
    var_dump($statistics);
} else {
    echo 'Please enter a valid email address';
}

I am baffled as to why this query is not returning any results. The code is supposed to be returning someone's entry in a survey for confirmation that it was received and the '$email' is the email address of the person who did the survey. But it appears not to be returning anything (nothing is returned from a var_dump of the $result variable) and I get the following message"

The relevant code:

$result = $mysqli->query("SELECT distinct * FROM interests, equipment, classes, housing, newSkills, selfIdentity, areYouA, locationIdeas, involved, contact WHERE interests.contactEmail='$email' AND equipment.contactEmail='$email' AND classes.contactEmail='$email' AND housing.contactEmail='$email' AND newSkills.contactEmail='$email' AND selfIdentity.contactEmail='$email' AND areYouA.contactEmail='$email' AND locationIdeas.contactEmail='$email' AND involved.contactEmail='$email' AND contact.contactEmail='$email';");
var_dump($result);
$statistics = $result->fetch_array(MYSQLI_ASSOC);
foreach ($statistics as $k => $v) {
    echo $k . ' : ' . $v;
}

The Error Message: : Call to a member function fetch_array() on a non-object in on line

When I do a dummy search for a test email using straight sql in MySQL Workbench, I get the results I am looking for.

Straight SQL:

SELECT distinct *
FROM interests, equipment, classes, housing, newSkills, selfIdentity, areYouA, locationIdeas, involved, contact 
WHERE interests.contactEmail='bob@t.qdTT' AND equipment.contactEmail='bob@t.qdTT' AND classes.contactEmail='bob@t.qdTT' AND housing.contactEmail='bob@t.qdTT' AND newSkills.contactEmail='bob@t.qdTT' AND 
selfIdentity.contactEmail='bob@t.qdTT' AND areYouA.contactEmail='bob@t.qdTT' AND locationIdeas.contactEmail='bob@t.qdTT' AND involved.contactEmail='bob@t.qdTT' AND contact.contactEmail='bob@t.qdTT';

Results:

'bob@t.qdTT', NULL, NULL, 'on', 'on', NULL, NULL, 'dbf gbcvb', 'bob@t.qdTT', NULL, NULL, NULL, 'on', NULL, NULL, NULL, NULL, 'on', 'dfxbv ', 'bob@t.qdTT', 'on', NULL, NULL, 'on', NULL, NULL, NULL, NULL, NULL, 'erdgettrf', 'bob@t.qdTT', 'yes', 'rtfhthtrfgh', 'bob@t.qdTT', 'no', 'ehfnsrtehtertdf', 'bob@t.qdTT', 'other', 'dherdgdfxbdfb', 'bob@t.qdTT', 'on', 'on', 'on', 'on', 'on', 'on', 'on', 'on', 'on', 'on', 'on', 'on', 'on', 'rhgergbdbgdfbdf', 'bob@t.qdTT', 'sfgdregbegerg', 'bob@t.qdTT', 'no', 'aegsdgdr', 'estgdfbdfxbdfx ', 'bob@t.qdTT', 'Asghd', '378767'

Answer 1

This is a known feature / bug with mysqli. You have to free all the results from the queries you've done so far before you can make more.

After this...

if (!($mysqli->multi_query($query))) {
    echo "Error: (" . $mysqli->errno . ") " . $mysqli->error;
}

Try...

while($mysqli->more_results())
{
    $mysqli->next_result();
    $discard = $mysqli->store_result();
}

While INSERT and UPDATE don't return result sets, they do return statuses that will be considered results to mysqli