mysql_fetch_array() not working

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PHP: mysql_fetch_array() expects parameter 1 to be resource, boolean given

I'm trying to send emails to addresses in a database. the addresses are in two parts: a username (aka 'number') and @something.com (aka 'carrier'). so the email address should be 'number' . 'carrier'

my code is below. the mail function works on its own (not in the loop). the error message i am getting is "mysql_fetch_array() expects parameter 1 to be resource, boolean given in home/username/public_html/script.php on line 15)

#!/usr/bin/php
<?
    require_once("PHPMailer/class.phpmailer.php");

    // connect to database
    mysql_connect("localhost", "username", "password");
    mysql_select_db("username_finalproject_Alerts");

    // prepare query
    $sql = "SELECT number, carrier FROM Alerts";
    // execute query
    $result = mysql_query($sql);
    while ($row = mysql_fetch_array($result))
    {
        //generate email address
        $email = $row['number'] . $row['carrier'];

        // send mail

        $mail = new PHPMailer();
        $mail->IsSMTP();
        $mail->Host = "private stmp address";
        $mail->ContentType = 'text/plain';
        $mail->IsHTML(false);
        $mail->SetFrom("private email address");
        $mail->AddAddress($email);
        $mail->Subject = "TRY 1";
        $mail->Body = "testing 1 2 3";
        if ($mail->Send() === false)
            die($mail->ErrorInfo . "\n");
    }

?>

$result is empty and therefore you have this error. Probably you haven't mysql connection. Check mysql status and data connection. — Anthony, Dec 08 '11 at 05:23

Insecurefarm · Answer 1 · 2014-09-24T16:22:06.383

3

There must be something wrong with your query, can you test it elsewere?

Perhaps you can try to catch the error

if(!$result) { echo mysql_error(); }

edited Sep 24 '14 at 16:22

answered Dec 08 '11 at 05:23

Insecurefarm

391
5
16

score 0 · Answer 2 · answered Dec 08 '11 at 05:26

0

Unless your login to the database is "username" and your password is "password" I think you probably meant

mysql_connect("localhost", $username, $password);

Not sure if this code is verbatim from yours or you just changed those for privacy but thought it might be worth pointing out.

answered Dec 08 '11 at 05:26

TheOx

2,208
25
28

oh sorry, i just changed those for privacy. thanks for the suggestion, though. – user1084785 Dec 08 '11 at 13:35

score 0 · Answer 3 · answered Dec 08 '11 at 05:32

to make easy you query you can also use: CONCAT(number, carrier) which provides you the combine result of both rows in a single result,

the second as the all suggested the use mysql_query($sql) or die('Error in Query!<br>' . mysql_error());

so that if the error in your query then you will get the error details through mysql_error()

score 0 · Answer 4 · answered Dec 08 '11 at 05:47

check mysql_num_rows($rows)>0

    // connect to database
    mysql_connect("localhost", "username", "password");
    mysql_select_db("username_finalproject_Alerts");

    // prepare query
    $sql = "SELECT number, carrier FROM Alerts";
    // execute query
    $result = mysql_query($sql);
    if(mysql_num_rows($result)>0) {
    while ($row = mysql_fetch_array($result))
    {
        //generate email address
        $email = $row['number'] . $row['carrier'];

        // send mail

        $mail = new PHPMailer();
        $mail->IsSMTP();
        $mail->Host = "private stmp address";
        $mail->ContentType = 'text/plain';
        $mail->IsHTML(false);
        $mail->SetFrom("private email address");
        $mail->AddAddress($email);
        $mail->Subject = "TRY 1";
        $mail->Body = "testing 1 2 3";
        if ($mail->Send() === false)
            die($mail->ErrorInfo . "\n");
    }
}

?>

score -2 · Answer 5 · answered Dec 08 '11 at 05:37

-2

write like this

$email = $row[0] . $row[1];

answered Dec 08 '11 at 05:37

Sonal Khunt

1,876
12
20

mysql_fetch_array() expects parameter 1 to be resource It Means there's a problem in query – Mohit Bumb Dec 08 '11 at 05:45

mysql_fetch_array() not working

5 Answers5