CODEWITHSUNDEEP
phpmysqldatabase
Detective merry
Detective merry

Reputation: 394

PHP query result into variable

Hi everyone I am a beginner programmer trying to get query results into variable in PHP. For registeration purposes, I want to assign an available campaignID to variable $campaignID, so I can insert $campaignID along with other variables to the new user row. I only need help on getting my result of my SELECT Statement into the variable campaignID! I am aware that there are multiple questions asking the exact thing want like 1 & 2. However, I have been trying the solutions for hours now but to no avail and I need help to apply the answers to my situation.

<?php
header("Access-Control-Allow-Origin: *");
header("Content-Type: application/json; charset=UTF-8");
error_reporting(E_ERROR);
try{
    $mysqli = new mysqli("localhost", "my_user", "my_password", "world");

    $query = "SELECT campaignID from ebaycampaigns WHERE Taken='N' LIMIT 1";
    $result = $mysqli->query($query);

    if(!$result){
        echo "No data was selected";
    }

    if (mysqli_num_rows($result)==1){
        $row = mysql_fetch_array($result);
        echo "campaignID: " . $row[0];

        //or
        //$row = $result->fetch_array(MYSQLI_BOTH);
        //printf ("%s (%s)\n", $row['campaignID']);

    }

    else{
        echo "not found!";
    }

    $conn->close();
}
catch(Exception $e) {
    $json_out = "[".json_encode(array("result"=>0))."]";
    echo $json_out;

}

?>

None of the responses were echoed, $row['campaignID'], $row[0], $campaignID. I am also not conerned with SQL injections so far as this is a small project! So feel free to recommend any methods!

Upvotes: 0

Views: 1188

Answers (1)

Paul de Koning
Paul de Koning

Reputation: 532

First off change mysql_fetch_array to mysqli_fetch_array :)

nevermind the following line, iil just leave it in, I saw you used LIMIT.

And secondly change mysqli_num_rows($result)==1 to mysqli_num_rows($result) > 0

Upvotes: 1

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