CODEWITHSUNDEEP
phpandroidmysqlsqlselect
Babak Rostami
Babak Rostami

Reputation: 31

having trouble showing SELECT result sql

i try to get id that gmail = $gmail and use it for UPDATE table

but i can't use from id and showin error Undefined property: mysqli_result::$fetch_assoc

function addverificationcode ($gmail , $random) {

    $connection = mysqli_connect(DataBaseManager::HOST, DataBaseManager::USER, DataBaseManager::PASSWORD, DataBaseManager::DATABASENAME);
    mysqli_set_charset($connection, "utf8");
    $sqlQuery = "SELECT id FROM users WHERE gmail='$gmail'";
    $result = mysqli_query($connection, $sqlQuery);
    if ($result->num_rows > 0) {
            $sqlCommand = "UPDATE users
                      SET verificationcode  = '$random' 
                        WHERE id = $result->fetch_assoc()";
    }

    if(mysqli_query($connection, $sqlCommand)){
        return true;
    }else{
        echo("Error description: " . mysqli_error($connection));
        return false;
    }


    }

Upvotes: 1

Views: 27

Answers (2)

ashir haroon
ashir haroon

Reputation: 243

fetch_assoc gives you the array of the record :) try this code

 if ($result->num_rows > 0) {
   while($res = mysqli_fetch_array($result)){
            $id = $res['id'];     // you are breaking the array here and storing the array index in a new variable
            $sqlCommand = "UPDATE users
                      SET verificationcode  = '$random' 
                        WHERE id = $id";
       }
    }

now it will work. good luck :)

Upvotes: 0

Joni
Joni

Reputation: 111329

You have to use curly braces when calling a method in an object:

        $sqlCommand = "UPDATE users
                  SET verificationcode  = '$random' 
                    WHERE id = {$result->fetch_assoc()}";

An alternative is using concatenation:

        $sqlCommand = "UPDATE users
                  SET verificationcode  = '$random' 
                    WHERE id = " . $result->fetch_assoc();

Note that in this case you could combine the two SQL statements into just one, for example update .. where id = (select id from ...).

Also note that your code as posted is vulnerable to an SQL injection attack. See How can I prevent SQL injection in PHP?

Upvotes: 1

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