having trouble showing SELECT result sql

Question

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;
    }


    }

Joni · Answer 1 · 2019-02-06T16:55:20.677

1

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?

edited Feb 06 '19 at 16:55

answered Feb 06 '19 at 16:49

Joni

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still say Error description: Unknown column 'Array' in 'where clause' – Babak Rostami Feb 06 '19 at 16:55
That's because `fetch_assoc` returns an array @BabakRostami. – Joni Feb 06 '19 at 16:55
can you help me how can i get only one id?? ): @Joni – Babak Rostami Feb 06 '19 at 17:04

score 0 · Answer 2 · answered Feb 07 '19 at 11:00

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 :)

having trouble showing SELECT result sql

2 Answers2