PHP an MySQL - Fatal error: Function name must be a string

Question

I'm trying to show a value from a database table through PHP echo. The MySQL result is a double (10, 2).

<?php $link = new mysqli('127.0.0.1', '*******', '*******', '*******');
                     if ($link->connect_errno) {
                        die('Failed to connect to MySQL: (' . $mysqli->connect_errno . ') ' . $mysqli->connect_error);
                     }
                    $user = $_SESSION['user'];
                    $result = $link->query("SELECT * FROM users WHERE username='$user' AND active=1");
                    $numrows = $result->num_rows;
                    if($numrows == 0 || $numrows > 1)
                    {
                        $link->close();
                        session_destroy();
                        echo '<META HTTP-EQUIV="Refresh" Content="0; URL=**************">';
                        exit;   
                    }
                    else if($numrows == 1)
                    {
                        //$sid = $result(8);
                        echo '<strong>this is my string in which i want to show the result in' . $result(8) . 'rest of the string';}?>

Line where the error is show is the echo line (in the end). Can anyone point me out to what I am doing wrong here? Thank you.

$result is a link to the query (resource), you have to fetch the results first. — Waygood
– Waygood, Commented May 1, 2013 at 17:58

MatthiasLaug · Accepted Answer · 2013-05-01 18:04:29Z

3

you are calling $result(8) which is a method call in php. I think you meant

$dataRow = $result->fetch_array(MYSQLI_ASSOC);
// collect whatever you need from the array $dataRow array

since PHP is an interpreted language you can do such things as assign a value to a variable and call that variable

$func = 'myFunc';
$func(); // will call the function myFunc

edited May 1, 2013 at 18:04

answered May 1, 2013 at 17:58

MatthiasLaug

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3 Comments

Waygood Over a year ago

$result is a resource not an array

Krešimir Čulina Over a year ago

Fatal error: Cannot use object of type mysqli_result as array

Waygood Over a year ago

php.net/manual/en/mysqli.query.php Returns FALSE on failure. For successful SELECT, SHOW, DESCRIBE or EXPLAIN queries mysqli_query() will return a mysqli_result object. For other successful queries mysqli_query() will return TRUE.

mbarlocker · Accepted Answer · 2013-05-01 18:00:36Z

2

The $result variable is a MySQLi Result. You want to get a row from that result set. To do that, use fetch_assoc. This will give you an associative array with all of the fields of the table as keys.

$row = $result->fetch_assoc();
echo $row['username'];
echo $row['whatever'];

EDIT: It may be valuable to note that you are susceptible to the following security risks: SQL injection, cross-site scripting, and Cookie tampering.

answered May 1, 2013 at 18:00

mbarlocker

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1 Comment

Waygood Over a year ago

Since OP is using (8) they may want to fetch_row();

Moisés Alcocer · Accepted Answer · 2013-09-17 22:10:24Z

0

You are trying to access to an array value, you must use:

$result[8] and not $result(8)

Best regards!

answered Sep 17, 2013 at 22:10

Moisés Alcocer

1

Comments

Matoha · Accepted Answer · 2013-05-01 18:00:00Z

-1

Look at this - $result(8) (last row). A variable can't have arguments. You probably wanted $result[8] (9th element in array).

answered May 1, 2013 at 18:00

Matoha

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