CODEWITHSUNDEEP
phpmysqlsqldate
Yahoo
Yahoo

Reputation: 4187

Insert FB birthday into Mysql gives NULL

I Get NULL when i insert the Birthday and current time in MYSQL

$graph_url = "https://graph.facebook.com/me?access_token=".$params['access_token']."&format=json&fields=birthday";
$user = json_decode(file_get_contents($graph_url),true);
     echo("birthday is " . $user['birthday']);

SQL

$queryip = "insert into table (birthday,currenttime) values (".$user['birthday'].','.CURDATE().")";
                $resultip = mysql_query($queryip);

MYSQL Structure Both the birthday,currenttime fields are Type DATE

Upvotes: 1

Views: 1062

Answers (5)

PJunior
PJunior

Reputation: 2767

Try

$user['birthday'] = date("Y-m-d", strtotime($user['birthday']));

before the SQL insert statement.

Upvotes: 0

psycho brm
psycho brm

Reputation: 7664

In PHP:

function formatDateFbToSql($fbdate)
{
    // facebook user_birthday format (02/19/1988)
    $date = DateTime::createFromFormat('m/d/Y', $fbdate);
    // mysql format (1988-02-19)
    return $date->format('Y-m-d');
}

Upvotes: 2

Bruno
Bruno

Reputation: 122679

(Despite being a different language and RDBMS, this is the same problem as in this question.)

Firstly, SQL strings must be contained in '': you're clearly missing them in (".$user['birthday'].' .... Secondly, and more importantly, they must be escaped, to prevent SQL injection attacks (and to make sure the string is valid anyway).

In general, it's better to use prepared statements to pass the parameters to SQL query instead of inserting the variables in the query itself: this should take care of the escaping and conversion for you.

You'll also need to cast the birthday string into a date if necessary.

I'd recommend using mysqli and prepared statements instead. (There are a number of examples out there, perhaps this would help.)

EDIT:

Try something along these lines perhaps (assuming you've switched to mysqli):

$query = "insert into table (birthday,currenttime) values (STR_TO_DATE(?, '%m/%d/%Y'), CURDATE())";
if ($stmt = $mysqli->prepare()) {
    $stmt->bind_param('s', $user['birthday']);
    $stmt->execute();
    $stmt->close();
} else {
    die($mysqli->error);
}

(This might need to be birthday_date as Jeroen suggested too.)

Upvotes: 1

Jeroen
Jeroen

Reputation: 13257

You need to request the birthday_date column, not birthday, because it's always in the same format, DDMMYYY. After that, you need to add the following to your INSERT query:

STR_TO_DATE('" . mysql_real_escape_string ($user ['birthday_date']) . "', '%m/%d/%Y')

Upvotes: 2

Vaidotas Strazdas
Vaidotas Strazdas

Reputation: 716

Did you try replacing ".$user['birthday'].','.CURDATE()." into '".$user['birthday']."',CURDATE() ?

Upvotes: 0

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