Php catchable fatal error

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i have this error when inserting $diff into my database table

SQL

time_in(time)

but im having this error

Catchable fatal error: Object of class DateInterval could not be converted to string

the error is in my query

What I have tried:

$conn = mysqli_connect(server,username,password)
        or die (mysqli_error($conn));
mysqli_select_db($conn,database)
            or die (mysqli_error($conn));

if ($now->format("H:i") > "22:00") {
    $deadline = DateTime::createFromFormat("H:i", "22:00");
    $diff = $now->diff($deadline);
    echo "You are ".$diff->h." hours and ".$diff->i." minutes late";
} else if ($now->format("H:i") > "19:00") {
    $deadline = DateTime::createFromFormat("H:i", "19:00");
    $diff = $now->diff($deadline);
    echo "You are ".$diff->h." hours and ".$diff->i." minutes late";
} else if ($now->format("H:i") > "16:00") {
    $deadline = DateTime::createFromFormat("H:i", "16:00");
    $diff = $now->diff($deadline);
    echo "You are ".$diff->h." hours and ".$diff->i." minutes late";
} else if ($now->format("H:i") > "13:00") {
    $deadline = DateTime::createFromFormat("H:i", "13:00");
    $diff = $now->diff($deadline);
    echo "You are ".$diff->h." hours and ".$diff->i." minutes late";
} else if ($now->format("H:i") > "10:00") {
    $deadline = DateTime::createFromFormat("H:i", "10:00");
    $diff = $now->diff($deadline);
    echo "You are ".$diff->h." hours and ".$diff->i." minutes late";
} else if ($now->format("H:i") > "07:00") {
    $deadline = DateTime::createFromFormat("H:i", "07:00");
    $diff = $now->diff($deadline);
    echo "You are ".$diff->h." hours and ".$diff->i." minutes late";
}

$query = "INSERT INTO `time_in`(`late`,`date_in`)
                VALUES ('$diff',NOW())";
                if (!$query) {
                printf("Error: %s\n", mysqli_error($conn));
                exit();
                } 
        if($query = $conn ->query($query)){
            echo "<br>successful";
        } else {
            echo "not successful";
        }

Posted 18-Feb-17 2:33am

Member 12986530

Updated 18-Feb-17 3:08am

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Comments

Richard Deeming 18-Feb-17 9:00am

Your code is vulnerable to SQL Injection[^]. NEVER use string concatenation or iterpolation to build a SQL query. ALWAYS use a parameterized query.

Everything you wanted to know about SQL injection (but were afraid to ask) | Troy Hunt[^]
How can I explain SQL injection without technical jargon? | Information Security Stack Exchange[^]
Query Parameterization Cheat Sheet | OWASP[^]

PHP: mysqli::prepare[^]

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Peter Leow · Answer 1 · 2017-02-18T03:08:00

Solution 1

That is because

$diff

is a

DateInterval

object, not a string. To insert the string representation of the intervals in hours and minutes, do this:

$diff->format("%H:%i")

Do use PHP Prepared Statements[^] to avoid sql injection risk.

Posted 18-Feb-17 3:08am

Peter Leow

Updated 18-Feb-17 3:25am

v3

Comments

Member 12986530 18-Feb-17 9:15am

i change $diff->format("H:i") into $diff->format('H:i') because it giving me a error of Parse error: syntax error, unexpected 'H' (T_STRING) but when im changing it to single quote it give me this Notice: Undefined property: DateInterval::$format

Peter Leow 18-Feb-17 9:25am

It should be
$diff->format("%H:%i")

Member 12986530 18-Feb-17 9:25am

i change into

$diff->format("H:i")

because im getting this

Parse error: syntax error, unexpected 'H' (T_STRING)

so change it to

$diff->format('H:i')

but im still getting this

Notice: Undefined property: DateInterval::$format

Peter Leow 18-Feb-17 9:30am

Solution corrected.