CODEWITHSUNDEEP
phpmysqldatetimeinsert
DIM3NSION
DIM3NSION

Reputation: 1761

Object of class DateTime could not be converted to string

I have a table with string values in the format of Friday 20th April 2012 in a field called Film_Release

I am looping through and I want to convert them in datetime and roll them out into another table. My second table has a column called Films_Date, with a format of DATE. I am receiving this error

Object of class DateTime could not be converted to string

$dateFromDB = $info['Film_Release'];
$newDate = DateTime::createFromFormat("l dS F Y",$dateFromDB); //( http:php.net/manual/en/datetime.createfromformat.php)

Then I insert $newdate into the table through an insert command.

Why am I be getting such an error?

Upvotes: 69

Views: 307723

Answers (9)

Prid
Prid

Reputation: 1624

For those coming from Javascript like me: you can't simply echo the DateTime object and expect a nicely formatted default string as the output!

You have to use the format() function and input a custom format like $dateTimeObject->format('Y-m-d') to get a string representation of the date stored in the object. (Syntax reference)

There are some default constants you can use though, and this one is the closest to Javascript's default format:

echo (new DateTime())->format( DateTime::RSS );
// Fri, 29 Dec 2023 00:48:56 +0000

Alternatively, use print_r( $dateTimeObject ) for a quick glance at the datetime value :)

print_r( (new DateTime()) );
/*(
  [date] => 2023-12-29 00:55:38.670878
  [timezone_type] => 3
  [timezone] => UTC
)*/

Upvotes: 2

Kavindu Dissanayake
Kavindu Dissanayake

Reputation: 883

$Date = $row['Received_date']->format('d/m/Y');

then it cast date object from given in database

Upvotes: 1

Samuel RIGAUD
Samuel RIGAUD

Reputation: 1702

If you are using Twig templates for Symfony, you can use the classic {{object.date_attribute.format('d/m/Y')}} to obtain the desired formatted date.

Upvotes: 3

KennyKivi
KennyKivi

Reputation: 8

It's kind of offtopic, but i come here from googling the same error. For me this error appeared when i was selecting datetime field from mssql database and than using it later in php-script. like this: 

$SQL="SELECT Created
FROM test_table";

$stmt = sqlsrv_query($con, $SQL);
if( $stmt === false ) {
    die( print_r( sqlsrv_errors(), true));
}

$Row = sqlsrv_fetch_array($stmt,SQLSRV_FETCH_ASSOC);


$SQL="INSERT INTO another_test_table (datetime_field) VALUES ('".$Row['Created']."')";
$stmt = sqlsrv_query($con, $SQL);
if( $stmt === false ) {
    die( print_r( sqlsrv_errors(), true));
}

the INSERT statement was giving error: Object of class DateTime could not be converted to string

I realized that you CAN'T just select the datetime from database:

SELECT Created FROM test_table

BUT you have to use CONVERT for this field:

SELECT CONVERT(varchar(24),Created) as Created FROM test_table

Upvotes: 0

user4272288
user4272288

Reputation:

Try this:

$Date = $row['valdate']->format('d/m/Y'); // the result will 01/12/2015

NOTE: $row['valdate'] its a value date in the database

Upvotes: 20

Tim
Tim

Reputation: 159

Use this: $newDate = $dateInDB->format('Y-m-d');

Upvotes: 5

onlymybest
onlymybest

Reputation: 339

Check to make sure there is a film release date; if the date is missing you will not be able to format on a non-object.

if ($info['Film_Release']){ //check if the date exists
   $dateFromDB = $info['Film_Release'];
   $newDate = DateTime::createFromFormat("l dS F Y", $dateFromDB);
   $newDate = $newDate->format('d/m/Y'); 
} else {
   $newDate = "none"; 
}

or 

 $newDate = ($info['Film_Release']) ? DateTime::createFromFormat("l dS F Y", $info['Film_Release'])->format('d/m/Y'): "none"

Upvotes: 0

Jim D
Jim D

Reputation: 978

You're trying to insert $newdate into your db. You need to convert it to a string first. Use the DateTime::format method to convert back to a string.

Upvotes: 0

Jon
Jon

Reputation: 437336

Because $newDate is an object of type DateTime, not a string. The documentation is explicit:

Returns new DateTime object formatted according to the specified format.

If you want to convert from a string to DateTime back to string to change the format, call DateTime::format at the end to get a formatted string out of your DateTime.

$newDate = DateTime::createFromFormat("l dS F Y", $dateFromDB);
$newDate = $newDate->format('d/m/Y'); // for example

Upvotes: 99

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