Object of class DateTime could not be converted to string
PhpMysqlDatetimeInsertPhp Problem Overview
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?
Php Solutions
Solution 1 - Php
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
Solution 2 - Php
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
Solution 3 - Php
Use this: $newDate = $dateInDB->format('Y-m-d');
Solution 4 - Php
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.
Solution 5 - Php
You're trying to insert $newdate into your db. You need to convert it to a string first. Use the DateTime::format method to convert back to a string.
Solution 6 - Php
Check to make sure there is a film release date; if the date is missing you will not be able to format on a non-object.
if ($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"
Solution 7 - Php
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
Solution 8 - Php
$Date = $row['Received_date']->format('d/m/Y');
then it cast date object from given in database