CODEWITHSUNDEEP
phpdatetime
RGriffiths
RGriffiths

Reputation: 5970

Object of class DateTime could not be converted

I can see that there are many questions along these lines but I am very confused as to why the following does not work, taken straight from the PHP docs:

$tempDate = DateTime::createFromFormat('j-M-Y', '15-Feb-2009');
echo $tempDate;

The error:

PHP Catchable fatal error: Object of class DateTime could not be converted to string.

In fact every example in the docs gives me this error. Any ideas?

Upvotes: 1

Views: 11744

Answers (2)

axiac
axiac

Reputation: 72177

The error message:

PHP Catchable fatal error: Object of class DateTime could not be converted to string.

is self-explanatory. The statement:

echo $tempDate;

attempts to print a DateTime object. The echo() language construct expects a string, you pass it a DateTime object. The DateTime class does not implement the __toString() magic method and PHP doesn't know how to convert a DateTime object to string.

There are so many ways to represent a DateTime object as string and all of them are handled by the DateTime::format() method.

In fact every example in the docs gives me this error.

In fact, every example in the documentation of DateTime::createFromFormat() reads:

echo $date->format('Y-m-d');

which is a different thing that echo $date;.

Any ideas?

Read the documentation of DateTime and DateTime::format() carefully.

Upvotes: 1

Sebastian Brosch
Sebastian Brosch

Reputation: 43564

You can't echo the DateTime object directly. You have to use the format method to get the date and / or time part:

$tempDate = DateTime::createFromFormat('j-M-Y', '15-Feb-2009');
echo $tempDate->format('Y-m-d H:i:s');
// NOT echo $tempDate!!!

demo: http://ideone.com/IyqRWj

If you want to see the details of the object (for debug) you can use var_dump:

$tempDate = DateTime::createFromFormat('j-M-Y', '15-Feb-2009');
var_dump($tempDate);

Upvotes: 5

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