php从datetime对象比较01到1

Question

DateTime object outputs 01, 02, 03 etc when I use

$num = $dt->format('d');

to get the day number

Then I compare the $num value if it's the first day of the month like so:

if ($num == 1)

but the $num value is '01'.

Now php compares it as expected

var_dump($num == 1)

returns true

I was thinking, should this be enough for me or I should enclose the $num variable with an intval like so:

intval($num)

this way if it is 01 it will display '1'

Answer 1

Basically $num = $dt->format('d') returns a String.

If you have == as a Comparison Operators and the two values are not from the same data type, PHP try to match them.

So in your case ($num == 1) you are comparing a String with a literal Integer

Therefore PHP is trying to converting the Sting into a Integer (only for the comparison).
In the end you are already comparing two Integers.

The type conversion does not take place when the Comparison Operator is === or !== as this involves comparing the type as well as the value.

So with $num = '1';, a comparison like $num === 1 would always return false.

If you like to strip of leading zeros but don't convert the data type, I would use:
$num_as_str = ltrim($num,'0');

If you like to convert the variable to an Integer use this:
$num_as_int = intval($num);

From the PHP manual:

String conversion to numbers

When a string is evaluated in a numeric context, the resulting value and type are determined as follows.

The string will be evaluated as a float if it contains any of the characters '.', 'e', or 'E'. Otherwise it will be evaluated as an integer.

The value is given by the initial portion of the string. If the string starts with valid numeric data, this will be the value used. Otherwise, the value will be 0 (zero). Valid numeric data is an optional sign, followed by one or more digits (optionally containing a decimal point), followed by an optional exponent. The exponent is an 'e' or 'E' followed by one or more digits.