在通过php检索所有日期后，如何检查我的日期是否有某一年？

I am retrieving my date like this in my php query:

DATE_FORMAT(items.r_date, '%M %D, %Y') as r_date

And then displaying it like this, for example:

January 1st, 1934

However, if the year is 1111, I would like to make the date say something custom.

Such as (I know this isn't real code, just trying to explain what I want to do:

if ($row['r_date'] CONTAINS '*1111*') { $r_date = "my custom message"; } else $r_date = $row['r_date'];

Could anyone help me solve this?

Thank you for your time.

A literal implementation of your pseudo-code uses strpos:

if (false !== strpos($row['r_date'], '1111')) {
    $r_date = "my custom message";
} else {
    $r_date = $row['r_date'];
}

However, the "only check the last four digits" answer by Dagon, or the alternate one below using substr_compare, may yield fewer false positives:

if (0 === substr_compare(rtrim($row['r_date']), '1111')) 
    $r_date = "my custom message";
} else {
    $r_date = $row['r_date'];
}

Note I've used trim here in the event your date has spurious white-space on the end.

if u only want to check by its year, then :

    $mydate = date('Y', strtotime($row['r_date']))
    if ($mydate == '1111' or $mydate == '1970') {
    echo "something";
    }else{
    echo "else"
    }

note:
Y = for 4 digit year.
y = for 2 digit year.
1970 is the lower year on system, so if your year is lower than 1970 system will make year to 1970 by default.

strtotime

You can use DateTime Class. I didn't test but it should be something similar, just change the date format:

$date = DateTime::createFromFormat("m-d-Y", "02-01-1111");
if($date->format("Y")=="1111")
    echo "your custom message";

i would use substr() to extract the last 4 characters being your year

if (substr($row['r_date'], -4) != '1111') { 
    $r_date = $row['r_date']; } 
else { 
    $r_date = "my custom message"; 
}