PHP taking user input and formatting output (dates)

Question

I'm new to php and need some help, I am trying to take the user input and format it so that the output shows the last day of the month. Here's what I have so far:

<?php
    //form for user-input
    echo "<form name='form' method='post' action='array.php'>";
    echo "<p>Enter Date in ' Month/Day/Year ' format<p>";
    echo "<input type='text' id='date-input' name='date-input' placeholder='Enter date' />";
    echo "</form>";

    //grab user-input
    $input=$_POST["date-input"];

    //output in correct format
    echo $input->format("m/t/Y");


?>

When I have just echo input variable without the format function and date format then it displays what the user inputs, but how it's set right now; nothing displays and I get this line:

Fatal error: Call to a member function format() on string in C:\xampp\htdocs\php-sessions\session-3\array.php on line 190

Line 190 is my echo input line.

$input is a string not an object. try using echo $input; and if you want date you may need to convert the string to date. Take a look at strtotime — bansi
– bansi, Commented Dec 14, 2016 at 4:08
When you say "output shows the last day of the month" does that mean you want it to output how many days are in the month provided? — scorgn
– scorgn, Commented Dec 14, 2016 at 4:11
Yes basically, which is why im trying to format using "t" for total days, I just want to output the number of days in the given month for what the user inputs as a date. So say the user types 2/05/2017. I want the output to display 2/28/2017 I also want it to count for leap years. — Matt Picca
– Matt Picca, Commented Dec 14, 2016 at 4:14
@bansi, I want to try an avoid strtotime due to the fact it will not work after year 2038 — Matt Picca
– Matt Picca, Commented Dec 14, 2016 at 4:16
try using DateTime. But don't forget DateTimeZone or you may get unexpected result — bansi
– bansi, Commented Dec 14, 2016 at 4:17

bansi · Accepted Answer · 2016-12-14 04:31:40Z

2

Try this. You may need to create a date first from your input string. Used createFromFormat

$date = DateTime::createFromFormat('m/d/Y', $input);
echo $date->format('m/t/Y');

Note: Remember to use timezone parameter also with all datetime functions.

answered Dec 14, 2016 at 4:31

bansi

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1 Comment

Matt Picca Over a year ago

This worked perfectly! I will look up now how to implement DateTimeZone and or use parameters for timezone. Thank you!

Naga · Accepted Answer · 2016-12-14 04:08:44Z

0

something like replace

//output in correct format
    echo $input->format("m/t/Y");

with

//output in correct format
    echo date("m/d/Y",strtotime($input));

answered Dec 14, 2016 at 4:08

Naga

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4 Comments

Matt Picca Over a year ago

Just saw the edit, but I want to try and avoid strtotime due to it not working after year 2038

Naga Over a year ago

which means you are passing bad input. look into this sample and test it echo $input=date('d M Y'); echo "<br />".date("m/d/Y",strtotime($input));

Naga Over a year ago

Make sure month should be string (ex, Jan or January) not numeric.

Matt Picca Over a year ago

I used echo "echo <br />".input=date("m/t/Y",strtotime($input)); and it worked just like I want, but again I would like to use another method other than strtotime

Soni Vimalkumar · Accepted Answer · 2016-12-14 04:47:56Z

0

Maybe Match your requirements,

<?php

$date = "2040-11-23";

$Year = DateTime::createFromFormat("Y-m-d", $date)->format("Y");

$YearWithDay = DateTime::createFromFormat("Y-m-d", $date)->format(" Y-m, t \d\a\y.");

function is_leap_year($year)
{
   return ((($year % 4) == 0) && ((($year % 100) != 0) || (($year % 400) == 0)));
}

echo $YearWithDay."<br>";
echo (is_leap_year($Year))?"This Year is Leap year": "";

?>

OUTPUT

2040-11, 30 day.
This Year is Leap year

answered Dec 14, 2016 at 4:47

Soni Vimalkumar

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