Convert Date Formats in PHP: From yyyy-mm-dd to dd-mm-yyyy

By the end of this page, you will understand how PHP date formatting works, how to convert a date string into another format, and when to use DateTime, strtotime(), or date() for safe and readable date handling.

In PHP, dates are often stored as strings, but formatting functions work best when PHP first understands that string as a date value.

The key idea is:

1. Parse the original string into a date object or timestamp.
2. Format that parsed date into the layout you want.

For example, if you have this string:

'2024-01-31'

that is just text until PHP interprets it as a date.

Why this matters

In real programs, dates often come from:

• databases
• forms
• APIs
• CSV files
• user input

These sources may provide dates in one format, while your application needs to display them in another.

In PHP, the most reliable modern approach is to use DateTime.

Two common ways to convert dates

1. Using DateTime

This is the preferred approach because it is clearer and more robust.

$date = new DateTime('2024-01-31');
echo $date->format('d-m-Y');

Output:

Think of a date string like a sentence written on paper:

2024-01-31

At first, PHP just sees characters.

To work with it properly, PHP needs to understand the text as a real calendar date. That is the parsing step.

After that, formatting is like choosing how to print the same date on screen:

• Y-m-d → 2024-01-31
• d-m-Y → 31-01-2024
• d/m/Y → 31/01/2024

So the process is:

• input text
• parse into a date
• output in a new format

Same date, different display.

Using DateTime

$input = '2024-01-31';
$date = new DateTime($input);
echo $date->format('d-m-Y');

Output:

31-01-2024

Explanation

• new DateTime($input) parses the string into a date object.
• format('d-m-Y') prints the date in day-month-year format.

Using strtotime() and date()

$input = '2024-01-31';
echo date('d-m-Y', strtotime($input));

Output:

31-01-2024

Explanation

Consider this code:

$input = '2024-01-31';
$date = DateTime::createFromFormat('Y-m-d', $input);
echo $date->format('d-m-Y');

Here is what happens step by step:

1. Store the original string

$input = '2024-01-31';

The variable contains plain text.

2. Parse the string as a date

$date = DateTime::createFromFormat('Y-m-d', $input);

PHP reads the string using this pattern:

• Y → year = 2024
• m → month = 01
• d → day = 31

Date format conversion appears in many real applications.

Common examples

Displaying database dates to users

Databases often store dates as:

2024-01-31

But users may expect:

31-01-2024

Formatting API responses

An external API may return ISO-style dates, and your app may need to display them in a local format.

Exporting reports

You might store dates in one format internally but export invoices or CSV files in another format.

Processing form input

A user may submit one format, while your backend stores a standardized format.

Logging and dashboards

Developers may prefer machine-friendly formats in storage and human-friendly formats in dashboards.

Practical rule

• Store dates in a standard format such as Y-m-d when possible.
• Display dates in whatever format users need.

In real PHP projects, developers usually avoid manually rearranging date strings like this:

// Avoid this
$parts = explode('-', $input);
echo $parts[2] . '-' . $parts[1] . '-' . $parts[0];

That works only when the input is always perfect, and it does not validate the date.

Common patterns in real code

Validation before formatting

$input = '2024-01-31';
$date = DateTime::createFromFormat('Y-m-d', $input);

if ($date === false) {
    echo 'Invalid date';
} else {
    echo $date->format('d-m-Y');
}

Guard clause pattern

{
     = ::(, );

     ( === ) {
         ;
    }

     ->();
}

1. Using the wrong format letters

A very common mistake is mixing up lowercase and uppercase letters.

echo date('d-m-y');

This prints a 2-digit year, not a 4-digit year.

Use this instead:

echo date('d-m-Y');

2. Assuming date() can directly reformat any string

Broken idea:

$input = '2024-01-31';
echo date('d-m-Y', $input);

Why it fails:

• date() expects a timestamp as its second argument.
• $input is a string, not a Unix timestamp.

Correct approach:

echo date('d-m-Y', strtotime());

DateTime vs strtotime() + date()

Quick reference

Convert Y-m-d to d-m-Y

$input = '2024-01-31';
echo (new DateTime($input))->format('d-m-Y');

Strict parsing

$input = '2024-01-31';
$date = DateTime::createFromFormat('Y-m-d', $input);
echo $date->format('d-m-Y');

Using strtotime()

$input = '2024-01-31';
echo date('d-m-Y', strtotime($input));

Common format characters

How do I change a date from yyyy-mm-dd to dd-mm-yyyy in PHP?

Use DateTime:

echo (new DateTime('2024-01-31'))->format('d-m-Y');

Can I use date() directly on a string?

No. date() expects a Unix timestamp as its second argument, not a raw date string.

Should I use strtotime() or DateTime?

For simple cases, both work. In most real projects, DateTime is preferred because it is clearer and more reliable.

What is the best way when I know the exact input format?

Use DateTime::createFromFormat().

$date = DateTime::createFromFormat('Y-m-d', '2024-01-31');

Why not just split the string with ?

• DateTime — the main PHP class for parsing and formatting dates.
• strtotime() — converts many date strings into Unix timestamps.
• date() — formats timestamps into readable strings.
• DateTime::createFromFormat() — useful for strict parsing of known input formats.
• Unix timestamp — the numeric value often used by date().
• Timezones in PHP — important when working with full dates and times.
• Form validation — useful when dates come from user input.
• String parsing — related because dates often arrive as strings.