PHP Create datetime variable from a string variable

Question

I want to create a new datetime variable from a string variable using datetime function.
So, I have:
$sqlData=$day . "." . $month . "." . $year;
echo $sqlData;

$local=new datetime('${sqlData}');
echo $local;

The problem appear on echo $local;

Error msg: Failed to parse time string (${sqlData}), unexpected character'

`'${sqlData}'` should be `"${sqlData}"` – Nick Jan 09 '19 at 12:08 — Nick, Jan 09 '19 at 12:08

Andreas · Accepted Answer · 2019-01-09T12:18:56.917

4

Why do you think you need the { and ' in DateTime?
$sqlData is a string, and DateTime wants a string. So why not just give it the string?

$sqlData=$day . "." . $month . "." . $year;
echo $sqlData;

$local=new datetime($sqlData);
var_dump($local);

Then echo won't work since it's an object, so var_dump/print_r or var_export

https://3v4l.org/nq8iY

To echo the string you need to format the object.

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

edited Jan 09 '19 at 12:18

answered Jan 09 '19 at 12:09

Andreas

23,610
6
30
62

Is not working, `Object of class DateTime could not be converted to string` – BOB Jan 09 '19 at 12:12
Did you just try to echo the object even though I explicitly wrote you can't echo an object? What exactly are you expecting? The date? The object itself? – Andreas Jan 09 '19 at 12:12
yes, sorry for that but I need to echo this variable – BOB Jan 09 '19 at 12:14
As a string? Then why bother with DateTime at all. Just place your day, month and year variables as you like and echo them. I don't understand why you create a DateTime object you don't use – Andreas Jan 09 '19 at 12:15
I'll need to use later as an object – BOB Jan 09 '19 at 12:16
https://3v4l.org/f8mNH you need to format the object. – Andreas Jan 09 '19 at 12:18
Ofcourse I format that object but the problem was solved after I removed `Var_dump($local);` . Thank you – BOB Jan 09 '19 at 12:20
I don't see how a commented line can ever be a problem, but ok. – Andreas Jan 09 '19 at 12:21
Not the commented line.... – BOB Jan 09 '19 at 12:22

score 2 · Answer 2 · answered Jan 09 '19 at 12:09

This below worked for me, you need to set the format and return it as a new variable.

   <?php
    $date = 05; $month = 05; $year = 1990;
    $sqlData="$date.$month.$year";
    $local=new datetime($sqlData);
    $date = $local->format('Y/m/d');
    echo $date;

PHP Create datetime variable from a string variable

2 Answers2