CODEWITHSUNDEEP
phpdate-formatsqlsrv
japu soft dev
japu soft dev

Reputation: 80

Adding Days to a date returned through sqlsrv_fetch_array in PHP

Initially when i was using php_msssql extension the following line was working

$test2_date = date('d-M-y', strtotime( "".$res["test_date"]." +".$noofdays." days" ));

$res is array.

Then i changed to php_sqlsrv_53_ts_vc9 extension due to which the above line was not working because "date" doesn't work in this extension so i used "date_format" which works in sqlsrv. My code line was like shown below

$test2_date = date_format(strtotime( "".$res["test_date"]." +".$noofdays." days" ), 'd-M-y');

but the strtotime is not working and hence i get the below error:-

Catchable fatal error: Object of class DateTime could not be converted to string.

what should i use to add days to the date?

Upvotes: 0

Views: 190

Answers (2)

Zhorov
Zhorov

Reputation: 29943

If you want to retrieve date and time types (datetime, date, time, datetime2, and datetimeoffset) as strings, you have following options:

Set connection option ReturnDatesAsStrings to false and format returned date field value. This is by default.

$server = 'server\instance,port';
$cinfo = array(
    "ReturnDatesAsStrings"=>false,
    "Database"=>'database',
    "UID"=>"user",
    "PWD"=>"password"
);
...
$test2_date = date_format(date_add($res["test_date"], date_interval_create_from_date_string($noofdays." days")), 'd-M-y')
...

Set connection option ReturnDatesAsStrings to true.

$server = 'server\instance,port';
$cinfo = array(
    "ReturnDatesAsStrings"=>true,
    "Database"=>'database',
    "UID"=>"user",
    "PWD"=>"password"
);
...
$test2_date = date_format(date_add(date_create($res["test_date"]), date_interval_create_from_date_string($noofdays." days")), 'd-M-y')
...

Additional information can be found here.

Upvotes: 0

Lovepreet Singh
Lovepreet Singh

Reputation: 4840

date_format take date object as first parameter. So need to create date object with date_create first. Code should be:

$date = date_create($res["test_date"]);
date_add($date, date_interval_create_from_date_string(" +" . $noofdays . " days"));
echo date_format($date, "d-M-y");

UPDATE: If you have date object already then date_create step can be skipped and code will be like:

date_add($res["test_date"], date_interval_create_from_date_string(" +" . $noofdays . " days"));
echo date_format($date, "d-M-y");

Upvotes: 1

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