For instance:
$sql = "SELECT * FROM db";
$query = sqlsrv_query($conn, $sql);
while($row = sqlsrv_fetch_array($query)){
echo "$row[date_column]";
}
will crash
Most of the answers I've found are assuming you want to order your query by a datetime, but I just want to turn the datetime object into a string after I have all the rows.
I don't really understand how to use the php date() function, I've tried:
echo date("m/d/Y",$row[date_column]); //prints nothing
$string=$row["date_column"]->format('Y-m-d H:i:s')
while($row=sqlsrv_fetch_array($rs,SQLSRV_FETCH_NUMERIC))
{
$logdate=date_format($row[4],"Y-m-d H:i:s");
}
here, $row[4] contains the date object.
This would surely help.
First of all, if "date_column" is the column name, it should be in quotes, like this:
echo $row["date_column"];
Secondly, PHP has a built-in function to turn MYSQL datetime into a unix timestamp, which PHP date() can use. Try this:
echo date("m/d/Y", strtotime($row["date_column"]));
Cheers!
echo date("m/d/Y", strtotime($row["date_column"]); is the date: 12/31/1969 for every entry. Also, that is not the correct date for any of the entries.