SETTING VARIABLE DATES IN PHP SCRIPT

[chewson]

Attempting to set a variable = NOW() + INTERVAL 90 DAY;

I can run SELECT DATE_ADD(NOW(), INTERVAL 90 DAY using MYSQL, but don't know how to move result to a variable.

[MasterACE14]

SELECT DATE_ADD(NOW(), INTERVAL 90 DAY AS variablename

[chewson]

THANK YOU, BUT STILL HAVE A PROBLEM WITH MY SCRIPT "Parse error: syntax error, unexpected T_LNUMBER in /home/chewson1/public_html/nurse_certification/TEST_DTE.php on line 11

THE SCRIPT IS:

        <h1>TEST DATE PROCESSING</h1>

        <?php

       

      $sql = "SELECT  "2011-10-12" + INTERVAL 90 DAY as date90";

      $res = mysqli_query($mysqli, $sql); 

     

      if ($res) {

      $DATE_90 = date90;

      echo $DATE_90                       

  }

?> 

[codefossa]

Why query SQL to get 90 days from the current time?

 

$then = strtotime("90 days");

echo date('F d, Y', $then);

[MasterACE14]

need to add to the query 'FROM table_name', you currently have incorrect syntax.

 

SELECT

 

Why query SQL to get 90 days from the current time?

 

$then = strtotime("90 days");

echo date('F d, Y', $then);

Or this method is much better if that is all you're trying to achieve.

[chewson]

Thank you very much Kira.  That worked, but now how do I use  variable in a MYSQL SELECT statement for example

$query = "SELECT * FROM A_record WHERE A_dte  <= $dte90";.  Dte90 was calculated using your routine.

[MasterACE14]

$query = "SELECT * FROM A_record WHERE A_dte  <= '{$dte90}'";

[PFMaBiSmAd]

You can do this all in your query statement - SELECT * FROM your_table WHERE A_dte  <= DATE_ADD(NOW(), INTERVAL 90 DAY)

 

However, what exactly are you trying to match, because you are forming a date 90 days into the future, then finding the rows that have a date less-than or equal to that value?