Inserting date into my sql database with php

Question

I tried inserting a date into a mysql database. but for some reasons it was processing the information before putting it into the database here is my code

define("FRIENDS",  "friends");

if (!$update) { 
     $date = date("m-d-Y");
        $q = "INSERT INTO " . FRIENDS . " VALUES ('$friend2', $date, $status)"; 
    }

this is what i see in my database: -2025 instead of 15-05-2015. i seems like its being subtracted.

And when i use

$date = date("m/d/Y");

it divides out, leaving 0.000165425971712158 in my database.

must be $friends right? or that should be insert into friends — Danyal Sandeelo, May 15 '15 at 20:34

castis · Accepted Answer · 2015-05-16T15:53:40.667

6

Quote your values.

$q = "INSERT INTO " . FRIENDS . " VALUES ('$friend2', $date, $status)";

ends up looking like

"INSERT INTO tbl_friends VALUES ('friend', 15-05-2015, status)"

mysql does the small equation it finds in your sql query; 15 minus 5, minus 2015

quote your values and it becomes a string instead of an expression.

$q = "INSERT INTO " . FRIENDS . " VALUES ('$friend2', '$date', '$status')";

edited May 16 '15 at 15:53

answered May 15 '15 at 20:33

castis

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*"15 minus 5, minus 2015"* - Yep, sold me on that one ;-) – Funk Forty Niner May 15 '15 at 21:27

score 2 · Answer 2 · answered May 15 '15 at 20:34

Your date format is wrong for MySQL storage. Create your date using:

$date = date('Y-m-d');

~~There is also probably an error in your SQL, unless you have a PHP constant set for "FRIENDS".~~

EDIT: nevermind, you updated your question showing you have the constant set

Inserting date into my sql database with php

2 Answers2