concatenated php string into mysql

[wev]

here i am again...

 

so i have this algorithm that computes the current academic year for our school..

it outputs the correct academic year on a browser..

but when i try to insert it into my database, the data inserted goes wrong..

 

it just keeps on inserting -1 instead of the string 2011-2012

 

here's my code:

<?php

$year = date('Y');
$currentyear = $year;
$lastyear = $year - 1;
$nextyear = $year + 1;

if (date('m') < 6)
{
 $current_ay = $lastyear."-".$currentyear;
}
else
if (date('m') >= 6)
{
 $current_ay = $currentyear."-".$nextyear;
}

echo $current_ay;
  $insertSQL = "INSERT INTO tbl_elections (election_id) values ($current_ay)";

  mysql_select_db($database_organizazone_db, $organizazone_db);
  $Result1 = mysql_query($insertSQL, $organizazone_db) or die(mysql_error());

?>

 

 

*the database column to be inserted into is of varchar data type

 

[DavidAM]

Is the column for the current academic year actually called "election_id" or do you have the wrong column name in that INSERT statement?

[Muddy_Funster]

a classic example of missing quotes for string input.  you are telling the DB that you want to enter 2011 - 2012 it's taking that as an arithmetic calculation and entering the result.  wrap quotes around your variable name and your all good.

[wev]

@david.. yes. my column name is correct..

 

@muddy_funster.. i did wrap it in quotes and it inserted a data into my database..but it only stored "2011". it did not include the "-2012"

 

 

--almost there!

 

[wev]

wrapping my variable with quotes actually inserted two(2) new rows to my database..one with zero "0" value,and another with "2011" value

 

[Muddy_Funster]

like this:

$insertSQL = "INSERT INTO tbl_elections (election_id) values ('$current_ay')";

[wev]

yeah..that's what i did..

figured out that the problem was with the datatype of the election_id column on my database..it was on int when it was supposed to be a varchar..

 

got it working now..thank you so much for your help..