Check the Date comparison ?

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hi,
I am facing an unusual problem when comparing two dates.Pleas have a look.
Database table

Warranty_renewal_date (Date type= DATE)
2010-02-12
2010-02-12
2013-06-11

My code in php

PHP

$sql4="select count(*) as Under_warrenty from machine_details where warranty_renewal_date >=".$current_date;
$result4 = mysql_query($sql4) or die('Query failed: ' . mysql_error());
if ($result4) {
  while ($row4 = mysql_fetch_assoc($result4)) {
      $totlapoll4=$row4["Under_warrenty"];
      
	    }
}

 
$sql5="select count(*) as OutOf_warrenty from machine_details where warranty_renewal_date <=".$current_date ;
$result5 = mysql_query($sql5) or die('Query failed: ' . mysql_error());
if ($result5) {
  while ($row5 = mysql_fetch_assoc($result5)) {
      $totlapoll5=$row5["OutOf_warrenty"];
      
	    }
}

In this case what i should get
$totlapoll4=1
$totlapoll5=2
But i am getting 3 and 0.
Can anyone suggest whats wrong here.?
Thanks

Posted 7-May-13 20:16pm

Rajeshkrathor

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2 solutions

Solution 1

echo sql query and check it by execute that in MySQL

Check Date format passed from PHP it should be yyyy-MM-dd

and in first query you are using >= then in second query you should use < only instead of <= it's just for correction.

Happy Coding!
:)

Posted 7-May-13 20:24pm

Aarti Meswania

Updated 7-May-13 20:38pm

v3

Comments

[no name] 8-May-13 2:31am

hi Aarti,
Thanks for your reply but i have done all getting same ans in both Mysql and in PHP.
that's why i post here to find out the bug.

Aarti Meswania 8-May-13 2:38am

check updated solution

[no name] 8-May-13 2:42am

hey, That was not reason. its just a matter of '' quote.

Aarti Meswania 8-May-13 2:48am

good you find it
and how? by echo/print query and then trying it in mysql - so, it show you error and you correct it :)

[no name] 8-May-13 3:05am

return 3 and 0 and Incorrect date value: '2000' for column 'warranty_renewal_date' at row 1 message and by using with '' colon giving the correct ans with no message.

Aarti Meswania 8-May-13 3:15am

exactly that was reason.
And it is simplest way to get exact reaason for error and fix it by testing query in database which was prepared in PHP.

[no name] 8-May-13 3:16am

Ok.
thanks
i was just checking same query again and again in both PHP and MYSQL.

Aarti Meswania 8-May-13 3:23am

:)when ever you face problem like this always check query in database prepared from php code it can give you exact hint so you can fix it easily
Have a nice time! :)

[no name] 8-May-13 3:26am

Sure!
Thanks for your advice..

Aarti Meswania 8-May-13 3:29am

Welcome!

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Rajeshkrathor · Accepted Answer · 2013-05-07T20:43:00

hi,
This solution code.

PHP

$sql4="select count(*) as Under_warrenty from machine_details where warranty_renewal_date >='$current_date'";
$result4 = mysql_query($sql4) or die('Query failed: ' . mysql_error());
if ($result4) {
  while ($row4 = mysql_fetch_assoc($result4)) {
      $totlapoll4=$row4["Under_warrenty"];
      
	    }
}

 
$sql5="select count(*) as OutOf_warrenty from machine_details where warranty_renewal_date <'$current_date'" ;
$result5 = mysql_query($sql5) or die('Query failed: ' . mysql_error());
if ($result5) {
  while ($row5 = mysql_fetch_assoc($result5)) {
      $totlapoll5=$row5["OutOf_warrenty"];
      
	    }
}

What a silly thing!