What is wrong?

[Anon]

I want to filter out records in the database which are registered in a given year.

I have the following option menu and it works...

<select name = years>";
$value = 2000;
for ($i =0; $i <= 100; $i++){
$value = $value + 1;
if($i== $years){$sel = "Selected";}
if($i!= $years){$sel = "";}
echo "<option value='".$i."'$sel>$value</option>";
}
echo "</select><br>";

Assume that there is a submit button called FilterByMonth.

if(isset($FilterByMonth)){
$query = "select * from table where YEAR(Datum) = $years";
.....
}

I know that the variable $years returns 1, 2, 3, 4, ....

As a result I couldn't get the correct out put as YEAR(Datum) belongs to 2000, 2001, 2002, ...

Is there any hint???

Thank you
solomon

[wchris]

Perhaps:

$query = "select * from table where YEAR(datum) = '$years'";

[Anon]

Dear WChris,

Thank you very much for your response.
$years stores values like 1, 2, 3, 4, ...

YEAR(Datum) returns 2000, 2001, 2002,...

How one can compare YEAR(Datum) with $years;

Example 2000 = 1

Still I couldn't solve it.

Solomon

[wchris]

I meant that you have to write $years in '' .

But I have to say, that I never saw YEAR(datum) in a SELECT-statement ... is YEAR the name of a method?

[Anon]

You are right but since the values of $years are integer putting them in '' does nothing.

Datum is my field in the table. YEAR() is a predefined function(method) in MYSQL.

Example:

YEAR('2003-12-12') returns 2003.
Look here,

http://www.mysql.com/doc/D/a/Date_calculations.html

Did yo understand my problem?

Now I have stopped there....

Solomon

[wchris]

Now I really understood your problem.
1.)
$max_year = 2100;
$min_year = 2000;
$value = $min_year;

for (; $value <= $max_year; $value ++) {

if($value == $years) {
print "<option value=/"$value/" Selected>$value</option>";
}
else {
print "<option value=/"$value/">$value</option>";
}

} // end for

2.)
Do you have a field in your dbase with the date?

--> Yes:
SELECT * from table WHERE date = '$year' ORDER BY date

[wchris]

OOps I think I made a misstake, it should be \"$value\" instead of /"$value/"

(but not really sure :-))

[Anon]

Try this instead of yours "select":
<select name = years>";
for ($i = 2000; $i <= 2100; $i++)
{
if($i== $years){$sel = "Selected";}
if($i!= $years){$sel = "";}
echo "<option value='".$i."'$sel>$value</option>";
}
echo "</select><br>";

[Anon]

Hallo WChris,

Great!!!!!!!!!!

Now every thing is going on. I have learnt a lot from you.

I appreciate your help.

Keep it up,

Solomon

[wchris]

I hope you wont think that I am a little bit confuse, but if I wrote my messages you inserted your new one already ...

[Anon]

Thank you Metin Arman,

It works if $value is replaced by $i as follows

echo "<option value='".$i."'$sel>$i</option>";

Again thank you

[metin]

It works if $value is replaced by $i as follows

Sorry,
missed that one 😉