how can I get these results out of my table ?

[doctor_james]

I've created a table to save my website visits , the fields are id,Ip,date,page,broswer .

By each visit , a Record are added to the table so each day can have several records. Now I wanted to get this Information out of these Information:

 

1) Hits for the busiest day .

2) Hits for this week.

3) Hits for this month.

4) Hits for this year.

 

How can I do it ? Is this possible with the information I've collected ?

Thanks in advance .

 

 

 

[refiking]

sure you can get those results

//Assuming your db table's name is table and your timestamp is named stamp:
// For all the hits of the month
$sql = mysql_query("SELECT * FROM `table` WHERE MONTH(`stamp`) = '10' AND YEAR(`stamp`) = '2008'");

// For all the hits of the year
$sql = mysql_query("SELECT * FROM `table` WHERE YEAR(`stamp`) = '2008'");

 

Can't think of the best route to get the busiest day right now, but yeah if you get the data you say you are, that's all you need to pull these records

 

[Maq]

Check out Count(*).  It allows you to collect the number of records for your desired condition.  There's plenty of examples online.

[doctor_james]

Thanks for the replies .. I used your method but It returns no result.

[trq]

You might want to post your current code.

[doctor_james]

$query="SELECT DISTINCT(ip) FROM stats WHERE YEAR('date')='2008'";
$h=mysql_query($query);
$hits=mysql_num_rows($h);

[Stooney]

Try changing

('date')

to this:

(`date`)

 

[doctor_james]

Try changing

('date')

to this:

(`date`)

 

 

It worked , thanks alot.

But there's still one question left. How can I get the hits for the busiest day according to the table I introdiced above .

Thank You all .

[trq]

Did you look at mysql's COUNT function?

[doctor_james]

Did you look at mysql's COUNT function?

 

Yes, I have.. I want to count the records of the busiest day But I really have no Idea how to do it .

[xylex]

SELECT COUNT(*) FROM stats GROUP BY DAYNAME(`date`);

[Maq]

SELECT COUNT(*) FROM stats GROUP BY DAYNAME(`date`);

 

Then if you grab the first record it should have the highest count, being the busiest day.