Help with a small code. . .?

Aliz__

Hey everyone, im new to this site but I am somewhat familiar with php. . .

Anyways I am having trouble with this script. Is what it is "supposed" to do is pull top searches from a mysql database.
but it isn't working. this is the error I am getting. . .

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource

and here is my script. . .

<?
//connection info
$DBHost     = "";		//MySQL-Server
$DBName     = "";		//MySQL-Database
$DBUser     = "";		//MySQL-Username
$DBPassword = "";		//MySQL-Password

mysql_connect($DBHost, $DBUser, $DBPassword) or
   die("Could not connect: " . mysql_error());
mysql_select_db($DBName);

// config
$file_path = "search.php";
$show = 3;

$getlogs = "SELECT word COUNT(*) as quantity FROM stats";
$result = mysql_query($getlogs);

$i = 0;
while ($row = mysql_fetch_array($result)) {
$logs[$i] = substr($logs[$i], 1, -1);
$count[$i] = $row['quantity'];
$i++; if ($i == $show) {break;}
}

echo "<div>Top Searches</div>";
echo "<div>";
for ($i=0; $i<$show; $i++) {
echo "$count[$i]";
}
echo "</div>\n";
?>

does anybody see a problem with this or a different way I should do this?

Thanks in advance!

spighita

$getlogs = "SELECT word COUNT(*) as quantity FROM stats";

change to

$getlogs = "SELECT word, COUNT(*) as quantity FROM stats";

I'm not sure though, but looks odd.

Aliz__

hmmmm nope. tried it but still same error. . .anyone else? thanks for trying though

darkside-x

count(*) is only ever going to return one number. are you sure you want to use that? if so, syntax is more along the lines of:

SELECT COUNT(*) FROM stats

dont think you can feed it a field name, on the grounds that it doesnt actually need to work with fields.

I don't really understand what you're doing here, feel free to explain it to me 🙂

Aliz__

when you asked me to explain more I almost forgot what I was even doing with the script. so many things going on at the same time. . .

But, I have a search and when somebody searches for a keyword it ads it to the database with the count 1. each time the word is searched for the count goes up 1 and so on (still working on this program too hehe)

but the script above is supposed to display the top keyword searches that were made. . .pulling the info from the database.

Final outcome = I want to have a small table with ummm lets say the top 5 searches made.

Does that make anymore sense. I hope so 🙂 If not let me know and I will try to explain it in another way.

Thanks.

darkside-x

So the search enters the count into the database?
If you've done that in a field your best bet is to SELECT word FROM stats ORDER BY count DESC and then display the first 5 results.

COUNT(*) is going to return a count of all rows, no matter what, so I honestly can't see how it'll be helpful in this situation...

even in a new line is added with every search, and...
Gah!

Long story short, the error you're getting is a result of the query not returning anything valid, if my similar experience is anything to go by.

Sorry I can't be more helpful.

Aliz__

ok now this is what I have

<? 
//connection info 
$DBHost     = "localhost";        //MySQL-Server 
$DBName     = "undangle_searchStats";        //MySQL-Database 
$DBUser     = "undangle_all";        //MySQL-Username 
$DBPassword = "encrypt";        //MySQL-Password 

mysql_connect($DBHost, $DBUser, $DBPassword) or 
   die("Could not connect: " . mysql_error()); 
mysql_select_db($DBName); 

// config 
$file_path = "search.php"; 
$show = 3; 

$getlogs = "SELECT word FROM stats ORDER BY quantity DESC "; 
$result = mysql_query($getlogs); 

$i = 0; 
while ($row = mysql_fetch_array($result)) { 
$logs[$i] = substr($logs[$i], 1, -1); 
$count[$i] = $row['word']; 
$i++; if ($i == $show) {break;} 
} 

echo "<div>Top Searches</div>"; 
echo "<div>"; 
for ($i=0; $i<$show; $i++) { 
echo "$count[$i]"; 
} 
echo "</div>\n"; 
?>

It is now displaying the keywords from my database like I want it to.
but what do I need to add so it will only display the top 5 keywords searched for. so that would mean i need to tell it to display the 5 keywords with the 5 highest number of hits. . .???

Aliz__

i can get it to display 5 or however many but how do i get it to display the top 5

darkside-x

Since you have ORDER BY x DESC, it will start at the highest and finish at the lowest values of quantity

This means that it is preordered for you, and therefore that the first 5 elements in $count are the top 5 searches.

So, I guess, if you display the first five, you've got your top 5.

Aliz__

awesome, Thanks for all the help!

darkside-x

rad!