[Resolved] Creating Array with info from database

[pipedreamer]

Hello.

I'll first just say that I always research heavily before asking a question - generally I find the answer that way and don't even need to post. But with this one I need a bit more help.

I am pulling numbers(weight) from a single column in the the database between a certain timeframe. I need to get these numbers into a variable array for use elsewhere. I can print the numbers during the while ($array = mysql_fetch_array($result)) { }, but printing them is not what I need, I need them in a named variable.

Following is my code:

............

mysql_select_db("$dbName", $db) or die ("Couldn't select the database.");
$sql = "SELECT weight FROM weight WHERE userID='$userID' AND date>='$date1' AND date<='$date' ORDER BY 'date' ASC" or die(mysql_error());
$result = mysql_query($sql);
while ($array = mysql_fetch_array($result)) {
$weight = $array["weight"];
print "$weight</br>";
}

...........

When I print the array $weight within the loop, I get:

128
126
125
(...etc)

Just as I should. But how do I get those numbers into an array that I can use outside of the {} tags?

Thanks in advance for the help. I get a little topsy turvy when working with arrays...

Jamie

[WinterDragon]

Ok try this...

#Create an empty array...
$weight = array();
mysql_select_db("$dbName", $db) or die ("Couldn't select the database."); 
$sql = "SELECT weight FROM weight WHERE userID='$userID' AND date>='$date1' AND date<='$date' ORDER BY 'date' ASC" or die(mysql_error()); 
$result = mysql_query($sql); 
while ($array = mysql_fetch_array($result)) { 
#Add the result onto the end of the array...
$weight[] = $array["weight"]; 
print "$weight</br>"; 
} 
for($i=0;$i<sizeof($weight);$i++) {
echo $weight[$i] . "<br>";
}

That should work!

[pipedreamer]

Excellent! It does work.

Thanks for the new perspective, WinterDragon, I don't think I would ever have tried it like that...

[WinterDragon]

No problem! Just ask if you need anymore help, I live here. 🙂 (just kidding, I live in Houston)