Recalling random values from mysql table

Question

I'm fairly new to php and mysql and I'm on the home stretch of finishing my page but I've been banging my head on the keyboard all day trying to figure out how to fix this problem. I've set up a php script to run as a cron even every 24 hours. The script assigns a random number between 10 and 30 to each field in my table. That works fine and every time I load the script the values change.

The problem I'm having is when I try to use those values. The result keeps printing as the word Array instead of the number in the table. So I'll give you some snippets of the code I'm running here. This is the cron event.

?php
$con = mysql_connect("localhost","username","pass");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}

mysql_select_db("dbname", $con);

$random= rand(10, 30);

mysql_query("UPDATE winners SET pool= '$random'");

mysql_close($con);

And here is the script to call up the values.

php?
$db = mysql_connect('localhost','username','pass') or die("Database error");
mysql_select_db('dbname', $db);


$query = "SELECT pool FROM winners";
$result = mysql_query($query) or die(mysql_error());

while ($row = mysql_fetch_array($result)) 

if ( $row % 2 )
{ 
echo "<h4>Result 1</h4>";
echo "$row";
echo "<br />";
}
else
{ 
echo "<h4>Result 2</h4>";
echo "<br />";
}

I've tried every possible variation I can think of to this code but all I can get is it to echo either Array or Resource #9. Any insight into this would be greatly appreciated.

Community · Accepted Answer · 2023-11-17 19:42:58Z

1

You do not want to echo the content of $row, which contains all data returned for the current line you've fetched from the database when calling mysql_fetch_array().

Instead, you want to access the content of $row's pool item :

echo $row['pool'];

You should probably take a closer look at the manual page for mysql_fetch_array(), and the examples it contains.

Note that you'll probably also want to modify the condition in the following line :

if ( $row % 2 )

You probably don't want the test to be done on $row, but on an item it contains.

edited Nov 17, 2023 at 19:42

CommunityBot

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answered Feb 13, 2012 at 7:09

Pascal MARTIN

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2 Comments

user1206214 Over a year ago

Thanks that actually fixed the array problem but I'm still having a problem getting the proper results from the if statement. I tried changing it to if ( $row['pool'] % 2 ) but it doesnt seem to be working. The if statement there is querying whether the value is divisible by 2 right?

user1206214 Over a year ago

There is only one row in the table, this should be really easy, I guess I've been staring at a screen for too long. I'm trying to test is the value in column 'pool' is divisible by 2 and it keeps telling me that 23 is divisible by 2.

Naveen Kumar · Accepted Answer · 2012-02-13 07:58:44Z

0

$query = "SELECT pool FROM winners";

$result = mysql_query($query) or die(mysql_error());
while ($row = mysql_fetch_array($result)) 
{
if ( $row['pool'] % 2 )
{ 
echo "<h4>Result 1</h4>";
echo "$row['pool']";
echo "<br />";
}
else
{ 
echo "<h4>Result 2</h4>";
echo "<br />";
}
}

Hope this helps.

answered Feb 13, 2012 at 7:58

Naveen Kumar

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1 Comment

user1206214 Over a year ago

Well, that works. I've tried it before too I just confused where I put result 1 and result 2. Time for me to go to bed and thanks for the help everyone

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