[SOLVED] Problem with "SELECT"

[R1der]

I am not sure if this is to do with the "SELECT" query but its only returning 1 result from the database. What have i done wrong? lol

 

$get_horse = mysql_query("SELECT * FROM hracing ORDER BY id");

$horses = mysql_fetch_array($get_horse);

echo"<br><Br>";
  echo "<table width=100%>";
                echo "<tr>";
                echo "<td width=30%><b>Horse</b></td>";
                echo "<td width=30%><b>Odds</b></td>";
                echo "<td width=30%><b>Bet Cost</b></td>"; 
                echo "<td width=30%><b>Bet</b></td>";       
                echo "</tr>";



echo "<td>$horses[name]</td>";
echo "<td>$horses[odds]</td>";
echo "<td>$horses[cost]</td>";
echo "<td><a href=hracing.php?action=$horses[name]>Bet</a></td>";
echo "</table>";
include("bottom.php");

 

Thanks for your time

[lur]

mysql_fetch_array() fetches one result row, to iterate over all the results:

 

while ($horse = mysql_fetch_array($get_horse)) {
print_r($horse);
}

[redarrow]

change array to assoc that it.

[Psycho]

mysql_fetch_array() only selects ONE record at a time from a query result. You need to create a loop to grap each record in order:

 

<?php

echo"<br><Br>";
echo "<table width=100%>";

echo "<tr>";
echo "<td width=30%><b>Horse</b></td>";
echo "<td width=30%><b>Odds</b></td>";
echo "<td width=30%><b>Bet Cost</b></td>"; 
echo "<td width=30%><b>Bet</b></td>";       
echo "</tr>";

$get_horse = mysql_query("SELECT * FROM hracing ORDER BY id");

while ($horses = mysql_fetch_array($get_horse)) {

   echo "<tr>";
   echo "<td>$horses[name]</td>";
   echo "<td>$horses[odds]</td>";
   echo "<td>$horses[cost]</td>";
   echo "<td><a href=hracing.php?action=$horses[name]>Bet</a></td>";
   echo "</tr>";

}

echo "</table>";
include("bottom.php");

?>

[R1der]

Thanks guys your the best