Can't insert data into mysql after user chooses from an arra
Posted: Mon Jul 05, 2010 7:24 am
Good afternoon to you all,
Please could someone kindly help me out with the below? Basically I'm pulling information back from mysql database in the form of football teams and then I've managed to put a drop down next to each of them in order for the user to choose the score. I then can't insert that value into my predictions table:
Basically zero values are being pushed across into the predictions table. Therefore connection is ok, but I'm obviously not doing something right
Thanking you in advance for any help you could give me.
Thanks,
Dave
Please could someone kindly help me out with the below? Basically I'm pulling information back from mysql database in the form of football teams and then I've managed to put a drop down next to each of them in order for the user to choose the score. I then can't insert that value into my predictions table:
Code: Select all
<?php
$goals = array ("0", "1", "2","3", "4", "5","6", "7", "8","9", "10");
$conn = mysql_connect("localhost","root","swarve") or die(mysql_error());
mysql_select_db("predictions");
$sql = processInput();
function processInput(){
//extract information from above
$WeekNo = $_POST["WeekNo"];
$sql = "SELECT name
FROM `fixtures_table`
INNER JOIN teams_table ON Home_Team_ID = Team_ID
WHERE WeekNo = '$WeekNo'
ORDER BY ID";
return $sql;
} // end processInput
$result = mysql_query($sql, $conn) or die(mysql_error());
while($row = mysql_fetch_assoc($result)){
foreach ($row as $name => $value){
print "$value";
echo "<select>";
foreach($goals as $goal)
{
echo "<option name='$goal'>$goal</option> <br />\n";
}
echo "</select>";
} // end foreach
print "<br> <br />\n";
} //end while
?>
I'm then trying to insert this into my predictions_table:
<?php
$conn = mysql_connect("localhost","root","swarve") or die(mysql_error());
mysql_select_db("predictions");
$GoalInsert = $_POST['$goal'];
$sql = "INSERT INTO predictions_table (,Home_Prediction,Away_Prediction)VALUES ('$GoalInsert','$GoalInsert')";
$result = mysql_query($sql, $conn) or die(mysql_error());
?>Thanking you in advance for any help you could give me.
Thanks,
Dave