What is wrong with my INSERT INTO sql statement?

[Xman51]

Hello All,

I don't understand what is wrong with my INSERT INTO sql statement. Does anyone see the error in my code?

<?php

Code: [Select]

$InsName= $_POST['pName'];
$InsRace= $_POST['pRace'];
$InsGender= $_POST['pGender'];
$InsEyes= $_POST['pEyes'];
$InsEyeLeftColor= $_POST['pEyeLeftColor'];
$InsEyeRightColor= $_POST['pEyeRightColor'];
$InsHeight= $_POST['pHeight'];
$InsWeight= $_POST['pWeight'];
$InsAge = $_POST['pAge'];
$InsHair= $_POST['pHair'];
$InsHairColor= $_POST['pHairColor'];
$InsSize= $_POST['pSize'];
$InsShirt= $_POST['pShirt'];
$InsShirtColor= $_POST['pShirtColor'];
$InsPants= $_POST['pPants'];
$InsPantsColor= $_POST['pPantsColor'];
$InsShoes= $_POST['pShoes'];
$InsShoesColor= $_POST['pShoesColor'];
$InsArmor= $_POST['pArmor'];
$InsArmorColor= $_POST['pArmorColor'];
$InsWeapon= $_POST['pWeapon'];
$InsHitPoints="10";
$InsAC="10";
$InsInitiative="0";
$InsExperience="0";
$InsLevel="0";


//Open the connection
$conn=odbc_connect('Combo','xman51','area51');
if (!$conn)
{
exit("Connection Failed: " . $conn);
}
$sql="INSERT INTO Members VALUES('$InsName', '$InsRace', '$InsEyes', '$InsHairColor', '$InsHair',

'$InsHeight', '$InsWeight', '$InsGender', '$InsAge', '$InsSize', '$InsShirt', '$InsShirtColor',

'$InsPants', '$InsPantsColor', '$InsShoes', '$InsShoesColor', '$InsHitPoints', '$InsArmor',

'$InsArmorColor', '$InsWeapon', '$InsAC', '$InsInitiative', '$InsExperience', '$InsLevel');";

echo $sql;
?>
Thanks in Advance,

Xman51

[Jorge Solis]

What's the error output? Probably quantity of data doesn't match quantity of columns

Jorge

[Xman51]

jorge,

There is no error output,..it simply does not input the values into the database. I was thinking that the values didn't match the quantity of columns either, but I double checked and it is correct.

I don't know what else to do...

Thanks for the reply, I look forward to your response,

Xman51

[Jorge Solis]

Copy&paste your insert into phpMyAdmin (or whatever you use to manage mysql) and check if it works

Jorge

[Musicman]

Hi,

looking t the code I find some
$sql = ....
followed by
echo $sql
Should not there be some statement to execute the sql, too?

Musicman