pingeyeg
asked on
Update statement causing the Resource #3 to display
I have created an update statement, but for some reason I am getting the Resource #3 display when I don't feel I should be. I know that this message is wanting the mysql_fetch_array function, but I don't feel I need it.
$conn = mysql_connect("localhost", "username", "password") or die(mysql_error());
mysql_select_db("353322_photos") or die(mysql_error());
if($_POST['Submit']) {
$id = $_POST['ID'];
$pA = $_POST['packageA'];
$pB = $_POST['packageB'];
$pC = $_POST['packageC'];
$printA = $_POST['8x10'];
$printB = $_POST['5x7'];
$printC = $_POST['4x6'];
$wallet = $_POST['wallet'];
$update = "update pricing set packageA = '$pA', packageB = '$pB', packageC = '$pC', 8x10 = '$printA', 5x7 = '$printB', 4x6 = '$printC', wallet = '$wallet' where ID = '$id'";
$result = mysql_query($update) or die(mysql_error());
if($_POST['Submit']) {
$result = "Your pricing changes have now been made and have reflected in the cart.";
} else {
$result = "There seems to have been an issue with your update. Please contact <a href=\"mailto:brannon@goodboyweb.com\">Goodboy</a> for further assistance.";
}
}
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ASKER
I'm still getting: Resource id #3
Is there more code? What I gave is the "fix" to what you posted.
ASKER
I must not have refreshed the page correctly because now I get a new error:
You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ''packageA' = '20.00', 'packageB' = '24.00', 'packageC' = '18.00', '8x10' = '12.0' at line 1
There is more code on the page, but the other code doesn't have anything to do with this update statement.
You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ''packageA' = '20.00', 'packageB' = '24.00', 'packageC' = '18.00', '8x10' = '12.0' at line 1
There is more code on the page, but the other code doesn't have anything to do with this update statement.
if($_POST['Submit']) {
$id = $_POST['ID'];
$pA = $_POST['packageA'];
$pB = $_POST['packageB'];
$pC = $_POST['packageC'];
$printA = $_POST['8x10'];
$printB = $_POST['5x7'];
$printC = $_POST['4x6'];
$wallet = $_POST['wallet'];
$update = "update pricing set 'packageA' = '$pA', 'packageB' = '$pB', 'packageC' = '$pC', '8x10' = '$printA', '5x7' = '$printB', '4x6' = '$printC', 'wallet' = '$wallet' where ID = '$id'";
mysql_query($update) or die(mysql_error());
if(mysql_affected_rows() > 0) {
$result = "Your pricing changes have now been made and have reflected in the cart.";
} else {
$result = "There seems to have been an issue with your update. Please contact <a href=\"mailto:email.com\">Company</a> for further assistance.";
}
}
You have apostrophes around your fields names:
...'packageB' =...
I do NOT. You need backtick ( the character underneath the ~ character ), not apostrophe. Copy and paste from my post.
...'packageB' =...
I do NOT. You need backtick ( the character underneath the ~ character ), not apostrophe. Copy and paste from my post.
ASKER
Still getting the same thing. Resource id #3
ASKER
What's weird is the fact that the page IS updating yet I am still getting Resource id #3 instead of the $result.
ASKER
Well I finally figured it out. I was already using that variable $result as my result for another area so apparently it was getting confused.
:)