help with mysql

corruption

I'm trying to begin a database from my fraternity. I'm having a few problems even though i've read through the stuff at php.net and mysql's website i still can not input data into the database.

<-------------------------------index.php---------------------------------->

<html>
<head>
<title>Member Registration</title>
</head>
<body>

<table>
<form action="register.php" name="registration" method="post">
<tr><td>First Name</td>
<td><input type="text" name="fname" size="25"></td></tr>
<tr><td>Last Name</td>
<td><input type="text" name="lname" size="25"></td></tr>
<tr><td>Username</td>
<td><input type="text" name="uname" size="25"></td></tr>
<tr><td>Password</td>
<td><input type="password" name="pword" size="25"></td></tr>

<tr><td><input type="submit" value="Register"></td></tr>
</form>
</table>
</body>
</html>

and the problem lies herein as the connection script works and i can post the values to the page as the script is correct at the bottom of this next page
<----------------------------------register.php------------------------------>

<?php
require_once 'conn.php';

INSERT INTO test ('fname', 'lname', 'uname', 'pword')
            VALUES ($fname, $lname, $uname, $pword);


echo $_POST['fname'];
print ("<p>");
echo $_POST['lname'];
print ("<p>");
echo $_POST['uname']; 
print ("<p>");
echo $_POST['pword']; 
?>

KevinMG

try this...

$sql = "INSERT INTO `test`(`fname`,`lname`,`uname`,`pword`) VALUES ('$fname','$lname','$uname','$pword')";
$result = mysql_query($sql);
if(!$result) {
echo mysql_error();
} else {
echo "Success...";
}

corruption

Thanks for the help- new problem now
the script worked well i'll have to store that one away.
However the script did not enter any information into the database. The database knows there should be a new record but it is blank when i view it with phpmyadmin.

Actually i just noticed that it doesn't post the last field.

try here
http://www2.truman.edu/~d1514

cgraz

Your variables are probably empty. Try using globals.

Instead of $fname, use $_POST["fname"]

Instead of $lname, use $_POST["lname"]

You can even doublecheck to see if your variables have values by printing $sql.

Cgraz

The_Chancer

I have come across a couple of irregularities when inputting and now use

$sql= ....... VALUES ('".$_POST['foo']."','".$_POST['bar']."')

Which works for both M$SQL and MySQL

corruption

each of these are good suggestions. But they produce the same results. A new row of empty data containers. I tried putting in
print($sql); in order to print that piece of information but it didn't print. But when i use echo to echo each variable on its own it prints properly.

The_Chancer

Works just fine for me...

$sql = "INSERT INTO `test`('fname','lname','uname','pword') VALUES ('".$_POST['fname']."','".$_POST['lname']."','".$_POST['uname']."','".$_POST['pword']."')";
$result = mysql_query($sql) or die (mysql_error);
if(mysql_affected_rows($result)>1) {
echo "Success... with ".$sql;
}
else {
echo "No success with ".$sql;
}

corruption

unfortunately this still posts a blank entry.

<----------------------------------register.php------------------------------->

<?php
require_once 'conn.php';


//enters blank values still

$sql = "INSERT INTO `test`('fname','lname','uname','pword','address','phone','number') 
                   VALUES ('".$_POST['fname']."','".$_POST['lname']."','".$_POST['uname']."','".$_POST['pword']."','".$_POST['address']."','".$_POST['phone']."','".$_POST['number'])"; 
$result = mysql_query($sql) or die (mysql_error); 
if(mysql_affected_rows($result)>1) { 
echo "Success... with ".$sql; 
} 
else { 
echo "No success with ".$sql; 
}

//don't mess with this it works

$result = mysql_query($sql);

if(!$result) { 
echo mysql_error(); 
}
 else { 
echo "Success..."; 
}

//This part works, don't mess with it

print ("<p>");
echo $_POST['fname'];
print ("<p>");
echo $_POST['lname'];
print ("<p>");
echo $_POST['uname']; 
print ("<p>");
echo $_POST['pword'];
print ("<p>");
echo $_POST['address'];
print ("<p>");
echo $_POST['phone'];
print ("<p>");
echo $_POST['number']; 
?>

cgraz

print $sql . "<br>";

Copy and paste the output of that.

Cgraz

corruption

i'd copy and paste something but there isn't anything to copy and paste

cgraz

Then you did something wrong.

$sql = "INSERT INTO  test('fname','lname','uname','pword','address','phone','number') VALUES ('".$_POST['fname']."','".$_POST['lname']."','".$_POST['uname']."','".$_POST['pword']."','".$_POST['address']."','".$_POST['phone']."','".$_POST['number'])"; 

print $sql . "<br>";

Even if your variables are empty, it would atleast print

INSERT INTO test ('fname','lname','uname','pword','address','phone','number') VALUES ('','','','','','','',)

Cgraz

corruption

www2.truman.edu/~d1514
gives no response

<?php

$sql_=_"INSERT INTO___test('fname','lname','uname','pword','address','ph
one','number') VALUES ('".$_POST['fname']."','".$_POST['lname']."','".$_POST['uname']."','".$_POST['pword']."','".$_POST['address']."','".$_POST['phone']."','".$_POST['number']."')";
print $sql . "<br>";

$result = mysql_query($sql) or die (mysql_error); 
if(mysql_affected_rows($result)>1) { 
echo "Success... with ".$sql; 
} 
else { 
echo "No success with ".$sql; 
}


?>

corruption

ok so i changed some things a little bit here's what i got.

INSERT INTO___test('fname','lname','uname','pword','address','phone','number') VALUES ('ron','nelson','corruption','outlaws','119 1/2 W Washington','349-3933','929')
mysql_error

corruption

yet some more changes.

here's the new code

<?php
include 'conn.php';

// Define post fields into simple variables
$fname = $_POST['fname'];
$lname = $_POST['lname'];
$uname = $_POST['uname'];
$pword = $_POST['pword'];
$address = $_POST['address'];
$phone = $_POST['phone'];
$number = $_POST['number'];

/* Let's strip some slashes in case the user entered
any escaped characters. */

$fname = stripslashes($fname);
$lname = stripslashes($lname);
$uname = stripslashes($uname);
$pword = stripslashes($pword);
$address = stripslashes($address);
$phone = stripslashes($phone);
$number = stripslashes($number);


// Enter info into the Database.
$sql = mysql_query("INSERT INTO test (fname, lname, uname, pword, address, phone, number)
		VALUES('$fname', '$lname', '$uname', '$pword', '$address', '$phone', '$number')") or die (mysql_error());
print($sql);
//don't fuck with this it works
$result = mysql_query($sql); 
if(!$result) { 
echo mysql_error(); 
} else { 
echo "Success..."; 
}
//This part works, don't fuck with it
print ("<p>");
echo $_POST['fname'];
print ("<p>");
echo $_POST['lname'];
print ("<p>");
echo $_POST['uname']; 
print ("<p>");
echo $_POST['pword'];
print ("<p>");
echo $_POST['address'];
print ("<p>");
echo $_POST['phone'];
print ("<p>");
echo $_POST['number']; 
?>

and the output looks like this

1You have an error in your SQL syntax. Check the manual that corresponds to your MySQL server version for the right syntax to use near '1' at line 1

ron

nelson

corruption

outlaws

119 1/2 w washington

349-3933

929

corruption

okay so that code inserts the data but still has a mysql error what should i do? should i worry about the mysql error or what?