PHP - Problem Double Insert

I have problem with this code. It does absolutely nothing. When INSERT is over it should redirect to index.php but it does nothing. There is no error, when the submit is clicked the page just refresh itself and because of echo function it write all the values. What seems to be the problem (I going slightly mad )

Code: [Select]
<?php require_once("public/includes/session.php"); ?>
<?php require_once("public/includes/connection.php"); ?>
<?php require_once("public/includes/functions.php"); ?>
<?php include_once("public/includes/form_functions.php");
   include_once("public/includes/header.php");


if (isset($_POST['submit'])) { // Form has been submitted.
$errors = array();


$required_fields = array('nik', 'lozinka', 'ime', 'prezime', 'adresa', 'grad', 'postanskiBroj', 'fiskni', 'moblini', 'email');
$errors = array_merge($errors, check_required_fields($required_fields, $_POST));
$username = trim(mysql_prep($_POST['nik']));
$password = trim(mysql_prep($_POST['lozinka']));
$hashed_password = sha1($password);
$ime = trim(mysql_prep($_POST['ime']));
$prezime = trim(mysql_prep($_POST['prezime']));
$adresa = trim(mysql_prep($_POST['adresa']));
$grad = trim(mysql_prep($_POST['grad']));
$postanskiBroj = trim(mysql_prep($_POST['postanskiBroj']));
$fiskni = trim(mysql_prep($_POST['fiksni']));
$moblini = trim(mysql_prep($_POST['mobilni']));
$email = trim(mysql_prep($_POST['email']));

echo $username . $hashed_password . $ime . $prezime . $adresa . $grad . $postanskiBroj . $fiskni . $moblini . $email;

if ( empty($errors) ) {
$query = " INSERT INTO `gume`.`korisnik` (`id`, `korisnicko_ime`, `lozinka`, `ime`, `prezime`, `adresa`, `grad`, `postanskiBroj`, `fiksni_telefon`, `mobilni_telefon`, `email`)
  VALUES
  (NULL, '$username', '$hashed_password', '$ime', '$prezime', '$adresa', '$grad', '$postanskiBroj', '$fiskni, $moblini', '$email' )";
$result = mysql_query($query, $connection) or die(mysql_error);
if ($result) {
redirect_to("index.php");
} else {
$message = "The user could not be created.";
$message .= "<br />" . mysql_error();
}
} else {
if (count($errors) == 1) {
$message = "There was 1 error in the form.";
} else {
$message = "There were " . count($errors) . " errors in the form.";
}
}
} else { // Form has not been submitted.
$username = "";
$password = "";
}
?>

<div id="telo">
   <div id="kreiranjeNaloga">
<script src="SpryAssets/SpryValidationTextField.js" type="text/javascript"></script>
<link href="SpryAssets/SpryValidationTextField.css" rel="stylesheet" type="text/css" />
<script src="SpryAssets/SpryValidationPassword.js" type="text/javascript"></script>
<script src="SpryAssets/SpryValidationConfirm.js" type="text/javascript"></script>
<link href="SpryAssets/SpryValidationPassword.css" rel="stylesheet" type="text/css" />
<link href="SpryAssets/SpryValidationConfirm.css" rel="stylesheet" type="text/css" />
<p>Polja sa * su obavezna</p>
<form action="new_user.php" method="post">
  <span id="sprytextfield1">
  <label>Korisnicko ime: </label>
  <input type="text" name="nik" id="nik"  size="40" value=""/>
  *<span class="textfieldMinCharsMsg">Korisnicko ime ne moze imati manje od 5 karaktera</span><span class="textfieldMaxCharsMsg">Korisnicko ime moze imati najvise 30 karaktera.</span></span><br />
  <span id="sprypassword1">
  <label>Lozinka:</label>
    <input type="password" name="lozinka" id="lozinka"  size="40" value=""/>
  *<span class="passwordMinCharsMsg">Sifra mora sadrzati najmanje 5 karaktera.</span><span class="passwordMaxCharsMsg">Sifra moze imati najvise 30 karaktera.</span></span>

  <br />
  <span id="spryconfirm1">
  <label>Potvrdite lozinku:</label>
    <input type="password" name="password1" id="password1" size="40" value=""/>
 
  <span class="confirmRequiredMsg">*</span>Obe lozinke moraju da budu iste.</span>
  <br />
  <span id="sprytextfield2">
  <label>Ime:</label>
    <input type="text" name="ime" id="ime" size="40" value=""/>
 
  *
  </span>
  <br />
  <span id="sprytextfield3">
  <label>Prezima</label>
    <input type="text" name="prezime" id="prezime"size="40" value="" />
  *</span>
  <br />
  <span id="sprytextfield4">
  <label>Adresa:</label>
    <input type="text" name="adresa" id="adresa" size="40" value=""/>
    *
 
  </span>
  <br />
  <span id="sprytextfield7">
  <label>Grad:</label>
    <input type="text" name="grad" id="grad" size="40" value="" />
    *
 
  </span> <br />
 
  </span><span id="sprytextfield9">
  <label>Postanski Broj: </label>
    <input type="text" name="postanskiBroj" id="postanskiBroj" size="10" value=""/>
    *
 
  <span class="textfieldInvalidFormatMsg">Postanski broj nije pravilno upisan.</span></span><br />
  <span id="sprytextfield5">
  <label>Broj fiksnog telefona: </label>
  <input type="text" name="fiksni" id="Broj fiksnog telefona" size="40" value="" />
  *<span class="textfieldInvalidFormatMsg">Broj telefona nije pravilno upisan</span></span>
  <br />
  <span id="sprytextfield6">
  <label>Broj mobilnog telefona: </label>
  <input type="text" name="mobilni" id="mobilni" size="40" value=""  />
 *<span class="textfieldInvalidFormatMsg">Broj telefona nije pravilno upisan</span></span><br />
  <span id="sprytextfield10">
  <label>Email:</label>
  <input type="text" name="email" id="email" size="40" value="" />
  *<span class="textfieldInvalidFormatMsg">Email adresa nija pravilno upisana.</span></span><br />
  <input name="submit" type="submit" id="submit" value="Kreiraj korisnika" />
</form>

  </div>
</div>
<?php include("public/includes/footer.php"); ?>

I have done a simple insert query on a comment system but it works once and when i go back and try add another record it just doesnt work at all.

Code: [Select]
<?php 

include ("class.database.connection.php");   // the database connection class
include ("settings.config.php");  // hostname, password ect

$name = $_POST["UserName"];
$email = $_POST["UserEmail"];
$comment = $_POST["UserComment"];

$sql="INSERT INTO comments (name, email, comment)VALUES('$name', '$email', '$comment')";
$result=mysql_query($sql);


?>
if anyone can help me with this i would be a happy chappy!

This php file cant echo in the 2nd php code, can anyone help me about this?
1st php code to connect into the 2nd php code
Code: [Select]
<?php
$Total = $Total + $Amount;
}
?>

</tr>
<tr>
    <td colspan="3" align="Right">Total</td>
    <td align ="Center"><?php echo number_format($Total,2);?></td>
</tr>
2nd php code
Code: [Select]
<?php
$db = mysql_connect("localhost", "root", "");
mysql_select_db("vinnex",$db);
$TransNo = $_POST['TransNo'];
$Username = $_POST['Username'];
$Date = $_POST['Date'];
$Total = $_GET['Total'];
echo $Total;
$sqltransaction = " INSERT INTO transaction (TransNo, Username, Date, Total)
VALUES ('$TransNo', '$Username', '$Date', '$Total')";
$resulttransaction = mysql_query($sqltransaction);
?>

Hey guys for some reason this code is not working i can't see a problem myself could someone please have a look and point the issue out to me. what i mean by it not working is it won't insert into the database or show a mysql_error.

thanks in advance

Code: [Select]
$guestip = $_SERVER['REMOTE_ADDR'];
$time = date('G:i');
$date = date("y-m-d");

$query = mysql_query("SELECT * FROM IP_Address") or die(mysql_error());
while($row = mysql_fetch_assoc($query)){
   
    if($guestip != $row['ip']){
        //insert into db.
        mysql_query("INSERT INTO IP_Address(id, ip, date, time) VALUES(NULL,'$questip','$date','$time')") or die(mysql_error());
        echo "inserted in to database"; echo mysql_error();
    }else{
        // add hit count and update time and date.
        echo "already in db";
    }
}

hi,
i was having difficulty and i'm quite confused how to insert these values to the database before it will become as it is, based on this article:
http://roshanbh.com.np/2008/01/populate-triple-drop-down-list-change-options-value-from-database-using-ajax-and-php.html

what i did was putting/saving the values directly in the database (mysql). but now i want those values to be coming from the user's input:the country,state and city so the system will just get those values, store it to the database with the fields of each table having the same content like that from article above..i really need some help here..

Hi,

I am trying to insert the contents of a csv file into a table, this is my code:


public function InsertCSVFileToDB(){

$has_title_row = true;
$not_done = array();

        if(is_uploaded_file($_FILES['csvfile']['tmp_name'])){
            $filename = basename($_FILES['csvfile']['name']);

            if(substr($filename, -3) == 'csv'){
                $tmpfile = $_FILES['csvfile']['tmp_name'];
                if (($fh = fopen($tmpfile, "r")) !== FALSE) {
                    $i = 0;
                    while (($items = fgetcsv($fh, 10000, ",")) !== FALSE) {

                        if($has_title_row === true && $i == 0){ // skip the first row if there is a tile row in CSV file
                            $i++;
                            continue;
                        }

                        $sql = "INSERT INTO ConfPaper SET
                                CPRid = ".$items[0].",
                                Pid = ".$items[1].",
                                CPtitle = '".mysql_real_escape_string($items[2])."',
                                CPabstract = '".mysql_real_escape_string($items[3])."',
                                CPspage = ".mysql_real_escape_string($items[4]).",
                                CPepage = ".mysql_real_escape_string($items[5]).",
                                CPlastesited = now()";
                        if(!mysql_query($sql)){
                            $not_done[] = $items;
                        }
                        $i++;
                    }
                }
                // if there are any not done records found:
                if(!empty($not_done)){
                    echo "<strong>There are some records could not be inserted</strong><br />";
                    print_r($not_done);
                }
            }
            else{
                die('Invalid file format uploaded. Please upload CSV.');
            }
        }
        else{
            die('Please upload a CSV file.');
        }
    }


This is the csv file:

http://www.prima.cse.salford.ac.uk:8080/~ibrarhussain/ConfPaper.csv

But i keep getting this:

Quote

Array ( => Array ( => 9 [1] => 1 [2] => CSV1 [3] => 4 [4] => 4 [5] => 01625 584412 ) [1] => Array ( => 9 [1] => 1 [2] => CSV2 [3] => 14 [4] => 24 [5] => 01625 584412 ) )

Any ideas what the problem might be?

Hope someone can help..

Thanks

Hi

My problem is that I can't insert my information into my database. I was able to insert the information into my database last night but when I tried to use validation it's not working any more.

Here is my code for the sign up
Code: [Select]
<?php
require_once("validation.php");
?>

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml" dir="ltr">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8" />
<title>yensdesign.com - Validate Forms using PHP and jQuery</title>
<link rel="stylesheet" href="css/general.css" type="text/css" media="screen" />
</head>
<body>
<a id="logo" title="Go to yensdesign.com!" href="http://www.yensdesign.com"><img src="css/images/logo.jpg" alt="yensdesign.com" /></a>
<div id="container">
<h1>Registration process</h1>

<?if( isset($_POST['send']) && (!validateName($_POST['name']) || !validateEmail($_POST['email']) || !validatePasswords($_POST['pass1'], $_POST['pass2']) || !validateMessage($_POST['message']) ) ):?>
<div id="error">
<ul>
<?if(!validateName($_POST['name'])):?>
<li><strong>Invalid Name:</strong> We want names with more than 3 letters!</li>
<?endif?>
<?if(!validateEmail($_POST['email'])):?>
<li><strong>Invalid E-mail:</strong> Stop cowboy! Type a valid e-mail please :P</li>
<?endif?>
<?if(!validatePasswords($_POST['pass1'], $_POST['pass2'])):?>
<li><strong>Passwords are invalid:</strong> Passwords doesn't match or are invalid!</li>
<?endif?>
<?if(!validateMessage($_POST['message'])):?>
<li><strong>Ivalid message:</strong> Type a message with at least with 10 letters</li>
<?endif?>
</ul>
</div>
<?elseif(isset($_POST['send'])):?>
<div id="error" class="valid">
<ul>
<li><strong>Congratulations!</strong> All fields are OK ;)</li>
</ul>
</div>
<?endif?>

<form method="post" id="customForm" action="index.php">
<div>
Name
<input id="name" name="name" type="text" />
<!--<span id="nameInfo">What's your name?</span>-->
</div>
<div>
E-mail
<input id="email" name="email" type="text" />
<!--<span id="emailInfo">Valid E-mail please, you will need it to log in!</span>-->
</div>
<div>
Password
<input id="pass1" name="password" type="password" />
<span id="pass1Info">At least 5 characters: letters, numbers and '_'</span>
</div>
<div>
Confirm Password
<input id="pass2" name="pass2" type="password" />
<span id="pass2Info">Confirm password</span>
</div>
<!--<div>
<label for="message">Message</label>
<textarea id="message" name="message" cols="" rows=""></textarea>
</div>-->
<div>
<input id="send"  type="submit" value="Send" />
</div>
</form>
</div>
<script type="text/javascript" src="jquery.js"></script>
<script type="text/javascript" src="validation.js"></script>
</body>
</html>
<?php

$host="localhost"; // Host name 
$username=""; // Mysql username 
$password=""; // Mysql password 
$db_name="test"; // Database name 
$tbl_name="emails"; // Table name

// Connect to server and select database.
mysql_connect("$host", "$username", "$password")or die("cannot connect"); 
mysql_select_db("$db_name")or die("cannot select DB");

// Get values from form 
$name=$_POST['name'];
$email=$_POST['email'];
$password=$_POST['password'];

// Insert data into mysql 
$sql="INSERT INTO $tbl_name(name, email, password)VALUES('$name', '$email', 'password')";
$result=mysql_query($sql);

// if successfully insert data into database, displays message "Successful". 
if($result){
echo "Successful";
echo "<BR>";
echo "<a href='insert.php'>Back to main page</a>";
}

else {
echo "ERROR";
}

// close connection 
mysql_close();
?>

This is the validation.php
Code: [Select]
<?php
function validateName($name){
//if it's NOT valid
if(strlen($name) < 4)
return false;
//if it's valid
else
return true;
}
function validateEmail($email){
return ereg("^[a-zA-Z0-9]+[a-zA-Z0-9_-]+@[a-zA-Z0-9]+[a-zA-Z0-9.-]+[a-zA-Z0-9]+.[a-z]{2,4}$", $email);
}
function validatePasswords($pass1, $pass2) {
//if DOESN'T MATCH
if(strpos($pass1, ' ') !== false)
return false;
//if are valid
return $pass1 == $pass2 && strlen($pass1) > 5;
}
function validateMessage($message){
//if it's NOT valid
if(strlen($message) < 10)
return false;
//if it's valid
else
return true;
}
?>

And this is the validation.js
Code: [Select]
/***************************/
//@Author: Adrian "yEnS" Mato Gondelle & Ivan Guardado Castro
//@website: www.yensdesign.com
//@email: yensamg@gmail.com
//@license: Feel free to use it, but keep this credits please!
/***************************/

$(document).ready(function(){
//global vars
var form = $("#customForm");
var name = $("#name");
var nameInfo = $("#nameInfo");
var email = $("#email");
var emailInfo = $("#emailInfo");
var pass1 = $("#pass1");
var pass1Info = $("#pass1Info");
var pass2 = $("#pass2");
var pass2Info = $("#pass2Info");
var message = $("#message");

//On blur
name.blur(validateName);
email.blur(validateEmail);
pass1.blur(validatePass1);
pass2.blur(validatePass2);
//On key press
name.keyup(validateName);
pass1.keyup(validatePass1);
pass2.keyup(validatePass2);
message.keyup(validateMessage);
//On Submitting
form.submit(function(){
if(validateName() & validateEmail() & validatePass1() & validatePass2() & validateMessage())
return true
else
return false;
});

//validation functions
function validateEmail(){
//testing regular expression
var a = $("#email").val();
var filter = /^[a-zA-Z0-9]+[a-zA-Z0-9_.-]+[a-zA-Z0-9_-]+@[a-zA-Z0-9]+[a-zA-Z0-9.-]+[a-zA-Z0-9]+.[a-z]{2,4}$/;
//if it's valid email
if(filter.test(a)){
email.removeClass("error");
emailInfo.text("Valid E-mail please, you will need it to log in!");
emailInfo.removeClass("error");
return true;
}
//if it's NOT valid
else{
email.addClass("error");
emailInfo.text("Stop cowboy! Type a valid e-mail please :P");
emailInfo.addClass("error");
return false;
}
}
function validateName(){
//if it's NOT valid
if(name.val().length < 4){
name.addClass("error");
nameInfo.text("We want names with more than 3 letters!");
nameInfo.addClass("error");
return false;
}
//if it's valid
else{
name.removeClass("error");
nameInfo.text("What's your name?");
nameInfo.removeClass("error");
return true;
}
}
function validatePass1(){
var a = $("#password1");
var b = $("#password2");

//it's NOT valid
if(pass1.val().length <5){
pass1.addClass("error");
pass1Info.text("Ey! Remember: At least 5 characters: letters, numbers and '_'");
pass1Info.addClass("error");
return false;
}
//it's valid
else{
pass1.removeClass("error");
pass1Info.text("At least 5 characters: letters, numbers and '_'");
pass1Info.removeClass("error");
validatePass2();
return true;
}
}
function validatePass2(){
var a = $("#password1");
var b = $("#password2");
//are NOT valid
if( pass1.val() != pass2.val() ){
pass2.addClass("error");
pass2Info.text("Passwords doesn't match!");
pass2Info.addClass("error");
return false;
}
//are valid
else{
pass2.removeClass("error");
pass2Info.text("Confirm password");
pass2Info.removeClass("error");
return true;
}
}
function validateMessage(){
//it's NOT valid
if(message.val().length < 10){
message.addClass("error");
return false;
}
//it's valid
else{
message.removeClass("error");
return true;
}
}
});

I also get that the name, email and password is undefined

I have an old site written for PHP 5.4 and under and trying (very trying) to get it to work with PHP 7x without much luck. Due to all the changes in 7 my code is one big error message, but one thing at a time. I cannot get the follow code to work at all, even though it worked in PHP 5.

Error:

QUERY 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 'viewuser.php?u=666' id='member'>THE PREDATOR [666] was added to the hit' at line 1
The Query was INSERT INTO gangevents VALUES('','20', UNIX_TIMESTAMP(),'THE  PREDATOR [666] was added to the hit list.')

I have tried at least 20+ different ways of doing this but just can't get the right syntax to get it inserted into MySQL, the code below is just the latest version.  If I echo the a href line out, it works perfect. 

I am sure it is something ridiculously simple, but I have been 4 hours and counting on this now.

Thanks

gangevent_add_2($gangdata['gangID'], "<a href='viewuser.php?u=".$r['userid']."' ".$csscode[$r['userlevel']-1].">".$r['username']."</a> [".$r['userid']."] was added to your hitlist");

function gangevent_add_2($gang, $text) { 
	global $db; $csscode; 
	$db->query("UPDATE users SET gangevent = gangevent + 1 WHERE gang={$gang}");
	$db->query("INSERT INTO gangevents VALUES('','$gang', UNIX_TIMESTAMP(),'$text')");
}

 

Hi...

Before I have no problem in using On Duplicate Key, but now i have because of using time function.

Before I only have insert query:

Code: [Select]
$result = mysql_query("INSERT INTO regular_sum_hours(EMP_NO, Hours) SELECT EMP_NO, sec_to_time(SUM(time_to_sec(Rendered))) FROM regular_dtr_total GROUP BY EMP_NO") or die(mysql_error());   

And now I think that I need to add a syntax for Update, so i revise my code:
Code: [Select]
$result = mysql_query("INSERT INTO regular_sum_hours(EMP_NO, Hours) SELECT EMP_NO, sec_to_time(SUM(time_to_sec(Rendered))) FROM regular_dtr_total GROUP BY EMP_NO ON DUPLICATE KEY EMP_NO = EMP_NO, Hours = sec_to_time(SUM(time_to_sec(Rendered)))") or die(mysql_error());

and I got a problem in this part:

 Hours = sec_to_time(SUM(time_to_sec(Rendered)))")


Thank you

Hello,

     I am trying phpfreaks out without much luck here. If anybody here truly knows how to validate checkboxes as a group please give some assistance. I had to remove the array from the 'name' attribute in my checkbox inputs in order to get the user's checked selections to be properly distributed into separate database columns (not all in one column as a string of text). I am now able to successfully get the entries into the database as "1" for selected and "0" for un-selected. Now however, since there is no array that is holding these input names my validation class method does not work. My validation is suppose to show an error if none of these checkboxes are selected.

      I had to change each input name to its own identity ( fruit_selection_apple, etc... ), then set my validation like (see code below). However, this is NOT OOP PHP. I had to hard code these values into the validation class. I am also now getting a notice, " Notice: Undefined index: fruit_selection ". Does anyone legitimately know their PHP enough to help with this issue?

public function check($source, $items = array()) {
			foreach($items as $item => $rules) {
				foreach($rules as $rule => $rule_value) {
										
					$value = trim($source[$item]);					
					$item = escape($item);

               $checkboxvalue =                 
                        (isset($_POST['fruit_selection_apple'])) ||
                        (isset($_POST['fruit_selection_orange'])) ||
                        (isset($_POST['fruit_selection_banana'])) ||
                        (isset($_POST['fruit_selection_pear'])) ||                        
                        (isset($_POST['fruit_selection_kiwi'])
                    )                              
                    if($rule === 'atleastone' && empty($checkboxvalue)) {
                        $this->addError("{$item} You must select at least one checkbox.");
                    }

i have created my own code of custom shopping cart   i have viewcart.php working great, now when i want to insert the orders from viewcart.php with list of like 5 items, how can i insert 5 names of products into my database 1 row   example   names are 1. jean 2. mond 3. richard 4. gwen   list above is the results of my while loop, now i want to insert those names to my database column[order_productname] so that i can identity what products are paid by my clients.   i tried fetch_array but if i assign variable to fetch array result, it only shows 1 which is "jean"   i wish this is possible   dforth  

Friends
this is a very strange problem i,m facing
i made a form by which you can enter comments to a topic
i get the comment title and comment topic and date and the name and e mail of the one who insert a comment
according to these variables


$id1=$row2['id'];
echo "$id1";
//Comment poster,s name
$name =strip_tags(@$_POST['coname']);
//Comment title
$title =strip_tags(@$_POST['comtitle']);
//ment poster,s email
$mail =strip_tags(@$_POST['comemail']);
//comment
$com =strip_tags(@$_POST['limitedtextarea']);
//comment,s date
$d= date("Y-m-d");
notice that i,m printing the topic id and i used the variable $id1 to get it inorder to use it to show the comments for this topic later and it shows the right id for the shown topic
then i used this code to insert these variable into the database
if(isset($_POST['add']) and $_POST['add']=='comm'){

$insertcomm =mysql_query("INSERT INTO comments
(com_name,com_title,com_mail,comment,com_date,tid)
VALUES
('$name','$title','$mail','$com','$d','$id1')") or die("comments were not inserted");
if(isset($insertcomm)){
echo "comment inserted ";}
}

Well, I made this yesterday then realised, I need to check is it exists...
I got this but when I go to accept the application and the member exists in the table it enters it anyway...

Code: [Select]
<?php
$member=$_POST['memberid'];
$status=$_POST['Status'];

$con = mysql_connect("host","user","pass");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("a2186214_hbclan",$con);
$sql="UPDATE application SET Status = '$status' WHERE ID = '$member'";
$sql1="INSERT INTO table_members(name) SELECT application.Name FROM application WHERE application.ID = '$member'";
if ($status == 'ACCEPTED')
{
if(mysql_num_rows(mysql_query("SELECT name FROM table_members WHERE name = '$member'")))
{
if(mysql_query($sql, $con) or die(mysql_error()))
{
echo 'Status Changed.<br /><a href="../applications.php">Return To Members List</a>';
}
else
{
die('Could not submit: ' . mysql_error());
}
}
else
{
if(mysql_query($sql, $con) or die(mysql_error()))
{
if(mysql_query($sql1, $con) or die(mysql_error()))
{
echo 'Status Changed.<br /><a href="../applications.php">Return To Members List</a>';
}
}
else
{
die('Could not submit: ' . mysql_error());
}
}
}
else
{
if(mysql_query($sql, $con) or die(mysql_error()))
{
echo 'Status Changed.<br /><a href="../applications.php">Return To Members List</a>';
}
else
{
die('Could not submit: ' . mysql_error());
}
}
mysql_close($con);

?>

Code: [Select]
$date = date('m-d-y');
$ip = $_SERVER['REMOTE_ADDR'];

mysql_query("INSERT INTO users VALUES ($username, $password, 0, $ip, $date)") or die(mysql_error());

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 '.60.116, 03-06-11)' at line 1


I'm not sure why I get this error. :/