display message to let user know that the value already exist - php

I want to display an error message if the insert already exists.
I've made name unique in the database and if I enter the same value, it does not insert. Just as I wanted. Here is my code:
$query = "INSERT INTO register(name) VALUES('$foo')";
mysql_query($query) or die('An Error occurred' . mysql_error());
Now I want to echo a message to let the user know that the value he has entered is already present in the database and will not be inserted. Anyone?

I take it you are collecting $foo from a form?
what I would do is an sql query of the table register collecting the name field then when you collect the name entered in the form and its posted you can run an if condition against the name field you have already gathered using the sql statement and if there is a name = to the name they enter on the field they can receive a message and exit before the sql injection into the register table.

The simpliest way:
$res = mysql_query($query):
if ($res)
echo 'Insertion ok';
else
echo 'error: name already exists';
A better way: do first a SELECT query to see if name exists or not.
Note: you should think about moving from mysql_* to mysqli_* or PDO

Try this :
$query = "INSERT INTO register(name) VALUES('$name')";
$user = mysql_query($query):
if ($user)
echo 'User Register';
else
echo 'User Already Exist';

As per Melon's comment, you should use mysqli.
// Create your connection
$mysqli = new mysqli($host, $user, $pass, $dbname);
// Do your query
$mysqli->query("INSERT INTO register(name) VALUES('$foo')");
if($mysqli->affected_rows == 0) {
// Your chosen method of alerting the user goes here
}

\\this first part collects name information from your table.
$name="SELECT name FROM register";
$name_query = mysqli_query($db_conx, $name);
$numrows = mysqli_num_rows($name_query);
if($numrows < 1){
echo "cannot find user";
header ("location: index.php");
}
while ($row = mysqli_fetch_array($name_query, MYSQLI_ASSOC)){
$name = $row["name"];
}
\\ this part gets the name from your form
if (isset($_POST['class']) && ($_POST['class'] !='')){
$foo = $_POST['foo'];
\\this part checks to see if the 2 values are equal
if ($name == $foo){
echo "This name has already been used try again";
\\this part then adds the name if its not already in the database
}else{
$query = "INSERT INTO register(name) VALUES('$foo')";
mysql_query($query) or die('An Error occurred' . mysql_error());
}
}
//then all you need to do is create your form to post the foo or name so it can be collected and passed through the querys.

Related

How I should fix if else statement didn't work

I want to test if user enter incorrect file number and ic, the message "Inccorect ref no and ic no" will shown, but it didn't work like that. It always shown the message "Thanks you for register attendance with us." Help me please..I'm new.
<?php
require "init.php";
$file_number = $_POST["file_number"];
$ic_no = $_POST["ic_no"];
$attendance = $_POST["attendance"];
//$ic_no = $_REQUEST['ic_no'];
$sql = "INSERT INTO reg_attend(file_number,ic_no,attendance) SELECT reg_meeting.file_number,reg_staff.ic_no,'".$attendance."' FROM reg_meeting,reg_staff WHERE (reg_meeting.file_number = '".$file_number."' AND reg_staff.ic_no = '".$ic_no."')";
$result = mysqli_query($conn,$sql);
$response = array();
if($result)
{
//var_dump($result);
$code = "reg_success";
$message = "Thanks you for register attendance with us.";
array_push($response,array("code"=>$code,"message"=>$message));
echo json_encode($response);
}
else
{
$code = "reg_failed";
$message = "Incorrect REF No. and IC No.";
array_push($response,array("code"=>$code,"message"=>$message));
echo json_encode($response);;
}
mysqli_close($conn);
?>
You need to validate your incoming data before you insert into your DB. It looks like currently, you will be inserting empty values into your reg_attend table. As this is likely successful, you will not be reaching the else.
If you need to validate that the passed in data exists, do a select first e.g.
SELECT reg_meeting.file_number,reg_staff.ic_no,'".$attendance."' FROM reg_meeting,reg_staff WHERE (reg_meeting.file_number = '".$file_number."' AND reg_staff.ic_no = '".$ic_no."')"
then check the results of that, before inserting.
As a side note, please please DO NOT use this code as shown. You should prefer to use query parameters with either PDO or mysqli. See https://www.php.net/manual/en/pdo.prepare.php

Data inserted into the database even though it already exists

I'm currently coding a registration script in PHP and my problem is that the data is still inserted into the database even though it already exists. It's probably some silly mistake or I need some else{} statement or I don't really know. The thing is that even though the email already exists in the database it stills enters it.
It does display the error just fine.
if(filter_var($email,FILTER_VALIDATE_EMAIL)){
$email = filter_var($email,FILTER_VALIDATE_EMAIL);
$email_check = mysqli_query($con, "SELECT email FROM database WHERE email='$email'");
$num_rows = mysqli_num_rows($email_check);
if($num_rows>0){
echo "The email is already in use.<br>";
}
$query = mysqli_query($con,"INSERT INTO database VALUES (NULL,'$username','$name','$email','$pwh','$date')");
}
?>
If the email is already in use it displays the echo "The email is already in use." just fine, yet it still inserts it. What am I missing? I already tried using 'else' variable yet nothing helped.
Your if only echo something, then you do the INSERT no matter what. Some solution :
if(filter_var($email,FILTER_VALIDATE_EMAIL)){
$email = filter_var($email,FILTER_VALIDATE_EMAIL);
$email_check = mysqli_query($con, "SELECT email FROM database WHERE email='$email'");
$num_rows = mysqli_num_rows($email_check);
if($num_rows>0){
echo "The email is already in use.<br>";
}
// ADD A ELSE SO YOU INSERT IF YOU HAVE NOTHING
else {
$query = mysqli_query($con,"INSERT INTO database VALUES (NULL,'$username','$name','$email','$pwh','$date')");
}
}
Now you can prevent it from your database too :
Add a UNIQUE INDEX on the column email from your table database
Use INSERT IGNORE now, so it will insert if the email is not used and ignore if email is already used
And last, use prepare statement and bind param to avoind SQL injection !
Hope it helps
Your if is fine, but you then proceed to always do the insert. This is because you have put it outside the if.
what you should do is :
if(!$num_rows <= 0){
<insert statement>;
}
else {
echo "The email is already in use.<br>";
}
write this statement inside else block
else
{
$query = mysqli_query($con,"INSERT INTO database VALUES (NULL,'$username','$name','$email','$pwh','$date')");
}

why i m giving success even the id and/or username does not exist in database

I m trying to code a deletation page, but when i type id and username and click delete, it gives me the message "success" even if the id and/or username not exist in database and if i type correct id and username is delete from database
with message "success" how to fix this please? thank you
<?php require ('server.php');?>
<?php
$uniqueid=$_GET['uniqueid'];
$username=$_GET['username'];
if(isset($_GET['uniqueid'], $_GET['username'])){
$sql= "DELETE FROM users WHERE uniqueid='$uniqueid' AND username='$username'";
if($sql)
echo "succces";
}
else {
echo 'ERROR';
}
mysqli_close($db);
?>
let's break down you code :
you are using mysqli, you may have to look into the use PDO instead
it have a way bigger database list interaction whereas mysqli only
got one.
you have no statement for running your sql statment, for now $sql is justplain text
the fix way for you code:
<?php
$uniqueid=$_GET['uniqueid'];
$username=$_GET['username'];
if(isset($_GET['uniqueid'], $_GET['username']))
{
$sql= $BD->("DELETE FROM users WHERE uniqueid='$uniqueid' AND username='$username'");
/*$BD is you connection to the database*/
if($sql) echo "succces";
else
{
echo 'ERROR';
}
}
else
{
echo 'no id or username';
}
$sql->closeCursor();?>

MySQL/PHP check for duplicate before INSERT

I'm trying to check for an existing entry in MySQL before executing the INSERT statement. If the user enters a name already in the database (field is set to unique) then they should be prompted to re-enter the name.
The problem I'm having is that if the new entry matches a record in any form then the error message displays and no INSERT happens.
For example, if the user enters DUMMY_NEW and there is a record DUMMY_OLD they aren't able to add the record even though DUMMY_NEW does not exist in the table.
I've searched and tried other answers already but can't seem to get this to work.
Code with extraneous bits removed for clarity:
//Create connection to database using mysqli
$conn = new mysqli($dbhost, $dbuser, $dbpass, $db);
//Set variables according to user input on previous form
$Server_Name = $_POST['Server_Name'];
//Check for duplicate server name - if exists inform user else run INSERT ($stmt)
$checkdup = "SELECT * FROM dcr_table WHERE SERVER_NAME = '".$Server_Name."'";
$dupresult = $conn->query($checkdup);
if($dupresult = 1)
{
print "<br>Error! <p></p>";
echo "" . $Server_Name . " already exists in the DCR";
print "<p></p>Please check the Server Name and try again";
}
else {
//Define the INSERT statement
$stmt = "INSERT INTO dcr_master (Server_Name, Description,..., ... , ... )";
//Execute the INSERT statement
$conn->query($stmt);
//Success and return new id
echo "<br><p></p>Record Added!<p></p>";
echo "New id: " . mysqli_insert_id($conn);
//Drop the connection
$conn->close();
};
Edit:
I'm aware of the injection vulnerability. The MySQL account only has SELECT, INSERT and UPDATE rights to the table. The end user must supply the password or submit will fail. This is small app with limited user access at the moment. MySQL escape strings will be implemented after current issue is resolved.
Edit 2:
Using Hobo Sapiens method does work in reporting an existing entry however a new (empty) row is still added to the table. The record ID still auto-increments so what I get is id#300 - record, id#301 - blank, id#302 - record. Is this a result of the IGNORE in the INSERT statement?
Your code creates a race condition if two people attempt to create the same ame at the same time and you're not handling the fallout properly.
If you have set the SERVER_NAME column to UNIQUE then you needn't check for the existence of a server name before you perform your INSERT as MySQL will do that for you. Use INSERT IGNORE ad check the number of affected rows after the query has executed to find out if it worked:
//Create connection to database using mysqli
$conn = new mysqli($dbhost, $dbuser, $dbpass, $db);
//Set variables according to user input on previous form
$Server_Name = $_POST['Server_Name'];
//Define the INSERT statement with IGNORE keyword
$stmt = "INSERT IGNORE INTO dcr_master (Server_Name, Description,..., ... , ... )";
if ($conn->query($stmt) === false) {
die("Database error:".$conn->error);
}
// Check for success
if ($conn->affected_rows == 0) {
print "<br>Error! <p></p>";
echo "" . $Server_Name . " already exists in the DCR";
print "<p></p>Please check the Server Name and try again";
} else {
//Success and return new id
echo "<br><p></p>Record Added!<p></p>";
echo "New id: " . $conn->insert_id;
}
This is an atomic operation so no race condition, and it involves only one call to the database.
I recommend you use either the OOP style or the procedural style for mysqli_*() but don't mix them. Usual warnings about SQL injection apply.
Use mysqli_num_rows
$row_cnt = $dupresult->num_rows;
if ($row_cnt > 0) {
echo "There is a matching record";
}else {
//insert into table
}
This statement:
if($dupresult = 1)
will always return 1. You should first retrieve the first query result (if any), like so:
$row=$dupresult->fetch_array(MYSQLI_NUM);
and then compare the result against NULL:
if(!$row)

user registration php

<?php
include"include/connection.php";
$checkusername=mysql_query("SELECT * FROM employer WHERE eusername='$username'");
if (mysql_num_rows($checkusername)==1)
{
echo "username already exist";
}
else
{
$query = "insert into employer(efname,elname,egender,eemail,eusername,epwd,eadd,ephone,ecity,ecountry) values ('".$_POST['first_name']."','".$_POST['last_name']."','".$_POST['gender']."','".$_POST['email']."','".$_POST['username']."','".$_POST['password']."','".$_POST['address']."','".$_POST['phone']."','".$_POST['city']."','".$_POST['country']."')";
$result = mysql_query($query) or die (mysql_error());
echo " Thanks for registration";
}
?>
This is my code for inserting registration form data into a database. This code adds the data but also gives a parse error, but does not give the error if the username already exists.
Notice: Undefined variable: username in C:\Program Files\EasyPHP5.3.0\www\register_hirer2.php on line 6
Thanks for registration
line 6 is:
$checkusername=mysql_query("SELECT * FROM employer WHERE eusername='$username'");
Well, your $username is undefined indeed.
Most probably you want to use $_POST['username'].
And of course this obligatory XKCD comic:
If the "data source" is an html form (supposedly using method="post") you have to use $_POST['username'] when register_globals is set to off (which is the default since ...ages). see http://docs.php.net/security.globals
Also have a read of http://php.net/manual/en/security.database.sql-injection.php
<?php
include"include/connection.php";
$query = "SELECT
*
FROM
employer
WHERE
eusername='". mysql_real_escape_string($username). "'
";
$checkusername=mysql_query($query) or die(mysql_error());
if (mysql_num_rows($checkusername)==1)
{
echo "username already exist";
}
else
{
$query = "INSERT INTO employer(efname,elname,egender,eemail,eusername,epwd,eadd,ephone,ecity,ecountry) values (". same mysql_real_escape_string() thing here for each parameter .")";
$result = mysql_query($query) or die (mysql_error());
echo " Thanks for registration";
}
?>
You can also use prepared statements. This way you don't need/can't forget using an escaping function.
edit and btw: you don't need the SELECT before the INSERT in order to make the username unique. Actually it will make things even harder since now you have to deal with race conditions. You'd have to lock the table between those two queries.
If you add an unique index for the username in your table MySQL will not allow the insertion of a doublet but instead return a specific error code which your script can fetch and handle without the need of dealing with race conditions.
define('ER_DUP_ENTRY', 1062);
$mysql = mysql_connect('..', '..', '..');
mysql_select_db('..', $mysql) or die(mysql_error($mysql));
$fields = array(
'efname'=>'first_name',
'elname'=>'last_name',
'egender'=>'gender',
'eemail'=>'email',
'eusername'=>'username',
'epwd'=>'password',
'eadd'=>'address',
'ephone'=>'phone',
'ecity'=>'city',
'ecountry'=>'country'
);
$sqlparams = array();
foreach($fields as $sql=>$form) {
if ( !isset($_POST[$form]) ) {
die('missing post parameter '. $form);
}
$sqlparams[$sql] = "'".mysql_real_escape_string($_POST[$form], $mysql)."'";
}
$query = '
INSERT INTO
employer
'. join(', ', array_keys($sqlparams)) .'
VALUES
('.join(',', $sqlparams).')
';
// table:employer has been defined with "unique key idxName (eusername)"
$result = mysql_query($query, $mysql);
if ( false!==$result ) {
echo " Thanks for registration";
}
else if ( ER_DUP_ENTRY===mysql_errno($mysql) ) {
echo 'username already exists';
}
else {
echo 'an error occurred';
}
That is because you do not define $username anywhere. It looks as though you want to use $_POST['username']
mysql_query("SELECT * FROM employer WHERE eusername='{$_POST['username']}'");
Also, your code is vulnerable to a SQL Injection
You never define $usernameanywhere, so it gives that error because you are trying to use a variable that it doesn't have a value for.
This is most likely because you've not defined the '$username' variable. I presume that you're relying on this being populated from the incoming GET/POST data (most likely via the depreciated register_globals) which is bad practice.
As such, you'll need to either populate $username via $_POST or $_GET.
More importantly, you should update the insert query to escape the incoming 'untrusted' data using mysql_real_escape_string (e.g.: mysql_real_escape_string($_POST['username']), etc.)
As #Yacoby said your code is vulnerable to an SQL Injection to prevent it you can use mysqli or PDO , if you would like to use mysqli use the following code:
<?php
include"include/connection.php";
$query = "SELECT
*
FROM
employer
WHERE
eusername='". mysql_real_escape_string($username). "'
";
$checkusername=mysql_query($query) or die(mysql_error());
if (mysql_num_rows($checkusername)==1)
{
echo "username already exist";
}
else
{
$query = $conn->prepare("INSERT INTO employer(efname,elname,egender,eemail,eusername,epwd,eadd,ephone,ecity,ecountry)) values ( ? , ? , ? , ? , ? , ? , ? , ? , ? , ?)"; // preparing the insert
$query->bind_param("ssssssssss" , $variable1 , $variable2 , $variable3 , $variable4 , $variable5 , $variable6 , $variable7 , $variable8 , $variable9 , $variable10); // binding parameters
$query->execute(); // sending the parameter values
$query->close(); // closing the query
$conn->close(); // closing the connection
if ($query) { // checking if the query has been executed with no errors
echo " Thanks for registration";
}
}
?>
BE SURE TO CHANGE THE $conn AND variables to whatever you want!

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