i Just want code for if veriable $absolute_url value is already exist in mysql then show me a message i just want this Thank you..
$_SESSION['varname'] = $absolute_url;
$mailingwork = mysql_connect("localhost","root","");
mysql_select_db("inbox",$mailingwork);
if(isset($absolute_url) && !empty($absolute_url))
{
mysql_query("Insert Into deliveredinbox(mailerss) value('$absolute_url')");
}
I think you just want to check if the $absolute_url already exists in the deliveredinbox table. If so, you can do a select query as follows to count the number of rows where the absolute url is equal to the url that you want to enter to the table.
if(isset($absolute_url) && !empty($absolute_url)){
$count=mysql_query("SELECT count(*) from deliveredinbox where mailerss='$absolute_url'");
mysqli_fetch_array($count);
if($count[0]>0){
//already exists
echo "The url already exists";
}
else{
mysql_query("Insert Into deliveredinbox(mailerss) value('$absolute_url')");
}
}
Try if (isset($absolute_url)) { your code here; }. It's not precisely the right tool, but I think that it will do the trick for you.
http://php.net/manual/en/function.isset.php.
Others are absolutely right about transitioning from old mysql calls in PHP. mysqli calls are more secure. PDO will save you time in coding and are even more secure.
Related
Ok, I'm confused. I have some code that searches a database table for a username, and then uses an if else statement to run some code depending on if the user is found or not. My code is below. The problem is that the code isn't even seeing the if else statement, and I have no idea why. Any help is appreciated.
$sqluser = "select * from users where username='" . $user ."'"; //Searching to see if the user is in the database
echo $sqluser . "<br><br>"; //writes out the select statement to make sure it is correct
$query = mssql_query($sqluser); //returns the results
$num_rows = mssql_num_rows($query); //gets the number of rows returned
echo $num_rows; //writes out the number of rows
if ($num_rows==0) //determines what happens next if the user exists or not
{
//displays an error box if the user doesn't exist
echo "<script type=text/javascript>";
echo "alert('That user doesn't exist. Please try again.')";
echo "</script>";
}
else
{
//will be code to run if the user does exist
echo "<script type=text/javascript>alert('Testing.')</script>";
}
I couldn't add a comment. So I will write this as an answer instead.
Since you state that the alert JavaScript is showing in the page source, this mean that the IF/ELSE statement in PHP is working fine. The problem is with the single quote. You have a single quote inside a single quoted alert function. Hence the JavaScript alert function cannot be executed.
echo "alert('That user doesn't exist. Please try again.')";
Try using this instead
echo "alert('That user doesn\'t exist. Please try again.');";
Please could someone give me some much needed direction...
I have a registration form, however I need to add a condition that if the username is already in the table, then a message will appear. I have a had a few goes except it just keeps adding to the SQL table.
Any help would be much appreciated. Here is my current code:
Thanks in advance!
<?php
session_start();session_destroy();
session_start();
$regname = $_GET['regname'];
$passord = $_GET['password'];
if($_GET["regname"] && $_GET["regemail"] && $_GET["regpass1"] && $_GET["regpass2"] )
{
if($_GET["regpass1"]==$_GET["regpass2"])
{
$host="localhost";
$username="xxx";
$password="xxx";
$conn= mysql_connect($host,$username,$password)or die(mysql_error());
mysql_select_db("xxx",$conn);
$sql="insert into users (name,email,password) values('$_GET[regname]','$_GET[regemail]','$_GET[regpass1]')";
$result=mysql_query($sql,$conn) or die(mysql_error());
print "<h1>you have registered sucessfully</h1>";
print "<a href='login_index.php'>go to login page</a>";
}
else print "passwords don't match";
}
else print"invaild input data";
?>
User kingkero offered a good approach. You could modify your table so that the username field is UNIQUE and therefore the table cannot contain rows with duplicate usernames.
However, if you cannot modify the table or for other reasons want to choose a different approach, you can first try to run a select on the table, check the results and act accordingly:
$result=mysql_query('SELECT name FROM users WHERE name="'.$_GET['regname'].'"');
$row = mysql_fetch_row($result);
You can then check $row if it contains the username:
if($row['name']==$_GET['regname'])
If this statement returns true, then you can show the user a message and tell him to pick a different username.
Please note
Using variables that come directly from the client (or browser) such as what might be stored in $_GET['regname'] and using them to build your SQL statement is considered unsafe (see the Wikipedia article on SQL-Injections).
You can use
$regname=mysql_escape_string($_GET['regname'])
to make sure that its safe.
Firstly, there is some chaos on the second line:
session_start();session_destroy();
session_start();
Why you doing it? Just one session_start(); needed.
Then you can find users by simple SQL query:
$sql="SELECT * FROM users WHERE name = '$regname'";
$result=mysql_query($sql) or die(mysql_error());
if (mysql_num_rows($result) > 0) {
//...echo your message here
}
When you got it, I suggest you to rewrite your code with use of PDO and param data binding, in order to prevent SQL injections and using of obsolete functions.
I'm trying to create a php scipt that checks if members are verified as they land on a page. If they are not they get redierected to login with an error message & instructions. So on the page I have this code:
<?php
if (loggedin()) {
$check_active = "SELECT active FROM members WHERE username == '$username'";
$active = mysql_query($check_active);
if ($active < 1) {
header("Location: login.php?verify=true");
} else {
exit();
}
}
?>
It is redirecting the user back to the login page but it doing it whether they are active or not. The values for active members are 0(not verified) & 1(verified). Is htere something wrong in the script I'm using?
Thank You
You'll need to handle the $active result and put it into a PHP variable/array. $active as it is in your code is simply a resource (see here) Try this:
$active = mysql_query($check_active); // run query and return resource
$row = mysql_fetch_assoc($active); // put resource data into php array
if ($row['active'] < 1) {
header("Location: login.php?verify=true");
} else {
exit();
}
Please don't use old mysql, use mysqli_ or read topic on http://php.net/manual/en/function.mysql-query.php.
Instead, the MySQLi or PDO_MySQL extension should be used. See also MySQL: choosing an API guide and related FAQ for more information.
You aren't fetching the data correctly after the query.
Right after you do the the mysql_query function. Try this:
$output = mysql_fetch_assoc($active);
$active_result = $ouput['active'];
Don't know if it might work.
And add a little more security for SQL injection there.
And use MySQLi instead since you aren't keen on preventing SQL injection yourself.
I have a code which kinda works, but not really i can't figure out why, what im trying to do is check inside the database if the URL is already there, if it is let the user know, if its not the go ahead and add it.
The code also makes sure that the field is not empty. However it seems like it checks to see if the url is already there, but if its not adding to the database anymore. Also the duplicate check seems like sometimes it works sometimes it doesn't so its kinda buggy. Any pointers would be great. Thank you.
if(isset($_GET['site_url']) ){
$url= $_GET['site_url'];
$dupe = mysql_query("SELECT * FROM $tbl_name WHERE URL='$url'");
$num_rows = mysql_num_rows($dupe);
if ($num_rows) {
echo 'Error! Already on our database!';
}
else {
$insertSite_sql = "INSERT INTO $tbl_name (URL) VALUES('$url')";
echo $url;
echo ' added to the database!';
}
}
else {
echo 'Error! Please fill all fileds!';
}
Instead of checking on the PHP side, you should make the field in MySQL UNIQUE. This way there is uniqueness checking on the database level (which will probably be much more efficient).
ALTER TABLE tbl ADD UNIQUE(URL);
Take note here that when a duplicate is INSERTed MySQL will complain. You should listen for errors returned by MySQL. With your current functions you should check if mysql_query() returns false and examine mysql_error(). However, you should really be using PDO. That way you can do:
try {
$db = new PDO('mysql:host=localhost;db=dbname', $user, $pass);
$stmt = $db->query('INSERT INTO tbl (URL) VALUES (:url)');
$stmt->execute(array(':url' => $url));
} catch (PDOException $e) {
if($e->getCode() == 1169) { //This is the code for a duplicate
// Handle duplicate
echo 'Error! Already in our database!';
}
}
Also, it is very important that you have a PRIMARY KEY in your table. You should really add one. There are a lot of reasons for it. You could do that with:
ALTER TABLE tbl ADD Id INT;
ALTER TABLE tbl ADD PRIMARY KEY(Id);
You should take PhpMyCoder's advice on the UNIQUE field type.
Also, you're not printing any errors.
Make sure you have or die (mysql_error()); at the end of your mysql_* function(s) to print errors.
You also shouldn't even be using mysql_* functions. Take a look at PDO or MySQLi instead.
You're also not executing the insert query...
Try this code:
if(isset($_GET['site_url']) ){
$url= $_GET['site_url'];
$dupe = mysql_query("SELECT * FROM $tbl_name WHERE URL='$url'") or die (mysql_error());
$num_rows = mysql_num_rows($dupe);
if ($num_rows > 0) {
echo 'Error! Already on our database!';
}
else {
$insertSite_sql = "INSERT INTO $tbl_name (URL) VALUES('$url')";
mysql_query($insertSite_sql) or die (mysql_error());
echo $url;
echo ' added to the database!';
}
}
else {
echo 'Error! Please fill all fileds!';
}
As PhpMyCoder said, you should add a unique index to the table.
To add to his answer, here is how you can do what you want to do with only one query.
After you add the unique index, if you try to "INSERT INTO" and it result in a duplicate, MySQL will produce an error.
You can use mysql_errno() to find out if there was a duplicate entry and tell the user.
e.g.
$sql = "INSERT INTO $tbl_name (URL) VALUES('$url')";
$result = mysql_query($sql);
if($result === false) {
if(mysql_errno() == $duplicate_key_error) {
echo 'Error! Already in our database!';
} else {
echo 'An error has occurred. MySQL said: ' . mysql_error();
}
}
mysql_error() will return the mysql error in plain english.
mysql_errno() returns just the numeric error code. So set $duplicate_key_error to whatever the code is (I don't know it off the top of my head) and you are all set.
Also note that you don't want to print any specific system errors to users in production. You don't want hackers to get all kinds of information about your server. You would only be printing MySQL errors in testing or in non-public programs.
ALSO! Important, the mysql functions are deprecated. If you go to any of their pages ( e.g. http://php.net/manual/en/function.mysql-errno.php) you will see recommendations for better alternatives. You would probably want to use PDO.
Anyone who wants to edit my answer to change mysql to PDO or add the PDO version, go ahead.
I'm writing a PHP code for my website. Currently, there's some problems with my code.
Here's my code. Ignore some Malay language used, I'd tried to translate most of them.
<?php
session_start();
include "../library/inc.connectiondb.php";
$txtUser = $_POST['txtUser'];
$txtPass = $_POST['txtPass'];
if(trim($txtUser) == "") {
echo "<b>User ID</b> is empty, please fill";
include "login.php";
}
else if(strlen(trim($txtPass)) <= 5) {
echo "<b>Password</b> is less then 6 characters, please fix";
include "login.php";
}
else {
$sqlPeriksa = "SELECT userID FROM admin WHERE userID='$txtUser'";
$qryPeriksa = mysql_query($sqlPeriksa, $sambung);
$hslPeriksa = mysql_num_rows($qryPeriksa);
if($hslPeriksa == 0) {
# If username doesn't exist
echo "<b>UserID</b> doesn't exist";
include "login.php";
}
else {
$sqlPassword = "SELECT passID FROM admin WHERE (userID='$txtUser' && passID='$txtPass')";
$qryPassword = mysql_query($sqlPeriksa, $sambung);
$hslPassword = mysql_num_rows($qryPassword);
if($hslPassword < 1) {
# If password is incorrect
echo "<b>Password</b> is incorrect";
include "login.php";
}
else {
# If login successful
$SES_Admin = $txtUser;
session_register('SES_Admin');
echo "LOGIN SUCCESSFUL";
# Redirect to index.php
echo "<meta http-equiv='refresh' content='0; url=index.php'>";
exit;
}
}
}
?>
The problem is this code allows me to login even if the password is wrong. I'd done some searches and it still doesn't solve my problem. I'm pretty sure that the problem is at line 27 onwards.
So, if anyone has a solution, please tell me quickly. I'm writing this code for my school, and it had to be finished before next year.
Edit
Ok, I'd already placed the mysql_real_escape_string in the code just like what many people told me. I don't know how this will help, but the mysql table for this was named "admin". It had 2 fields; userID and passID. To test the code, I'd inserted the value "admin" and "12345678" into the table.
This is where your problem is:
$sqlPassword = "SELECT passID FROM admin WHERE (userID='$txtUser' && passID='$txtPass')";
$qryPassword = mysql_query($sqlPeriksa, $sambung);
$hslPassword = mysql_num_rows($qryPassword);
You see, your mysql_query is executing $sqlPeriksa which is:
$sqlPeriksa = "SELECT userID FROM admin WHERE userID='$txtUser'";
Instead, your code should be like this:
$sqlPassword = "SELECT passID FROM admin WHERE (userID='$txtUser' && passID='$txtPass')";
$qryPassword = mysql_query($sqlPassword, $sambung);
$hslPassword = mysql_num_rows($qryPassword);
Please try this out and let us know what happens.
[edit/additional] : I strongly suggest that you look into the following:
Using PDO:
http://net.tutsplus.com/tutorials/php/why-you-should-be-using-phps-pdo-for-database-access/
Using stored procedures:
http://dev.mysql.com/doc/refman/5.0/en/create-procedure.html
Using PDO + stored procedures:
http://php.net/manual/en/pdo.prepared-statements.php (See example #4)
just plain troubleshoot is necessary. how many rows are returned? what are the values of userID and passID in the query that returns rows? put some breaks in and see what's going on. i don't see a problem, it but its hard to troubleshoot code posted here since it really can't be run without a db.
I don't see any reason this isn't working as you expected, I suspect the problem might be elsewhere. For example, I don't see you checking if a "SES_Admin" session is already registered. But at the very least you need to replace lines 5 and 6 with this, otherwise someone could potentially delete your entire user table, and do various other malicious things with your MySQL databases.
$txtUser = mysql_real_escape_string($_POST['txtUser']);
$txtPass = mysql_real_escape_string($_POST['txtPass']);
Please read the article on mysql_real_escape_string at http://php.net/manual/en/function.mysql-real-escape-string.php