I am trying to create a variable in my php code where I can use a value from my database to use user information. Basically, I'm trying to fetch the UserID from the database desikitchen where it matches the variable $dkuser.
Can anyone tell me where I'm going wrong?
$sqlStremail = "SELECT UserID
FROM User
WHERE Name = '$dkuser'";
$result = mysql_query($sqlStremail);
$row = mysql_fetch_assoc($result);
$variable = $row["UserID"];
$sql2 = "INSERT INTO desikitchen.Inventory (ItemName, ItemUser)
VALUES ('{$itemname1}', '{$sqlStremail}')";
$result = mysql_query($sql2);
You better Use MySQLi class library. Something like this:
$mysqli = new mysqli("",$mysqlUser,$mysqlPass,$mysqlTable);
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
print FALSE; exit();
}
$query = "SELECT UserID FROM User WHERE Name = '$dkuser'";
$result=$mysqli->query($query);
if($mysqli->query($query)==False) {print mysqli_error($mysqli); exit();}
if($result->num_rows>0)
{ $row = $result->fetch_array();
echo $row["UserID"];
} else
{ echo "User not found"; }
Related
I'm tring to retrive information from a database while a user send Login command from iOS app.
To test this function i'm launching my php page manually (ex. http://www.testdatabase.com/LoginFunctions.php) and forcing username programmatically.
The problem is that mysqli_query return NULL value. if i use "or die(mysql_error()" nothing happens. Even if i use mysqli_num_rows return 1, but $result is still empty.
So when mysql_fetch_assoc is been executed the programm crashes without showing any error.
Any idea? Thanks
<?php
// Create connection
$con=mysqli_connect("localhost","super","super","testdb");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$action = "login";
$username = "Peperoncino";
$response = array();
if ($action == "login")
{
$query = "SELECT psw AS pswrd,id FROM Activities WHERE nome = 'Peperoncino' LIMIT 1";
if ($result = mysqli_query($con, $query))
{
$values = mysql_fetch_assoc($result);
$password = $values['pswrd'];
$response["password"] = $password;
$response["message"] = "Get information from db";
}
else
{
echo "err";
}
echo json_encode($response);
}
// Close connections
mysqli_close($con);
?>
You are using the deprecated mysql_fetch function.Use the new one
<?php
// Create connection
$con=mysqli_connect("localhost","super","super","testdb");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$action = "login";
$username = "Peperoncino";
$response = array();
if ($action == "login")
{
$query = "SELECT psw AS pswrd,id FROM Activities WHERE nome = 'Peperoncino' LIMIT 1";
if ($result = mysqli_query($con, $query))
{
$values = mysqli_fetch_assoc($result);
$password = $values['pswrd'];
$response["password"] = $password;
$response["message"] = "Get information from db";
}
else
{
echo mysqli_error($conn);
}
echo json_encode($response);
}
// Close connections
mysqli_close($con);
?>
I want echo my DB results from Session but i get no results or errors:
$_SESSION['username'];
$link = mysqli_connect("$myHost", "$myUser", "$myPass", "$myDB");
// Check connection
if($link === false){
die("ERROR: Could not connect. " . mysqli_connect_error());
}
$username = mysqli_real_escape_string($link, $_SESSION['username']);
$sql = "SELECT * FROM users where username = $username";
$result = mysqli_query($link, $sql);
echo $result;
Anyone know why not? Session works.
Thanks
You should change your query, like this:
$sql = "SELECT user FROM yourtablename WHERE username = $username"
Where "user" is what you want to SELECT if you want to select all data, you can use "*", yourtablename is table name of table you want to select.
After your edits, your code should look like
$_SESSION['username'];
$link = mysqli_connect("$myHost", "$myUser", "$myPass", "$myDB");
// Check connection
if($link === false){
die("ERROR: Could not connect. " . mysqli_connect_error());
}
$username = mysqli_real_escape_string($link, $_SESSION['username']);
$sql = "SELECT * FROM users where username = $username";
if ($result = $link->query($sql)) {
while ($row = $result->fetch_row()) {
var_dump($row);
}
$result->close();
}
More info here
Notice: mysqli_real_escape_string it's not very security. A better option to protect against SQL injections is using prepared statements, more info here
I have an issue with ajax response. I am using custom query for fetch result from database. Json response always shows null value while query is running successfully. Here is my code:
if(isset($_POST)){
$arr = array();
//$query = mysql_query(" insert into login (user,pass) values ('".$_POST['firstname']."','".$_POST['lastname']."') ") or die('test');
$query = mysql_query(" select * from login where user = '".$_POST['firstname']."' && pass = '".$_POST['pass']."' ") or die('test');
$rows = mysql_num_rows($query);
while($fetch = mysql_fetch_array($query))
{
if($fetch)
{
$_SESSION['ID']= $fetch['id'];
$arr['id'] = $_SESSION['ID'];
}
else
{
$arr['failed']= "Login Failed try again....";
}
}
echo json_encode($arr);
}
#Amandhiman i did not get what is the use of if statement with in the while
if($fetch)
{
$_SESSION['ID']= $fetch['id'];
$arr['id'] = $_SESSION['ID'];
}
the mention code definitely works for you
if($rows>0)
{
while($fetch = mysql_fetch_array($query))
{
$_SESSION['ID']= $fetch['id'];
$arr['id'] = $_SESSION['ID'];
}
}else
{
$arr['failed']= "Login Failed try again....";
}
Try with the below code, (mysql is deprected), working for me with my test database table (Debug it, var_dump $result, $result->fetch_assoc(), $result->num_rows)
<?php
$servername = "localhsot";
$username = "yourser";
$password = "passyour";
// Create connection
$conn = new mysqli($servername, $username, $password);
// Check connection
if ($conn->connect_error)
die("Connection failed: " . $conn->connect_error);
if(isset($_POST)){
$arr = array();
$query = "select * from login where user = '".$_POST['firstname']."' && pass = '".$_POST['pass']."' ";
$result = $conn->query($query);
$rows = $result->num_rows;
while($fetch = $result->fetch_assoc())
{
if($fetch)
{
$_SESSION['ID']= $fetch['id'];
$arr['id'] = $_SESSION['ID'];
}
else
{
$arr['failed']= "Login Failed try again....";
}
}
echo json_encode($arr);
}
try this
if(isset($_POST)){
$arr = array();
//$query = mysql_query(" insert into login (user,pass) values ('".$_POST['firstname']."','".$_POST['lastname']."') ") or die('test');
$query = mysql_query(" select * from login where user = '".$_POST['firstname']."' && pass = '".$_POST['pass']."' ") or die('test');
$rows = mysql_num_rows($query);
if($rows>0)
{
while($fetch = mysql_fetch_array($query))
{
$_SESSION['ID']= $fetch['id'];
$arr['id'] = $_SESSION['ID'];
}
}else
{
$arr['failed']= "Login Failed try again....";
}
echo json_encode($arr);
}
First of all start session before playing with it on the top of page
session_start();
check your database connectivity.
Use print_r($arr) for testing your array();
first off all you are using session variable . to use session variable you need to initialize it by session_start()
you are using key in the array in this way it will return the last inserted record in the array . try this code
<?php
$servername = "localhsot";
$username = "yourser";
$password = "passyour";
// Create connection
$conn = new mysqli($servername, $username, $password);
// Check connection
if ($conn->connect_error)
die("Connection failed: " . $conn->connect_error);
if(isset($_POST)){
$arr = array();
$query = "select * from login where user = '".$_POST['firstname']."' && pass = '".$_POST['pass']."' ";
$result = $conn->query($query);
$rows = $result->num_rows;
while($fetch = $result->fetch_assoc())
{
if($fetch)
{
// if wants to use session then start session
session_start(); // else it will return null
$_SESSION['ID']= $fetch['id']; // i don't know why this ??
// $arr['id'] = $_SESSION['ID']; comment this
$arr[] = array('msg'=>'succsee','ID'=>$fetch['id']);
}
else
{
$arr[]= array('msg'=>'fail','ID'=>null);
}
}
echo json_encode($arr);
}
Here's my code. Actually it run without error but it always enter the first condition even though the value in mysql is more than 0 say for example is 10.
<?php
session_start();
$loginuser = $_SESSION['result'];
$con=mysqli_connect("localhost","root","","leavecalendar");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$sql="SELECT leavecount from employee WHERE username = $loginuser[username]";
$resource = mysql_query($sql);
if ($resource == 0) {
echo "ajsdjsdasjd"; <----always enters here
}
else
{
header("insert.php");
}
mysqli_close($con);
?>
First off, you are mixing api's. Dont mix mysqli_* and mysql_*. Just stick with mysqli_*.
$con = mysqli_connect("localhost","root","","leavecalendar"); // mysqli
$resource = mysql_query($sql); // mysql
Second, since you are using mysqli_* now, just use prepared statements.
$sql="SELECT leavecount from employee WHERE username = $loginuser[username]";
Third, dont compare the result set to zero, instead, check how many rows did the result return.
$resource = mysqli_query($con, $sql);
if ($resource == 0) {
Consider this example:
session_start();
if(isset($_SESSION['result'])) {
$mysqli = mysqli_connect("localhost","root","","leavecalendar");
$loginuser = $_SESSION['result'];
$stmt = $mysqli->prepare("SELECT leavecount from employee WHERE username = ?");
$stmt->bind_param("s", $loginuser);
$stmt->execute();
if($stmt->num_rows > 0) {
// if it has results
header("Location: insert.php");
} else {
// if it has NO results
}
}
Problems
1) You can not use
mysql_query($query)
For mysqli
2)
"SELECT leavecount from employee WHERE username = $loginuser[username]"
The username should be single quoted or double quoted.
For ex: "SELECT leavecount from employee WHERE username = 'dineshrawat'";
Solution
<?php
session_start();
$loginuser = $_SESSION['result'];
$con=mysqli_connect("localhost","root","","leavecalendar");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$sql="SELECT leavecount from employee WHERE username = $loginuser[username]";
$resource = mysqli_query($con, $sql);
if ($resource == 0) { // If there is no record in result set
echo "No record found";
}
else
{
header("insert.php");
}
mysqli_close($con);
?>
I am completely new to MYSQL and PHP, so i just need to do something very basic.
I need to select a password from accounts where username = $_POST['username']... i couldn't figure this one out, i keep getting resource id(2) instead of the desired password for the entered account. I need to pass that mysql through a mysql query function and save the returned value in the variable $realpassword. Thanks!
EDIT:
this code returned Resource id (2) instead of the real password
CODE:
<?php
$con = mysql_connect('server', 'user', 'pass');
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
echo '<br/> ';
// Create table
mysql_select_db("dbname", $con);
//Variables
//save the entered values
$enteredusername = $_POST['username'];
$hashedpassword = sha1($_POST['password']);
$sql = "SELECT password from accounts where username = '$enteredusername'";
$new = mysql_query($sql,$con);
echo "$new";
if (!mysql_query($sql,$con))
{
die('Error: ' . mysql_error());
}
mysql_close($con);
?>
It will be a lot better if you use PDO together with prepared statements.
This is how you connect to a MySQL server:
$db = new PDO('mysql:host=example.com;port=3306;dbname=your_database', $mysql_user, $mysql_pass);
And this is how you select rows properly (using bindParam):
$stmt = $db->prepare('SELECT password FROM accounts WHERE username = ?;');
$stmt->bindParam(1, $enteredusername);
$stmt->execute();
$result = $stmt->fetch(PDO::FETCH_ASSOC);
$password = $result['password'];
Also, binding parameters, instead of putting them immediately into query string, protects you from SQL injection (which in your case would be very likely as you do not filter input in any way).
I think your code looks something like this
$realpassword = mysql_query("SELECT password
from accounts where username = '$_POST[username]'");
echo $realpassword;
This will return a Resource which is used to point to the records in the database. What you then need to do is fetch the row where the resource is pointing. So, you do this (Note that I am going to use structural MySQLi instead of MySQL, because MySQL is deprecated now.)
$connection = mysqli_connect("localhost", "your_mysql_username",
"your_mysql_password", "your_mysql_database")
or die("There was an error");
foreach($_POST as $key=>$val) //this code will sanitize your inputs.
$_POST[$key] = mysqli_real_escape_string($connection, $val);
$result = mysqli_query($connection, "what_ever_my_query_is")
or die("There was an error");
//since you should only get one row here, I'm not going to loop over the result.
//However, if you are getting more than one rows, you might have to loop.
$dataRow = mysqli_fetch_array($result);
$realpassword = $dataRow['password'];
echo $realpassword;
So, this will take care of retrieving the password. But then you have more inherent problems. You are not sanitizing your inputs, and probably not even storing the hashed password in the database. If you are starting out in PHP and MySQL, you should really look into these things.
Edit : If you are only looking to create a login system, then you don't need to retrieve the password from the database. The query is pretty simple in that case.
$pass = sha1($_POST['Password']);
$selQ = "select * from accounts
where username = '$_POST[Username]'
and password = '$pass'";
$result = mysqli_query($connection, $selQ);
if(mysqli_num_rows($result) == 1) {
//log the user in
}
else {
//authentication failed
}
Logically speaking, the only way the user can log in is if the username and password both match. So, there will only be exactly 1 row for the username and password. That's exactly what we are checking here.
By seeing this question we can understand you are very very new to programming.So i requesting you to go thru this link http://php.net/manual/en/function.mysql-fetch-assoc.php
I am adding comment to each line below
$sql = "SELECT id as userid, fullname, userstatus
FROM sometable
WHERE userstatus = 1"; // This is query
$result = mysql_query($sql); // This is how to execute query
if (!$result) { //if the query is not successfully executed
echo "Could not successfully run query ($sql) from DB: " . mysql_error();
exit;
}
if (mysql_num_rows($result) == 0) { // if the query is successfully executed, check how many rows it returned
echo "No rows found, nothing to print so am exiting";
exit;
}
while ($row = mysql_fetch_assoc($result)) { //fetch the data from table as rows
echo $row["userid"]; //echoing each column
echo $row["fullname"];
echo $row["userstatus"];
}
hope it helps
try this
<?php
$con = mysql_connect('server', 'user', 'pass');
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
echo '<br/> ';
// Create table
mysql_select_db("dbname", $con);
//Variables
//save the entered values
$enteredusername = mysql_real_escape_string($_POST['username']);
$hashedpassword = sha1($_POST['password']);
$sql = "SELECT password from accounts where username = '$enteredusername'";
$new = mysql_query($sql,$con);
$row = mysql_fetch_array($new) ;
echo $row['password'];
if (!$new)
{
die('Error: ' . mysql_error());
}
mysql_close($con);
?>
<?php
$query = "SELECT password_field_name FROM UsersTableName WHERE username_field_name =".$_POST['username'];
$result = mysql_query($query);
$row = mysql_fetch_array($result);
echo $row['password_field_name'];
?>
$username = $_POST['username'];
$login_query = "SELECT password FROM users_info WHERE users_info.username ='$username'";
$password = mysql_result($result,0,'password');