I have a record i want to remove from the database. I have so far been able save to the mysql database now i have several information in several rows, now suppose the information isnt the needed one, i want to delete it from the database. Thats what i am trying to achieve here
I tried this
<?php
session_start();
require_once('inc/config.php');
if(!isset($_SESSION['username'])){
header('Location: signon.php');
}
?>
<?php
require_once('inc/config.php');
$con = mysqli_connect($host, $user, $pass, $db) or die ('Cannot connect: '.mysqli_error());
$sql = "SELECT * FROM education_info WHERE username = '" . $_SESSION['username'] . "'";
$result = mysqli_query($con,$sql);
$row = mysqli_fetch_array($result);
$id = $row['id'];
$username = $row['username'];
$sql2 = "DELETE FROM education_info WHERE id = $id" ;
$result = mysqli_query($con,$sql);
mysqli_close($con);
header("Refresh:0; url=EDWE.php");
?>
Only that the information still remains present in the database, How do i go about deleting it completely, if not needed?
You are passing wrong variable while executing delete query:
$sql2 = "DELETE FROM education_info WHERE id = $id" ;
$result = mysqli_query($con,$sql2); //<---pass $sql2
Related
I need to make a Sign in form for my website. And I have to use MySQLi because MySQL will cause decaprated on my try.
So, here's the index.php code:
<?php
session_start();ob_start();
$con=mysqli_connect("localhost","root","","oos");
if (mysqli_connect_errno()) echo "Failed to connect to MySQL: " . mysqli_connect_error();
if(isset($_POST['signin']))
{
$username = $_POST['userid'];
$pass = $_POST['password'];
$query1 = "select * from admintb where adID = '$username' and adPass = 'password' ";
$result1 = mysqli_query($con,$query1) or die;
$co=0;
while($row=mysqli_fetch_assoc($result1)) $co++;
if($co==1)
{
$_SESSION['a']=$username;
header("Location: main_menu.php");
}
} ?>
The problem is, when I make $username="admin" and $password = "admin", it will go to main_menu.php alright. But when I try to do as above, base on my database, it won't go to main_menu.php.
How can I sign in, go to the main_menu.php using ID from my database?
Sorry, I already checked it, it's a stupid mistake. Inside this snippet:
$username = $_POST['userid'];
$pass = $_POST['password'];
$query1 = "select * from admintb where adID = '$username' and adPass = 'password' ";
$result1 = mysqli_query($con,$query1) or die;
fix to this:
$query1 = "select * from admintb where adID = '$username' and adPass = '$pass' ";
I need to display a reply data on my page from my 'feedback' field in my mysql table. I want each user to have a different 'feedback' response stored per row and fetched when the user logs into a page through a session. I have set up my database but find it difficult forming the php code to view the feedback on my page...please can someone help../
<?php
session_start();
if ($_SESSION['username'])
{
$con = mysqli_connect('localhost','root','');
if (!$con)
{
die('Could not connect: ' . mysqli_error($con));
}
mysqli_select_db($con,"loginsession");
$username = $_SESSION['username'];
$sql="SELECT * FROM users WHERE username = $username";
$result = mysqli_query($con,$sql);
$feedback = mysql_query("SELECT feedback FROM users WHERE username='$username'");
echo $feedback;
}
else
header("Location: index.php");
?>
$feedback in this case is not a string, its a mysql resource. You need to fetch each row individually with something like:
echo "<PRE>";
while ($row = mysql_fetch_assoc($feedback)) {
print_r($row);
}
Also you should put $username through mysql_real_escape_string() or else your code may be vulnerable to SQL injection attacks.
Edit: (Disclaimer) The method you are using and my suggestion are very outdated and have been depreciated in php5.5 I suggest you look into prepared statements.
$sql = mysql_query("SELECT feedback FROM users WHERE username='{$username}' LIMIT 1");
$feedback = mysql_fetch_assoc($sql);
echo $feedback[0];
<?php
session_start();
if ($_SESSION['username'])
{
$con = mysqli_connect('localhost','root','');
if (!$con)
{
die('Could not connect: ' . mysqli_error($con));
}
mysqli_select_db($con,"loginsession");
$username = $_SESSION['username'];
$sql='SELECT feedback FROM users WHERE username = "'.$username.'"';
$result = mysqli_query($con,$sql);
while($row = mysqli_fetch_array($result))
{
echo $row['feedback'];
}
}
else
header("location: index.php");
?>
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');
I want to get a value from a MySQL database and put it into a PHP variable.
I tried this:
$data = mysql_query("SELECT userid FROM ao_user " .
"WHERE username = '{$this->_username}' " .
"AND password = '{$this->_password}' AND display = '{$this->_display}'");
The code says invalid username/password.
Here is the user login code:
<?php
$username = "Nynex71";
mysql_connect("localhost", "root", "test") or die(mysql_error());
mysql_select_db("test") or die(mysql_error());
$result = mysql_query("SELECT display FROM ao_user " .
"WHERE username = '{$username}'") or die(msyql_error());
$row = mysql_fetch_assoc($result);
echo $row['display'];
?>
and
public function getDisplay()
{
mysql_connect("localhost", "root", "test") or die(mysql_error());
mysql_select_db("test") or die(mysql_error());
$result = mysql_query("SELECT display FROM ao_user " .
"WHERE username = '{$this->_username}'");
$row = mysql_fetch_assoc($result);
$this->_display = $row['display'];
$_SESSION['display'] = $this->_display;
}
The program does not put any words into the PHP variable. What am I doing wrong and how do you do this?
mysql_query returns a result handle, not the value you selected. you have to first fetch a row, then retrieve the value from that row:
$result = mysql_query("SELECT ...") or die(msyql_error());
$row = mysql_fetch_assoc($result);
echo $row['userid'];
I am sure this is a stupid question but I can't figure out how to do it. I need to retrieve a large block of text that is stored in a database. I know how to connect to the database and submit the sql query. What I can't figure out how to do is have it store that text in a string so I can have it echo later on the page.
What I have so far
$db_name = "j_db";
$table_name = "contacts";
$connection = #mysql_connect("localhost", "*****", "****") or die(mysql_error());
$db = #mysql_select_db($db_name, $connection) or die(mysql_error());
$sql = "SELECT meaning
FROM $table_name
WHERE card = (php variable here)";
$result = #mysql_query($sql, $connection) or die(mysql_error());
Using mysql_fetch_array()
while( $row = mysql_fetch_array($result) ){
$meaning = $row["meaning"];
echo $meaning;
}
Look at the mysql_fetch_* functions. mysql_fetch_assoc is a popular choice.
<?php
$url= '' ;
$username = '' ;
$password = '' ;
$dbname = '' ;
$connection = mysql_connect('$url' , '$username', '$password');
mysql_select_db('$dbname' , $connection);
$result = mysql_query("SELECT textColumnName FROM tableName", $connection);
while($row = mysql_fetch_array($result)){
echo $row['textColumnName'];
}
?>