No Database Selected - PHP & MySQL - php

I am trying to connect to my database and when I run the code, I get an error. Can anybody tell me what is wrong also any errors in my PHP code? Thanks.
Error: No database selected
PHP Code:
include('.conf.php');
$prof = $_GET['profile'];
$prof = addslashes(htmlentities($prof));
$query = "SELECT * FROM aTable where id = '".mysql_real_escape_string($prof)."'";
$info_array = mysql_query($query, $con) or die (mysql_error()).;
while($row = mysql_fetch_array( $info_array ))
{
echo $row['num1'];
echo "</br>";
echo $row['num2'];
echo "</br>";
echo $row['num3'];
echo "</br>";
echo $row['num4'];
};
mysql_close($con);
.conf.php file:
<?php
$conf['db_hostname'] = "xxx";
$conf['db_username'] = "xxx";
$conf['db_password'] = "xxx";
$conf['db_name'] = "xxx";
$conf['db_type'] = "mysql";
$con = mysql_connect('xxx', 'xxx', 'xxx') or die (mysql_error());
$db = mysql_select_db("aTable", $con);
?>

I had that problem and solved it with prefixing the table name with the database name, so
select * from database.table

Unless you have the password incorrect and need to fix it, run a GRANT statement to grant access to your database user:
GRANT ALL PRIVILEGES ON aTable.* TO xxx#localhost IDENTIFIED BY 'password_for_xxx';
The above grants all privileges. It's often better to restrict to only what's needed. For example, if you only intend to SELECT, but not modify data,
GRANT SELECT ON aTable.* TO xxx#localhost IDENTIFIED BY 'password_for_xxx';
Update
Since you have identified the database name as dedbmysql, change your mysql_select_db() call to:
$db = mysql_select_db("dedbmysql", $con);
Following this, the GRANT statements above are probably unnecessary, as your host may have already worked out the correct database permissions.

Related

Cant connect to sql database using xampp as local server

I have a problem connecting to sql database:
<?php
$servername = "localhost";
$username = "";
$password = "";
$myDB = "udemy";
// Create connection
$link = mysqli_connect($servername, $username, $password, $myDB );
if (mysqli_connect_error()){
die ("There was an error connecting to the database");
}
$query = "SELECT * FROM users";
if (mysqli_query($link, $query)){
echo "Query was successfull"
}
?>
Error is showing when I try to connect database named "udemy"...
Set username and password or add a new user in your mysql server and try with those user details.
To create new user :-
To create user in MySQL/MariaDB 5.7.6 and higher, use CREATE USER syntax:
CREATE USER 'new_user'#'localhost' IDENTIFIED BY 'new_password';
then to grant all access to the database (e.g. my_db), use GRANT Syntax, e.g.
GRANT ALL ON my_db.* TO 'new_user'#'localhost';
Where ALL (priv_type) can be replaced with specific privilege such as
SELECT, INSERT, UPDATE, ALTER etc
Then to reload newly assigned permissions run:
FLUSH PRIVILEGES;
Now set newly created username and password in your code.
Hope it helped.

php & mySQL connection error

I managed to get to phpMyAdmin and created a table .Then, I created the users with their username & password. Note that phpMyAdmin is running from a free website hosting area. It looks like it's done but I get this error message:
Access denied for user 'apache'#'localhost' (using password: NO)
<?php
#session_start();
include(".login.php");
print_r($conf);
$con = mysql_connect($conf['db_hostname'], $conf['$username'], $conf['1000']) or die (mysql_error());
mysql_select_db("usersA", $con);
$prof = '000';
//$prof = $_GET['profile'];
$result = "SELECT * FROM usersA WHERE id = '$user_id'";
$q = mysql_query($result, $con);
?>
try naming file properly..
include("login.php"); instead of include(".login.php");
what is
print_r($conf);
use
$con=mysql_connect("ip(like http://1.1.1.1)","user_name","password");

PHP / MYSQL troubleshoot I can't see whats wrong

This is me experimenting with PHP I'm new to it. I'm trying to see if a given username is already found in a mysql database here is my current code:
<?php
// $uname is the username I am trying to see if is the database
$uname = "djm";
//server info
$servuser = "root";
$servpass = "";
$db = "toob";
$server = "127.0.0.1";
//connecting to server
$db_handle = mysql_connect($server, $servuser, $servpass);
$db_found = mysql_select_db("toob", $db_handle);
//checking to see if ocnnected
if ($db_found) {
print("connected");
//defining my sql statement
$sql = "SELECT username FROM users WHERE username = $uname";
$result = mysql_query($sql);
if ($result) {
print("Yes");
} else {
print("No");
}
} else {
print("Can't connect to server");
}
I always prints no, i have managed to make to print yes by replacing :
$sql="SELECT username FROM users WHERE username = $uname";
$result=mysql_query($sql);
with
$sql="SELECT username FROM users WHERE username = 'djm'";
$result=mysql_query($sql);
However i need if to work off variables.
You're missing the quotes:
It's a good idea to escape the input before passing them as a mysql query.
Try:
$sql = "SELECT username FROM users
WHERE username = '". mysql_real_escape_string($uname)."'";
Currently, it will say "yes" if the query is executed successfully. Replace if ($result) with if(mysql_num_rows($result) > 0) if you're trying to check if the record exists.
Unrelated: mysql_* functions are now deprecated, and I'd suggest switching to mysqli or PDO.
Firstly you should escape all user input:
$username = mysql_real_escape_string($uname);
then you need to wrap SQL values in quotes:
$sql="SELECT username FROM users WHERE username = '$username'";
A bit away from your question; something on mysql referenced from php's website: "This extension is deprecated as of PHP 5.5.0, and is not recommended for writing new code as it will be removed in the future."
You would be better off using mysqli
replace this
if ($result)
with
if(mysql_num_rows($result) > 0)

PHP Select fields from database where username equals X

im having problems in PHP with selecting Infomation from a database where username is equal to $myusername
I can get it to echo the username using sessions from the login page to the logged in page.
But I want to be able to select things like 'bio' and 'email' from that database and put them into variables called $bio and $email so i can echo them.
This is what the database looks like:
Any ideas?:/
You should connect to your database and then fetch the row like this:
// DATABASE INFORMATION
$server = 'localhost';
$database = 'DATABASE';
$dbuser = 'DATABASE_USERNAME';
$dbpassword = 'DATABASE_PASSWORD';
//CONNECT TO DATABASE
$connect = mysql_connect("$server", "$dbuser", "$dbpassword")
OR die(mysql_error());
mysql_select_db("$database", $connect);
//ALWAYS ESCAPE STRINGS IF YOU HAVE RECEIVED THEM FROM USERS
$safe_username = mysql_real_escape_string($X);
//FIND AND GET THE ROW
$getit = mysql_query("SELECT * FROM table_name WHERE username='$safe_username'", $connect);
$row = mysql_fetch_array($getit);
//YOUR NEEDED VALUES
$bio = $row['bio'];
$email = $row['email'];
Note 1:
Dont Use Plain Text for Passwords, Always hash the passwords with a salt
Note 2:
I used MYSQL_QUERY for your code because i don't know PDO or Mysqli, Escaping in MYSQL is good enought but Consider Using PDO or Mysqli , as i don't know them i can't write the code with them for you
Simplistic PDO examples.
Create a connection to the database.
$link = new PDO("mysql:host=$db_server;dbname=$db_name", $db_user, $db_pw, array(PDO::MYSQL_ATTR_INIT_COMMAND => "SET NAMES utf8"));
$link->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
Use the $link variable when creating (preparing) and executing your SQL scripts.
$stmt = $link->prepare('insert into `history` (`user_id`) values(:userId)');
$stmt->execute(array(':userId' => $userId));
Code below will read data. Note that this code is only expecting one record (with 2 data elements) to be returned, so I'm storing whatever is returned into a single variable (per data element), $webId and $deviceId.
$stmt = $link->prepare('select `web_id`, `device_id` from `history` where `user_id` = :userId');
$stmt->execute(array(':userId' => $userId));
while($row = $stmt->fetch()) {
$webId = $row["web_id"];
$deviceId = $row["device_id"];
}
From the picture I can see you are using phpMyAdmin - a tool used to handle MySQL databases. You first must make a connection to the MySql server and then select a database to work with. This is shown how below:
<?php
$username = "your_name"; //Change to your server's username
$password = "your_password"; //Change to your server's password
$database = "your_database" //Change to your database name
$hostname = "localhost"; // Change to the location of your server (this will prolly be the same for you I believe tho
$dbhandle = mysql_connect($hostname, $username, $password)
or die("Unable to connect to MySQL");
echo "Connected to MySQL<br>";
$selected = mysql_select_db($database, $dbhandle)
or die("Could not select examples");
?>
Then you can write something like this:
<?php
$bio = mysql_query("SELECT bio FROM *your_database_table_name* WHERE username='bob' AND id=1");
?>
and
<?php
$email = mysql_query("SELECT email FROM *your_database_table_name* WHERE username='bob' AND id=1");
?>
Where *your_database_table_name* is the table in the database you selected which you are trying to query.
When I was answering your question, I was referencing this site: http://webcheatsheet.com/PHP/connect_mysql_database.php. So it might help to check it out as well.

MySQL parameter resource error

Here is my error:
Warning: mysql_query() expects parameter 2 to be resource, null given...
This refers to line 23 of my code which is:
$result = mysql_query($sql, $connection)
My entire query code looks like this:
$query = "SELECT * from users WHERE userid='".intval( $_SESSION['SESS_USERID'] )."'";
$result = mysql_query($query, $connection)
or die ("Couldn't perform query $query <br />".mysql_error());
$row = mysql_fetch_array($result);
I don't have a clue what has happpened here. All I wanted to do was to have the value of the users 'fullname' displayed in the header section of my web page. So I am outputting this code immediately after to try and achieve this:
echo 'Hello '; echo $row['fullname'];
Before this change, I had it working perfectly, where the session variable of fullname was echoed $_SESSION['SESS_NAME']. However, because my user can update their information (including their name), I wanted the name displayed in the header to be updated accordingly, and not displaying the session value.
Your $connection variable is NULL that's what your error message is referring to.
Reason being is that you have not called mysql_connect. Once called it will assign you a resource where you can set it to the $connection variable, thus being non-null.
As an example:
$connection = mysql_connect('localhost', 'mysql_user', 'mysql_password');
// now $connection has a resource that you can pass to mysql_query
$query = "SELECT * from users WHERE userid='".
intval( $_SESSION['SESS_USERID'] )."'";
$result = mysql_query($query, $connection)
include the mysql connections on your class file, for example:
connections/mysql.php
<?
$hostname_MySQL = "localhost";
$database_MySQL = "database";
$username_MySQL = "user";
$password_MySQL = "password";
$MySQL = mysql_pconnect($hostname_MySQL, $username_MySQL, $password_MySQL) or trigger_error(mysql_error(),E_USER_ERROR);
mysql_select_db($database_MySQL,$MySQL);
?>
class.php
<?
include "Connections/MySQL.php";
class utils {
public function myFunction()
{
global $MySQL;
$sql = "select * from table";
$rs = mysql_query($sql, $MySQL) or die(mysql_error());
$filas = mysql_fetch_assoc($rs);
$totalFilas = mysql_num_rows($rs);
...
}
}
?>
You have two ways of doing this, you need to use mysql_connect to connect to your database, you can pass this to mysql_query if you desire, if you don't pass anything to mysql_query PHP uses the last link opened from mysql_connect
$conn = mysql_connect("localhost", "mysql_user", "mysql_password");
$sql = "SELECT id as userid, fullname, userstatus
FROM sometable
WHERE userstatus = 1";
$result = mysql_query($sql);
Have you connected to your database? If so please show this code too.
For now, just try removing the $connection variable, like this:
$result = mysql_query($query);
And see where that gets you.
$connection is assigned the value of the database connection resource id. You don't have that in your script, so the value of $connection is NULL, and that is why you are getting the error. You need to connect to the database before using mysql_query(). You should be okay after that.
You need to do:
$connection=mysql_connect('host','user','pass');
if($connection === false) {
echo "Error in connection mysql_error()";
}

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