database connection error fatal errors - php

having an issue with connecting to my DB (been a long long day) - so anyway I am creating a simple search query into my DB but getting scripting errors - so here is my code:
<?php
mysqli_connect("localhost", "my username", "my password");
mysqli_select_db("smudged");
$search = mysqli_real_escape_string(trim($_POST['searchterm']));
$find_image = mysqli_query("SELECT * FROM 'smd_images' WHERE 'img_description' LIKE'%$search%'");
while($row = mysqli_fetch_assoc($find_image))
{
$name = $row['name'];
echo "$name";
}
?>
Here is my error:
search.php on line 4 Warning: mysql_select_db(): A link to the server could not be established in /marble/search.php on line 4 Fatal error: Call to undefined function mysql_real_escape_sring() in /marble/search.php on line 6

Typo. Instead of mysql_real_escape_sring() you probably meant mysql_real_escape_string()

Related

ILLEGAL & UNINITIALIZED string offset error [duplicate]

This question already has answers here:
"Notice: Undefined variable", "Notice: Undefined index", "Warning: Undefined array key", and "Notice: Undefined offset" using PHP
(29 answers)
Illegal string offset Warning PHP
(17 answers)
Closed 5 years ago.
I'm having trouble fixing my errors, the functions are working fine but i need to get rid of the errors
i have this following errors:
Warning: Illegal string offset 'userID' in C:\xampp\htdocs\checkout.php on line 15
Notice: Uninitialized string offset: 0 in C:\xampp\htdocs\checkout.php on line 15
Fatal error: Call to a member function fetch_assoc() on boolean in C:\xampp\htdocs\checkout.php on line 19
and heres my code:
CHECKOUT.PHP
<?php
// include database configuration file
include 'dbConfig.php';
include 'login.php';
// initializ shopping cart class
include 'Cart.php';
$cart = new Cart;
// redirect to home if cart is empty
if($cart->total_items() <= 0){
header("Location: index.php");
}
// set customer ID in session
$_SESSION['sessCustomerID'] = $sessData['userID']; //this is the ID for the logged in user
// get customer details by session customer ID
$query = $db->query("SELECT * FROM users WHERE id =".$_SESSION['sessCustomerID']);
$custRow = $query->fetch_assoc();
?>
LOGIN.PHP
<?php
session_start();
$sessData = !empty($_SESSION['sessData'])?$_SESSION['sessData']:'';
if(!empty($sessData['status']['msg'])){
$statusMsg = $sessData['status']['msg'];
$statusMsgType = $sessData['status']['type'];
unset($_SESSION['sessData']['status']);
}
?>
<div class="container">
<?php
if(!empty($sessData['userLoggedIn']) && !empty($sessData['userID'])){
include 'user.php';
$user = new User();
$conditions['where'] = array(
'id' => $sessData['userID'],
);
$conditions['return_type'] = 'single';
$userData = $user->getRows($conditions);
?>
DBCONFIG.PHP
<?php
//DB details
$dbHost = 'localhost';
$dbUsername = 'root';
$dbPassword = '';
$dbName = 'dbblair';
//Create connection and select DB
$db = new mysqli($dbHost, $dbUsername, $dbPassword, $dbName);
if ($db->connect_error) {
die("Unable to connect database: " . $db->connect_error);
}
?>
Line 15:
$_SESSION['sessCustomerID'] = $sessData['userID'];
Both error messages that refer to this line are very clear.
Warning: Illegal string offset 'userID' in C:\xampp\htdocs\checkout.php on line 15
$sessData is a string (this is what the error says and line #3 of login.php confirms this statement).
Individual string characters can be accessed using the square bracket syntax (similar to arrays) but the offset must be an integer. The offset is a string in your code, and this is what the error says (a string is not a legal value for the offset).
Notice: Uninitialized string offset: 0 in C:\xampp\htdocs\checkout.php on line 15
Because it expects an integer for the offset and you use a string instead, the interpreter converts the string to number and the result is 0. The message says that the offset 0 does not exist in the string (and this is correct, as $sessData is ''). As a result, $_SESSION['sessCustomerID'] is initialized with the empty string.
For PHP the two messages above are just a warning and a notice (i.e. the script can continue) but in fact they reveal a serious error in your code.
The string offset and the way you use $sessData in login.php tell that $sessData must always be an array. It's unexplainable why do you set it to an empty string. Line #3 of login.php should read:
$sessData = !empty($_SESSION['sessData']) ? $_SESSION['sessData'] : array();
Lines 18 and 19:
$query = $db->query("SELECT * FROM users WHERE id =".$_SESSION['sessCustomerID']);
$custRow = $query->fetch_assoc();
The error:
Fatal error: Call to a member function fetch_assoc() on boolean in C:\xampp\htdocs\checkout.php on line 19
Information about this error was asked and answered dozen of times on [so]. I won't repeat the answer here.
mysqli::query() returns
FALSE when the query has syntax errors or refers to object that does not exist in the database. In this case, the query ends with WHERE id= because $_SESSION['sessCustomerID'] is empty, as explained above.
But $_SESSION['sessCustomerID'] is still empty on the first page load, even if you fix the initialization of $sessData. Apart from using prepared statements (see the linked answer for details), you should not issue a query to the database if you know in advance you won't get any result (this happens when $_SESSION['sessCustomerID'] is empty).

MySQLi warning with php7

I'm getting the following errors from php7 on a new Ubuntu server:
Notice: Trying to get property of non-object in ~/tLogServ.php on line 14
Warning: mysqli::query(): Couldn't fetch mysqli in ~/tLogServ.php on line 17
Fatal error: Uncaught Error: Call to a member function fetch_assoc() on null in ~/tLogServ.php:18
Stack trace:
0 {main}thrown in ~/tLogServ.php on line 18
Here's the tLogServ.php
<?php
header("Access-Control-Allow-Origin: *");
header("Content-Type: application/json; charset=UTF-8");
$request = json_decode( file_get_contents('php://input') );
$variable = $request->data;
// echo($variable);
$result = $conn->query("SELECT password FROM login where userID LIKE 'technician'");
$rows = $result->fetch_assoc();
$passHashed = $rows['password'];
if(password_verify($variable, $passHashed)){
$loginMess = "success";
}else{
$loginMess = "fail";
}
echo json_encode($loginMess);
$conn->close();
?>
and my connection script
<?php
DEFINE ('user', '$%^*(');
DEFINE ('pass', '^*&%*');
DEFINE ('host', '1*&^*&^');
DEFINE ('DB', '^*%*(&%^');
$conn = new mysqli(host, user, pass, DB) OR die('Fail Whale ' . mysqli_connect_error());
?>
The notice will go away with input, but I'm unsure about and the warning and the fatal error it causes.
This code works without issue on Ubuntu 14 with PHP5. I've uncommented
extension=php_mysqli.dll
in php.ini. This is obviously a compatibility issue, but I'm unsure if I need to re-write my code or if it's a matter of a simple setting that I can't find.
The issue was two stupid mistakes.
1)The user I referenced in my connection was not given proper rights.
and
2) 'Host' was defined as the IP of the localhost. For whatever reason this worked on Cat connection, but not on WiFI (both had static Ip's).
For others with the same/similar issue. Verify that your php.ini mysqli extension is enabled/uncommmented. As A.L was onto the issue was with mySQL settings not PHP/Ubuntu.

Connecting mysql database using PDO in external PHP file [duplicate]

This question already has answers here:
"Notice: Undefined variable", "Notice: Undefined index", "Warning: Undefined array key", and "Notice: Undefined offset" using PHP
(29 answers)
Closed 7 years ago.
I have an Index.php which has a form for fetching user details when that form is submitted it fires the data to a new program.php for validation in program.php I've linked db.php in which I've the connection to the database, code of db.php is given below:
<?php
$link=mysql_connect('localhost', 'root', '') or die ("mysql_connect_error()");
$dbselect=mysql_select_db('test',$link) or die ("Error while connecting the database");
?>
since using it this way sql injections are possible, so I tried changing it to code given below:
<?php
$hostname='localhost';
$username='root';
$password='';
try
{
$dbh = new PDO("mysql:host=$hostname;dbname=test",$username,$password);
$dbh->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION); // <== add this line
$dbh = null;
}
catch(PDOException $e)
{
echo $e->getMessage();
}
?>
but I am getting an error when I connect submit the form. Inside my program.php I have called db.php by include "db.php";. Since I am new to PDO, I am not sure where am I going wrong.
Updated program.php code
<?php
if($_POST)
{
include "link_db.php";
if ($_POST[admin_sign_up])
{
$fname=$_POST[fname];
$lname=$_POST[lname];
$id =$_POST[id];
$id_pass=$_POST[id_pass];
$sql="insert into admin_database(fname, lname, id, id_pass)
value ('$fname','$lname','$id','$id_pass')";
mysql_query($sql);
$error=mysql_error();
if(empty($error))
{
echo "<script>alert('Registration Successful...')</script>";
header("Location:index.php",true);
}
else
{
echo "Registration Failed...<br> Email Id already in use<br>";
echo "<a href='failed.php'>Click to SignUp again</a>";
}
}
if ($_POST[admin_login])
{
$id =$_POST[id];
$id_pass=$_POST[id_pass];
$sql="select * from admin_database where id = '$id' and id_pass= '$id_pass'";
$result=mysql_query($sql);
echo mysql_error();
$row=mysql_fetch_array($result);
$rowcnt=mysql_num_rows($result);
if($rowcnt==1)
{
session_start();
$_SESSION['id']=$id;
$_SESSION['fname']=$row['fname'];
$_SESSION['lname']=$row['lname'];
$_SESSION['varn']="Y";
echo "Login Successfully....";
header("Location:home.php",true);
}
else
{
$id =$_POST[id];
$id_pass=$_POST[id_pass];
$sql="insert into adminfailure(id, id_pass, date_time)
value ('$id','$id_pass',NOW())";
mysql_query($sql);
$error=mysql_error();
if(empty($error))
{
Echo "Invalid Login ID or Password....";
header("Location:fail.php",true);
}
else
{
echo "incorrect details";
}
}
}
if ($_POST[logout])
{
header("location:destroy.php",true);
}
}
?>
Updated Errors which I get
Notice: Use of undefined constant test_sign_up - assumed 'test_sign_up' in B:\XAMPP\htdocs\test\program.php on line 6
Notice: Undefined index: test_sign_up in B:\XAMPP\htdocs\test\program.php on line 6
Notice: Use of undefined constant test_login - assumed 'test_login' in B:\XAMPP\htdocs\test\program.php on line 32
Notice: Use of undefined constant id - assumed 'id' in B:\XAMPP\htdocs\test\program.php on line 35
Notice: Use of undefined constant id_pass - assumed 'id_pass' in B:\XAMPP\htdocs\test\program.php on line 36
No database selected
Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in B:\XAMPP\htdocs\test\program.php on line 41
Warning: mysql_num_rows() expects parameter 1 to be resource, boolean given in B:\XAMPP\htdocs\test\program.php on line 42
Notice: Use of undefined constant id - assumed 'id' in B:\XAMPP\htdocs\test\program.php on line 56
Notice: Use of undefined constant id_pass - assumed 'id_pass' in B:\XAMPP\htdocs\test\program.php on line 57
incorrect details
Notice: Use of undefined constant logout - assumed 'logout' in B:\XAMPP\htdocs\test\program.php on line 73
Notice: Undefined index: logout in B:\XAMPP\htdocs\test\program.php on line 73
In your code, you first create a connection to the database, then you set it to null.
Whenever you try to access the $dbh object after that, it will be null.
$dbh = new PDO("mysql:host=$hostname;dbname=test",$username,$password);
$dbh->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
$dbh = null; // <= Right here.
Remove the $dbh = null; line, and you should be able to use the object as intended.
The $dbh object it not just a "link" as you do in your mysql_* code, but it is a object that you use to call the database, this is not the same object that you use in your mysql_* calls.
i.e., You can not use the earlier mysql_* code and just pass the pdo object into the call instead of the mysql link.
So the code will differ a bit from your earlier code.
Example:
// Earlier code using `mysql_* API`:
$sql="select * from admin_database where id = '$id' and id_pass= '$id_pass'";
$result=mysql_query($sql);
$row=mysql_fetch_array($result);
// Would look something like this using PDO:
$statement = $dbh->prepare('SELECT * FROM admin_database WHERE id =:id AND id_pass =:idpass');
// Here you can either use the bindParam method, or pass the params right into the execute call:
$statement->execute(array('id' => $id, 'idpass' => $id_pass);
$row = $statement->fetch();
I'd recommend reading up on PDO in the docs if you have issues with converting the code.
Further recommendations:
When you are including a file like this, one you only want to be included once per script run, its always a good idea to make sure that it is only included once. This can be done by using the include_once keyword instead of just include. Now, if you use include, this will include the script if possible, if it cant, it will keep run the script, and the script will crash when you try to use the varaiables set in the file.
Instead of using include in this case, I would recommend using the require (or rather require_once) keyword. Which will include the file, and if it cant, stop execution of the script and display an error message (if you have error reporting on).
You have to change not only db.php but ALL the queries over your code. And always use prepared statements to pass variables to queries. Otherwise PDO won't protect you from injections.
At the moment I am writing tutorial on PDO, it is still incomplete but can give you basics you may start from.

mysqli_query not accepting my mysqli_connect and proper use of parameters of mysqli_query

i have set my configdb.php on a different page and include it on my other php pages..
here is my configdb.php
<?php
$hostname ="localhost";
$username ="root";
$password ="";
$db ="practicedb";
$connect = mysqli_connect($hostname,$username,$password) or die("cannot connect to server");
mysqli_select_db($connect,$db) or die("database not found!");
?>
these are the errors that i get:
Notice: Undefined variable: configdb in /Applications/XAMPP/xamppfiles/htdocs/practicesystem/add.php on line 14
Warning: mysqli_query() expects parameter 1 to be mysqli, null given in /Applications/XAMPP/xamppfiles/htdocs/practicesystem/add.php on line 14
Warning: mysqli_num_rows() expects parameter 1 to be mysqli_result, null given in /Applications/XAMPP/xamppfiles/htdocs/practicesystem/add.php on line 15
Notice: Undefined variable: configdb in /Applications/XAMPP/xamppfiles/htdocs/practicesystem/add.php on line 28
this is my add.php where i INSERT items into database from the $_POST method from a previous php page..
<?php
include "configdb.php";
$studid=$_POST['studid'];
$lastname=mysql_real_escape_string($_POST['lastname']);
$firstname= mysql_real_escape_string($_POST['firstname']);
$middlename= mysql_real_escape_string($_POST['middlename']);
$email=$_POST['email'];
$check = "SELECT * from studinfo where stud_id = '".$studid."'";
$qry = mysqli_query($configdb,$check);
$num_rows = mysqli_num_rows($qry);
if($num_rows > 0){
// Here we are checking if username is already exist or not.
echo "The person you have entered is already existing. Please try again.";
echo 'Try Again';
exit;
}
$query = "INSERT INTO studinfo (stud_id,lastname,firstname,middlename,email) VALUES ('".$studid."','".$lastname."','".$firstname."','".$middlename."','".$email."');";
//echo $query;
mysqli_query($configdb, $query);
echo "Thank You for Registration.";
echo 'Click Here to login you account.';
exit;
?>
i don't know and i am not sure what to put on the first parameter of mysqli_query..
i tried putting this code $con=mysqli_connect("localhost","root","","practicedb"); it worked but its not practical putting that on every php page where i should connect to the database...
Yet another question on a silly typo...
$connect = mysqli_connect( ...
vs.
$qry = mysqli_query($configdb,$check);
so the error message clearly says: Undefined variable: configdb

Fatal error: Call to undefined function mysqli_master_query()

I am having a problem when querying my database. I am getting the following error:
Fatal error: Call to undefined function mysqli_master_query() in /home/**/**/**/edit/add.php on line 4
Here is my PHP code..
<?php
$db = mysqli_connect("localhost","XXX","XXX","XXX") or die(mysql_error());
if(mysqli_master_query($db, "UPDATE users SET order =1 WHERE id =1")){
echo "Y";
}else{
echo "N";
}
?>
mysqli_master_query()
This function has been DEPRECATED and REMOVED as of PHP 5.3.0.
Which version you are using ?
mysqli_master_query() function has been DEPRECATED and REMOVED as of PHP 5.3.0.
Refer : http://php.net/manual/en/function.mysqli-master-query.php

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