ILLEGAL & UNINITIALIZED string offset error [duplicate] - php

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).

Related

How do I solve the "Undefined variable" error on php [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 6 years ago.
login.php
<?php
function login()
{
$con = mysqli_connect("localhost", "XXX", "XXX") or die('Could not connect to server');
mysqli_select_db('$con', "store") or die('Could not connect to database');
}
?>
validate.php
Line 10 - ERROR - Notice: Undefined variable: con in C:\wamp\www\store\admin\validate.php
Line 10 - Warning: mysqli_query() expects parameter 1 to be mysqli, null given in C:\wamp\www\store\admin\validate.php
Line 12 - Warning: mysqli_num_rows() expects parameter 1 to be mysqli_result, null given in C:\wamp\www\store\admin\validate.php
<?php
session_start();
include ("../mylibrary/login.php");
login();
$userid = $_POST['userid'];
$password = $_POST['password'];
$query = "SELECT userid, name from admins where userid = '$userid' and password = PASSWORD('$password')";
$result = mysqli_query($con, $query); (**Line 10)
if (mysqli_num_rows($result) == 0)(**Line 12)
{
echo "<h5>Sorry, your account was not validated.</h5><br>\n";
echo "Try again<br>\n";
} else
{
$_SESSION['store_admin'] = $userid;
header("Location: admin.php");
}
?>
I tried to figure out something wrong. Let me know thanks.
See this
mysqli_select_db('$con'
^ ^
Variable don't get parsed in single quotes.
Either remove them or use double " quotes.
Note: Make sure that all your POST arrays contain values and that the form you're using is indeed using a POST method and elements hold their respective name attributes.
You're also open to an SQL injection here.
Use a prepared statement.
References:
https://en.wikipedia.org/wiki/SQL_injection
https://en.wikipedia.org/wiki/Prepared_statement
Also consider using password_hash() to store your passwords:
http://php.net/manual/en/function.password-hash.php (Read the entire manual).
You are mixing up local and global variables.
$con is not available outside the login function.
If you add global $con; as the first line of the login function, its available outside of the function as well.
Have a look at the php manual: http://php.net/manual/en/language.variables.scope.php
PS: While its not the source of the problem, read Fred's answer too.

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.

unable to get data from sql using PDO [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.
Am using PDO method to get data from SQL But am unable to get data from SQL database since am new to PDO its little hard to understand
but i did the following code but Doesn't work can someone help me
CODE
<?php
$servername = "localhost";
$username = "sanoj";
$password = "123456";
$dbname = "test";
try {
$conn = new PDO("mysql:host=$servername;dbname=$dbname", $username, $password);
$conn->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
$stmt = $conn->prepare("SELECT * FROM members");
$stmt->execute();
// set the resulting array to associative
$membid=($_GET['memberID']);
$email=($_GET['email']);
$state=($_GET['username']);
}
catch(PDOException $e) {
echo "Error: " . $e->getMessage();
}
echo $membid;
echo $email;
echo $state;
$conn = null;
echo "</table>";
?>
ERROR I GET
Notice: Undefined index: memberID in C:\Users\sanoj\Documents\NetBeansProjects\godaddy optimized\data.php on line 14
Call Stack
# Time Memory Function Location
1 0.0020 132392 {main}( ) ..\data.php:0
( ! ) Notice: Undefined index: email in C:\Users\sanoj\Documents\NetBeansProjects\godaddy optimized\data.php on line 15
Call Stack
# Time Memory Function Location
1 0.0020 132392 {main}( ) ..\data.php:0
( ! ) Notice: Undefined index: username in C:\Users\sanoj\Documents\NetBeansProjects\godaddy optimized\data.php on line 16
Call Stack
# Time Memory Function Location
1 0.0020 132392 {main}( ) ..\data.php:0
Using $_GET variable won't make you get your data from your database using PDO. You were doing right up until $stmt->execute();. What you need to do next is:
while($result = $stmt->fetch(PDO::FETCH_ASSOC))
{
foreach($result as $key => $value)
echo $key.': '.$value.'<br/>';
echo '<hr/>';
}
To get your data.
$_GET array is to fetch your data from the URL or from a GET Ajax call.

database connection error fatal errors

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()

index error in database reading PHP script [duplicate]

This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
PHP: “Notice: Undefined variable” and “Notice: Undefined index”
So Im back with another php error, that I cant solve. But Im getting better:)
I have created a simple script that stores images in a database, I have no problems to store the file, but when Im reading the file i get an index error. It says
Notice: Undefined index: id in C:\wamp\www\gallery2\show.php on line
13
I cant really get what the problem is since Im thinking that everything is correct !?
the code for showing the images are
<?php
$username = "root";
$password = "";
$host = "localhost";
$database = "guestbook";
#mysql_connect($host, $username, $password) or die("Can not connect to database: ".mysql_error());
#mysql_select_db($database) or die("Can not select the database: ".mysql_error());
$id = $_GET['id'];
if(!isset($id) || empty($id)){
die("Please select your image!");
}else{
$query = mysql_query("SELECT * FROM tbl_images WHERE id='".$id."'");
$row = mysql_fetch_array($query);
$content = $row['image'];
header('Content-type: image/jpg');
echo $content;
}
?>
You are trying to access $_GET['id']. However, if no argument id is present in the querystring of the request, the index id will not be available in the $_GET superglobal. That's why you receive the notice.
So you should be doing something like this:
$id = !empty( $_GET[ 'id' ] ) ? $_GET[ 'id' ] : null;
It is telling you the value id does not exist in the $_GET array. Does the URL you are accessing have a ?id=something on it?
The variable...
$_GET['id'];
...doesn't contain a value, therefore it is not defined.
Check if your form mechanism leading to this $_GET is working, maybe var_dump it.
If your script should cope with an empty variable there, maybe check this variable first, something like this:
$id = "";
if (isset($_GET['id']) { $id = $_GET['id']; }

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