This question already has answers here:
Call to a member function prepare() on a non-object PHP Help [duplicate]
(8 answers)
Closed 6 years ago.
I have a following code
//dsn.php
//Object Oriented way
$servername = "localhost";
$username = "root";
$password = "password";
$dbname = "database";
//check connection
$conn = new mysqli($servername, $username, $password, $dbname);
if($conn->connect_error) {
die("could not connect:".$conn->connect_error);
}
//index.php
include 'dsn.php';
function a() {
$sql = "sql command";
$result = $conn->query($sql);
//working
$conn->close();
}
function b() {
$sql = "sql command";
$result = $conn->query($sql);
//not working
$conn->close();
}
This will display warning and notice that says:
Warning: mysqli::query(): Couldn't fetch mysqli
Notice: Trying to get property of non-object
Warning: mysqli::close(): Couldn't fetch mysqli
However this one works:
include 'dsn.php';
function a() {
$sql = "sql command";
$result = $conn->query($sql);
//working
$conn->close();
}
function b() {
include $dsn.php
$sql = "sql command";
$result = $conn->query($sql);
//working
$conn->close();
}
How do I use just one include file for DSN and use repeatedly on other functions?
EDIT
sorry I forgot to mention
function a($conn) {}
function b($conn) {}
I passed the variable $conn but it still displays the warning and notice I mentioned above
When you include a file, you can imagine that in the background it is just copy-pasting that code into the current document.
There are 2 problems with your code...
The $conn variable is not in scope inside function a or b.
Even if it was in scope and accessible, you are closing the connection after each query. A better way to do it is to open the connection, run all queries and close the connection when it is no longer needed.
The second piece of code you gave works because it is creating a new variable $conn inside of b(), but this is not ideal as it will create a new database connection every time you execute that function.
Something like this may suit your needs:
include 'dsn.php';
function a($conn) {
$sql = "sql command";
$result = $conn->query($sql);
return $result;
}
function b($conn) {
$sql = "sql command";
$result = $conn->query($sql);
return $result;
}
$aResult = a($conn);
$bResult = b($conn);
$conn->close();
Notice that we are only including 'dsn.php' once, and then passing around that existing connection to the functions that need it.
This is very simple. On page load, the connection file is included which makes $conn the connection object available to the remaining codes. The $conn is used by the functiona() and it is then closed at the end of the function. $conn->close(); destroys the database connection object means, $conn is no more object hence it should not be treated as object. But the Function b() is treatong it as database connection object and resulting into error.
But if you again include the connection file, inside the function b() then $conn becomes available to the function as local object. And works as it should.
Do not close the $conn() on the any function till you are dealing with DB.
function a() {
incDbConnectionFile();
$sql = "sql command";
$result = $conn->query($sql);
//working
$conn->close();
}
function b() {
incDbConnectionFile();
$sql = "sql command";
$result = $conn->query($sql);
//working
$conn->close();
}
function incDbConnectionFile() {
include 'dsn.php';
}
Related
I'm currently working on a project. It's almost done, there's only one big problem. I tested my code all the time with a xamp server on my computer, which worked perfectly fine. the goal is to run it (apache server, mysql database) on my raspberry pi. Now my project is finished, I came figured out the problem why my code doesn't work on my raspberry (at least not as I expected).
I turned on error reporting in PHP and came to this error message:
Notice: Trying to get property of non-object in /var/www/html/test.php on line 41
I use this function for all my SQL queries. Can someone provide a solution so I don't have to rewrite the whole code? Thanks in advance!
PS: this is just a piece of the code (the function where I pull the data out of the database + example of one of my queries)
<?php
// Enable debugging
error_reporting(E_ALL);
ini_set('display_errors', true);
$servername = "localhost";
$username = "root";
$password = "*****"; // I just dont want to give my sql database password its nothing wrong ;)
$dbname = "test";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
} else {
print_r("ok connection");
function sqlquery ($sql, $conn, $naamtabel) {
global $myArray;
global $stateLoop;
$stateLoop = "0";
$result = $conn->query($sql);
if ($result->num_rows > 0) { //line 41 in my code ==> do a while loop to fetch all data to an array
// output data of each row
while($row = $result->fetch_assoc()) {
$myArray[] = $row["$naamtabel"]; //alle data van kolom "tijd" in een array
}
$stateLoop = "1";
}
else { // if there are no results
}
}
$sql1 = "SELECT stopTijd FROM gespeeldeTijd WHERE naam = 'thomas' ORDER BY ID DESC LIMIT 1"; // get data with SQL query
sqlquery($sql1,$conn,"stopTijd");
if ( $stateLoop == "1") {
print_r("ok loop");
$date1 = $myArray["0"];
print_r($date1);
$myArray = [];
$stateLoop == "0";
}
}
?>
It pretty much looks like you have some sql error in your query; check if your field names in your database match those on the raspberry.
Seeing through your code it seems like you are pretty new to programming (which is no bad thing, I was once, too). So I made a few more modifications to your code showing you the prettiness of PHP
use "return" in function sqlquery instead of globals
check for errors after executing the code
use only one variable to check if data was loaded
I commented everything I changed
<?php
// Enable debugging
error_reporting(E_ALL);
ini_set('display_errors', true);
$servername = "localhost";
$username = "root";
$password = "*****";
$dbname = "test";
// Your function with some modifications
function sqlquery($sql, $conn, $naamtabel) {
$result = $conn->query($sql);
// Check for errors after execution
if(!$result)
die('mysqli error: '. htmlentities(mysqli_error($con)));
// If we have no data, we simply return an empty array
if($result->num_rows == 0)
return array();
// This is a variable we store the data we processed in
// We will return it at the end of our function
$myArray = null;
// Read all field data and store it $myArray
while($row = $result->fetch_assoc())
$myArray[] = $row[$naamtabel]; // if you use "$naamtabel" here, PHP first needs to interpret the string (= slower)
return $myArray;
}
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_error)
die("Connection failed: " . $conn->connect_error);
// Because we use "die" above we don't need an "else"-clause
print_r("ok connection");
$sql = "SELECT `stopTijd` FROM `gespeeldeTijd` WHERE `naam` = 'thomas' ORDER BY `ID` DESC LIMIT 1";
$data = sqlquery($sql, $conn, "stopTijd");
// $data will contain $myArray (see "return $myArray" in function sqlquery)
// Instead checking for $stateLoop being "1" we check if $data contains any values
// If so, we fetched some data
if(sizeof($data) >= 1) {
print_r("ok loop");
$date1 = $data[0]; // No "0", because we are trying to get index 0
print_r($date1);
$data = array(); // Are you sure this is nessecary?
} else {
echo 'No data returned from query!';
}
?>
Note: code tipped on my smartphone -> untested!
If you don't want to adapt the code I wrote, the important part for this question is:
if(!$result)
die('mysqli error: '. htmlentities(mysqli_error($con)));
Your error Notice: Trying to get property of non-object means "you are trying to get num_rows from $result, but $result is not an object, so it can't contain this property".
So to figure out why $result is not an object, you need to get the error from $conn->query - my code above probably won't fix your error, but it will display you one you can work with (+ it's too long for a comment)
If you have a more detailed error message and you can't solve it on your own, feel free to comment; I will update my answer!
I'm trying to execute an Insert query to write data into a Database. I'm using Mysqli and PHP.
The code looks OK for me. However, every time I go to the webpage to check if the form works, the query gets executed an a new row is created in the DB (empty).
I'm pretty sure there is something wrong with the last if statement. Could you advise?
BTW, the snippet is only for the PHP to execute the sql query, since the form is working just fine.
Thanks!
$servername = "localhost";
$username = "root";
$password = "mysqlpassword";
$dbname = "bowieDB";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$album = $_POST['album'];
$relyear = $_POST['relyear'];
$label = $_POST['label'];
$chart = $_POST['chart'];
$track1 = $_POST['track1'];
$track2 = $_POST['track2'];
$track3 = $_POST['track3'];
$track4 = $_POST['track4'];
$track5 = $_POST['track5'];
$sql = "INSERT INTO Albums (album, relyear, label, chart, track1, track2, track3, track4, track5)
VALUES ('$album', '$relyear', '$label', '$chart', '$track1', '$track2', '$track3', '$track4', '$track5')";
$result = mysqli_query($conn, $sql);
if ($conn->query($sql) === TRUE) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
$conn->close();
You are mixing Procedural and Object Orientated SQL interactions.
This is Procedural:
$result = mysqli_query($conn, $sql);
This is Object Orientated:
$conn->query($sql)
You can not use both with the same connection details, you should do one or the other throughout your code. The best one to use is Object Orientated approach, so rework the Procedural code to:
$result = $conn->query($sql);
if ($result) {
...
So actually you can simply remove the line starting $result = ... and let the IF statement query you already have handle itself.
Other notes:
Use MySQL error feedback such as checking if(!empty($conn->error)){print $conn->error;} after SQL statements. See example code below...
Use the following PHP error feedback too, set at the very top of your PHP page:
...
error_reporting(E_ALL);
ini_set('display_errors',0);
ini_set('log_errors',1);
you need to read up and be aware of SQL injection that can destory your database should someone POST data that also happens to be MySQL commands such as DROP.
Code for Comment:
if ($_SERVER['REQUEST_METHOD'] == "POST") {
//run SQL query you already have coded and assume
// that the form has been filled in.
$result = $conn->query($sql);
if ($result) {
//all ok
}
if(!empty($conn->error)) {
print "SQL Error: ".$conn->error;
}
}
use
1. if(isset($_POST['Submit'])){//your code here }
and
2. if($result){...
if you are using procedural method
This is a general example.
The application has 3 files.
conn.inc.php -- setting up the database connection
<?php
$db_host = "localhost";
$db_username = "root";
$db_pass = "";
$db_name = "hmt";
$conn = new mysqli($db_host, $db_username, $db_pass, $db_name);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
?>
func.inc.php -- file including functions
<?php
function load_module($module_name){
$sqlCmd = "SELECT content FROM modules WHERE name='$module_name' LIMIT 1";
$result = $conn->query($sqlCmd);
if ($result->num_rows > 0) {
while($row = $result->fetch_assoc()) {
$module_footer = $row["content"];
}
}else {
echo 'Error while loading module '.$module_name;
}
return $result;
$result->free_result();
}
?>
index.php -- the main page to display content
<?php
include 'conn.inc.php';
include 'func.inc.php';
if (!isset($_GET['page_name'])) { // if page_name is not set then reset it to the homepage
$page_name = 'module_footer';
}else{
$page_name = $_GET['module_footer'];
}
$module_content = load_module($page_name);
echo $module_content;
?>
Now my goal was to include functions inside the func.inc.php file and database into conn.inc.php, so as to keep separate and easier to read in the future.
My problem now is that the $conn variable declared in conn.inc.php cannot be used inside the function and it can't get my head around how to use it. I even tried using GLOBALS with no success.
The error for the files is this:
Notice: Undefined variable: conn in ./func.inc.php on line 4
Fatal error: Call to a member function query() on a non-object in ./func.inc.php on line 4
Which (I assume) is because the $conn variable is not in a global scope.
Now my question is. How can I keep the nested files but have the functions working? Is there a mistake in my approach or is it not possible to use a nested call to a mysql object?
Eventually you'll want to get into object-oriented coding but for now lets make what you have a little prettier.
When including files, you'll want to avoid things like global variables. They sound great, but end up being a pain when handling scope. So instead include a set of functions to call.
conn.inc.php
function getConnection(){
$db_host = "localhost";
$db_username = "root";
$db_pass = "";
$db_name = "hmt";
$conn = new mysqli($db_host, $db_username, $db_pass, $db_name);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
return $conn;
}
Then function.inc.php
<?php
function load_module($module_name){
$sqlCmd = "SELECT content FROM modules WHERE name='$module_name' LIMIT 1";
//Here is where we get our database connection.
$conn = getConnection();
$result = $conn->query($sqlCmd);
if ($result->num_rows > 0) {
while($row = $result->fetch_assoc()) {
$module_footer = $row["content"];
}
}else {
echo 'Error while loading module '.$module_name;
}
return $result;
$result->free_result();
}
And finally finish up with your index page the way it is. So now, any page that wants a database connection will need to include the conn.inc.php and simply call getConnection() to get a mysqli connection object.
Think of your program as being a series of individual functions working together. And eventually you'll get to it being a series of objects working together. Nothing should be just floating off in global space. Try to encapsulate everything in some sort of function or object to be called over and over with consistent results.
You'd have to do something like this:
$conn = '';
function connect() {
global $conn;
... do db stuff
}
But this is usually bad practice. A more common method is to use a singleton object to "carry" your db handle, and you new that singleton everywhere you need to do DB operations.
function do_something() {
$conn = new DBSingleton();
... do db stuff
}
Hi I have been learning php from this book PHP Solutions Dynamic Web Design Made Easy and gotten to the part where I have to work with mysqli api for databases.After writing a connection function and running the script I get this error:
This is my code:
function dbConnect($usertype , $connectionType = 'mysqli'){
$host = 'localhost';
$db = 'phpsols';
if($usertype == 'read'){
$user = 'psread';
$pwd = 'Aleczandru1989';
}elseif($usertype == 'write'){
$user = 'aleczandru';
$pwd = 'Aleczandru1989';
}else{
exit('Unrecognized type');
}
if($connectionType == 'mysqli'){
return new mysqli($host , $user , $pwd , $db) or die ('Cannot open database');
}else{
try{
return new PDO("mysql:host=$host;dbname=$db", $user, $pwd);
} catch (PDOException $e){
echo 'Cannot connect to database';
exit;
}
}
}
$conn = dbConnect('read');
$sql = 'SELECT * FROM images';
$result = $conn->query($sql) or die(mysqli_error()); //Line 5
$numRows = $result->num_rows;
Line 5 in this case refers to $result = $conn->query($sql) or die(mysqli_error());.
What Am I doing wrong here?
The $conn object you are attempting to create with dbConnect('read'); fails. If you would do a var_dump($conn); it probably shows it is not what you aspect it to be. The error is actually describing what is wrong. You try to access the query function with '->query(..' on $conn. But $conn has to be an object reference that actually has the query function. The points where this object will be created are:
return new mysqli($host , $user , $pwd , $db)
and
return new PDO("mysql:host=$host;dbname=$db", $user, $pwd);
Since you are showing a different error then
or die ('Cannot open database');
My guess it is actually gong wrong at
return new PDO("mysql:host=$host;dbname=$db", $user, $pwd);
And you will catch the exception. But the echo statement is not visible anymore due to the fatal error. You will have to do some debugging there!
I have no experience with PDO, but construction of the object seems ok. (but can this help you out: http://nl1.php.net/manual/en/class.pdo.php#84751) If the construction is ok, than check if your database engine is actually running :) ?
This question already has answers here:
mysql_fetch_array()/mysql_fetch_assoc()/mysql_fetch_row()/mysql_num_rows etc... expects parameter 1 to be resource
(31 answers)
Closed 11 months ago.
I get the error when trying to run this:
<?php
require_once('includes/DbConnector.php');
$connector = new DbConnector();
$result = $connector->query('SELECT title,content FROM staff_vacancies ORDER BY ordering LIMIT 0,100');
// Get an array containing the results.
// Loop for each item in that array
while ($row = $connector->fetchArray($result)){
echo $row['title'].'</h3>';
echo $row['content'];
}
?>
I have a linked file: DbConnector.php:
<?php
////////////////////////////////////////////////////////////////////////////////////////
// Class: DbConnector
// Purpose: Connect to a database, MySQL version
///////////////////////////////////////////////////////////////////////////////////////
require_once 'SystemComponent.php';
class DbConnector extends SystemComponent {
var $theQuery;
var $link;
//*** Function: DbConnector, Purpose: Connect to the database ***
function DbConnector(){
// Load settings from parent class
$settings = SystemComponent::getSettings();
// Get the main settings from the array we just loaded
$host = $settings['dbhost'];
$db = $settings['dbname'];
$user = $settings['dbusername'];
$pass = $settings['dbpassword'];
//the settings
$host = 'localhost';
$db = 'xxx';
$user = 'xxx';
$pass = 'xxx';
// Connect to the database
$this->link = mysql_connect($host, $user, $pass);
mysql_select_db($db);
register_shutdown_function(array(&$this, 'close'));
}
//*** Function: query, Purpose: Execute a database query ***
function query($query) {
$this->theQuery = $query;
return mysql_query($query, $this->link);
}
//*** Function: getQuery, Purpose: Returns the last database query, for debugging ***
function getQuery() {
return $this->theQuery;
}
//*** Function: getNumRows, Purpose: Return row count, MySQL version ***
function getNumRows($result) {
return mysql_num_rows($result);
}
//*** Function: fetchArray, Purpose: Get array of query results ***
function fetchArray($result) {
return mysql_fetch_array($result);
}
//*** Function: close, Purpose: Close the connection ***
function close() {
mysql_close($this->link);
}
}
?>
Does anyone know what the problem is?
Your query must have a problem which is causing $result to be an invalid resource.
Try checking for mysql_error() after the line on which you run your query.
Edit:
In fact, I would alter your DBConnector class function query() to something like the following, so that an identifiable error is thrown when you have a bad query:
function query($query) {
$this->theQuery = $query;
$queryId = mysql_query($query,$this->link);
if (! $queryId) {
throw new Exception(mysql_error().". Query was:\n\n".$query."\n\nError number: ".mysql_errno();
}
return $queryId;
}
I find this in a post, to me solved my problem.
Slds.
Yeah, Answer is simple, the query used is not a true result as it's a query inside of a getrow so to speak..
Here is the fix:
Find all lines that look like this:
mysql_fetch_array(mysql_query("...snip...";-) );
And just add a "#" in front of it so it looks like this:
#mysql_fetch_array(mysql_query("...snip...";-) );
Then do the same thing for the following lines..
Code:
mysql_num_rows(mysql_query("...snip...";-) );
Perform the same steps as above by adding the "#" to it so it looks like this:
#mysql_num_rows(mysql_query("...snip...";-) );
All this does it say "While doing xxx within yyy" Otherwise, it's dead due to missing result value. It's a PHP thing..
Works like a charm, took me 5 mins to rip the whole code apart and slap it all into Modernbill, Shares the same database and works perfectly for me.
This error means your query failed. mysql_query() returns false if an error occurred, you are then passing false to mysql_fetch_array() which is triggering the error message.
Your query could be failing due to a missing/wrong table or field. To see the detailed error, print out the result of mysql_error().
The mysql_* library is deprecated. It is recommended to upgrade to MySQLi or PDO.
// Load settings from parent class
$settings = SystemComponent::getSettings();
// Get the main settings from the array we just loaded
$host = $settings['dbhost'];
$db = $settings['dbname'];
$user = $settings['dbusername'];
$pass = $settings['dbpassword'];
//the settings
$host = 'localhost';
$db = 'xxx';
$user = 'xxx';
$pass = 'xxx';
Did you mean to reassign the connection vars? OR was that a few lines of stub code you forgot to take out? Or just an example to show what $settings contains?
Please provide the error from mysql_error(). Without that I can only guess... try escaping your field names?
$result = $connector->query('SELECT `title`,`content` FROM `staff_vacancies` ORDER BY `ordering` LIMIT 0,100');
Your query must have a problem which is causing $result to be an invalid resource.
Use this
<?php
require_once('includes/DbConnector.php');
$connector = new DbConnector();
$result = $connector->query('SELECT title,content FROM staff_vacancies ORDER BY ordering LIMIT 0,100');
// Get an array containing the results.
// Loop for each item in that array
if($result){
while ($row = $connector->fetchArray($result)){
echo $row['title'].'</h3>';
echo $row['content'];
}
}
?>