I have a a form that pulls data from a database(mysql to be specific) and echos the data into the value section of <input> tags. It doesn't seem to be working I have coded a view section of my website to do the same thing but from a different table in my database. I use the same code to make making changes easy and if another developer works on my site in the future. Anyway it doesn't seem to be working I'm not sure why though.
The full error I get:
Warning: mysqli_query() expects parameter 1 to be mysqli, null given in /home/caseol5/public_html/jj/admin/news_update.php on line 9
Here is line 9 that the error is referring to:
$result = mysqli_query($link,$sql);
I know that both of those function are not null as I did:
echo $link
echo $sql
before that line after I started feting the error and they both are not null.
Here is the full code segment:
$nid = $_GET['nid'];
include ("../sql/dbConnect.php");
$sql = "SELECT * FROM jj_news WHERE news_id = $nid";
echo "<p>The SQL Command: $sql </p>";
echo "<p>Link: $link </p>";
$result = mysqli_query($link,$sql);
if (!$result)
{
echo "<h1>You have encountered a problem with the update.</h1>";
die( "<h2>" . mysqli_error($link) . "</h2>") ;
}
$row = mysqli_fetch_array($result);
$ntitle = $row['news_title'];
$ntline = $row['news_titleline'];
$ndesc = $row['news_desc'];
$nother = $row['news_other'];
I have looked into mysqli_query and I can't find anything I'm missing. I have also tired breaking the code down (and running parts of it and it gives the same error. My guess is it something small that I missed. I've looked at other question on this site that do that are a little similar but none seem to help. I've been looking at this for a while now and need another pair of eyes.
Update
As requested the contents of my dbconnect.php file:
$hostname = "localhost";
$username = "caseol5_jjoes";
$database = "caseol5_jj_site";
$password = "password1";
$link = mysqli_connect($hostname, $username, $password, $database);
$link = mysqli_connect($hostname,$username,$password,$database) or die("Error " . mysqli_error($link));
if (!$link)
{
echo "We have a problem!";
}
As clearly stated in the error message, mysqli_querydocs expects the first parameter to be a mysqli resource. In your case, this parameter is called $link but it holds a null value. A proper mysqli resource is normally obtained from connecting with the database by making use of mysqli_connectdocs
I expect the ../sql/dbConnect.php file holds the logic to connect with the database. Verify whether the $link variable is indeed initialized there. If it's not there, try to find an occurrence of mysqli_connect - maybe the resource is set to a different variable.
Without knowing what exactly is in ../sql/dbConnect.php, your problem right now is that you do not have a valid mysqli resource to use for mysqli_query.
Related
My automation code encountered a subject-like problem when calling a particular function.
It occurs randomly during execution, not immediately after execution, and after the problem occurs, the error "mysql_num_rows() expect parameters 1 to be resource, boolean given in" occurs and normal behavior is not performed.
The OS is Ubuntu 18.04.
PHP version is 5.6.40.
Mysql version is 5.7.38.
The problematic function code.
$conn = mysql_connect("127.0.0.1","ID","PW");
if($conn ==NULL)
{
echo "<script> alert(\" Error : DB 연결 에러입니다. 불편을 드려 죄송합니다. 관리자에게 문의 하십시요\"); </script>";
return $conn;
}
mysql_select_db("mysql");
mysql_query("set session character_set_connection=utf8;");
mysql_query("set session character_set_results=utf8;");
mysql_query("set session character_set_client=utf8;");
$dbname = "test_table";
$str = "select * from $dbname where 1";
$leak_result = mysql_query($str);
$leak_num1 = mysql_num_rows($leak_result);
Please give me a solution.
This error is appearing because of the maximum number of database connection requests have made.
Your Ubuntu 18 with PHP server it's making a lot of requests.
The "PID=3629" you read it's the Process Id that generates the error.
The method you are using mysql_query it's old and deprecated.
Now PHP uses mysqli method like this
$connect = mysqli_connect( $host, $user, $pass, $DBname );
$query = 'SELECT .... '; //here your query
$result = mysqli_query($connect,$query);
Like said in comments, if there's an error mysqli_query gives back a false result
PHP OFFICIAL SITE
If you want a more complete answer on the error use the mysqli_error() function.
Good luck
I am working on a login script with prepared statements in PHP procedural mysqli syntax. Here is my current code:
<?php
include "/ssincludes/functions.php";
$host = HOST;
$username = USER;
$password = PASSWORD;
$db_name = DATABASE;
$table = TABLEU;
//These includes and constants are fine I checked them all
$con = mysqli_connect($host, $username, $password, $db_name);
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$myusername='test';
$mypassword='password1';
$sql="SELECT * FROM $table WHERE user_name=? and password=?";
$result=mysqli_prepare($con, $sql);
mysqli_stmt_bind_param($result, 'ss', $myusername, $mypassword);
mysqli_execute($result);
mysqli_stmt_fetch($result);
$row_cnt = mysqli_num_rows($result);
echo $row_cnt;
?>
The error returned is: Warning: mysqli_num_rows() expects parameter 1 to be mysqli_result, object given
I thought I took out all instances of OO PHP in my script? Also I understand that this may mean my query is incorrect so I ran it on MySQL in the database and all seems to be fine there:
So I am lost as to what the problem could be. I read many similar posts (maybe I'm missing one that is exactly similar to mine) and none seem to handle the problem. I appreciate your time and help.
P.S. I understand the security issues with plain text passwords and using "password1". I plan to use better security practices as I build this but I just want to get prepared statements down first.
You should use
mysqli_stmt_execute
mysqli_stmt_num_rows
Instead of the mysqli_execute and mysqli_num_rows.
although this question has been asked (and answered) many times, I didn't find a solution to the problem.
Here is my code:
<?php
#session_start();
include("./include/config.php");
include("./include/db_connect.php");
include("functions.php");
if (!isset($_GET['artikelID'])){$_GET['artikelID'] = "";}
if (!isset($_SESSION['UserID'])){$_SESSION['UserID'] = "";}
$sql = "SELECT kundenID FROM kunden WHERE username = '".$_POST['myusername']."' AND password = '".md5($_POST['mypassword'])."' ";
$result = mysqli_query($connect, $sql) OR die("<pre>\n".$sql."</pre>\n".mysqli_connect_error()); // this is line 13
$row = mysqli_fetch_assoc($result);
if (mysqli_num_rows($result)==1){
doLogin($row['kundenID'], isset($_POST['Autologin']));
header("location:cart.php?action=add&artikelID=".$_GET['artikelID']."&id=". $_SESSION['UserID'] ." ");
}
else {
header("location:k_login.php?error=TRUE ");
}
include("./include/db_close.php");
?>
mysqli_connect_error() shows me the absolute correct sql-query; the sql-query is tested with a tool named mysql-front and brings exactly one (and the correct one) result, which is 'kundenID'.
I have tested many things (like $_SESSION['connect'] or $_GLOBALS['connect'] instead of $connect in db_connect.db), but with no result.
Can anyone please help me?
-- Update --
Why does nobody answer?
Is the description of the problem unclear?
The db-connection is established like this:
<?php
error_reporting(E_ALL);
$connect = mysqli_connect($dbserver,$dbuser,$dbpass,$dbname);
// Check connection
if (mysqli_connect_errno()){
echo "Zeile ".__LINE__.": Datenbankverbindung ist fehlgeschlagen ! " . mysqli_connect_error();
exit();
}
?>
All the db-variables are known in the checklogin-script (tested). All the $_POST-variables are also known in the checklogin-script (tested). I even tried a hard-coded sql-query (with the real data of the test-record in the db).
The result is still the same: mysqli_connect_error() reports the correct query - but then nothing more happens.
I have spent more than 10 hours in the meantime. I really would appreciate, if someone could help me.
Couldn't fetch mysqli means that PHP is unable to identify the contents of your $connect variable as a valid mysqli connection. Try adding some error handling into "./include/db_connect.php" to get an idea of what happened to the mysqli connection that is preventing you from using it.
I've got a pretty standard call to a MySQL database and for some reason I can't get the code to work. Here's what I have:
$mysqli = mysqli_connect("localhost","username","password");
if (!$mysqli)
{
die('Could not connect: ' . mysqli_error($mysqli));
}
session_start();
$sql = "SELECT * FROM jobs ORDER BY id DESC";
$result = $mysqli->query($sql);
$num_rows = mysqli_num_rows($result);
Now, first, I know that it is connecting properly because I'm not getting the die method plus I added an else conditional in there previously and it checked out. Then the page displays but I get the errors:
Warning: mysqli_num_rows() expects parameter 1 to be mysqli_result, boolean given in blablabla/index.php on line 11
Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given in blablabla/index.php on line 12
I've double-checked my database and there is a table called jobs with a row of "id" (it's the primary row). The thing that confuses me is this is code that I literally copied and pasted from another site I built and for some reason the code doesn't work on this one (I obviously copy and pasted it and then just changed the table name and rows accordingly).
I saw the error and tried:
$num_rows = $mysqli_result->num_rows;
$row_array = $mysqli_result->fetch_array;
and that fixed the errors but resulted in no data being passed (because obviously $mysqli_result has no value). I don't know why the error is calling for that (is it a difference in version of MySQL or PHP from the other site)?
Can someone help me track down the problem? Thanks so much. Sorry if it's something super simple that I'm overlooking, I've been at it for a while.
You didn't selected the database
$mysqli = mysqli_connect("localhost","username","password","database");
The problem is you haven't selected the database.
use this code for select database.
$mysqli = mysqli_connect("localhost","username","password");
mysqli_select_db("db_name",$mysqli);
You have to select database in order to fire mysql queries otherwise it will give you error.
I believe that schtever is correct, I do not think you are selecting the database. It isn't in the code snip and if you search online you see other people with similar errors and it was because the database wasn't selected. Please let us know if you selected a database before anything else is checked. Thanks.
Try this:
session_start();
$mysqli = new mysqli(DB_HOST, DB_USERNAME, DB_PASSWORD, DB_NAME);
if ($mysqli->connect_errno)
{
echo "Failed to connect to MySQL: (" . $mysqli->connect_errno . ") " . $mysqli->connect_error;
$mysqli->close();
}
$query ="SELECT * FROM jobs ORDER BY id DESC";
$values = $mysqli->query($query);
if($values->num_rows != 0)
{
while($row = $values->fetch_assoc())
{
//your results echo here
}
}
else
{
//if no results say so here
}
See this manual for mysqli_connect you can select the database right in this function.
This question already has answers here:
Warning: mysqli_error() expects exactly 1 parameter, 0 given error
(4 answers)
Closed 4 years ago.
When trying to return a simple set of results from my database table 'checklist' I receive the following error;
"Warning: mysqli_error() expects exactly 1 parameter, 0 given"
The code of my list.php file is as follows;
<?php
require_once('/includes/connection.inc.php');
// create database connection
$conn = dbConnect('read');
$sql = 'SELECT * FROM checklist ORDER BY created DESC';
$result = $conn->query($sql) or die(mysqli_error());
?>
<?php while($row = $result->fetch_assoc()) { ?>
<?php echo $row['created']; ?>
<?php echo $row['title']; ?>
<?php } ?>
The contents of my connection.inc.php file (for reference) is as follows;
<?php
function dbConnect($usertype, $connectionType = 'mysqli') {
$db = 'projectmanager';
$host = 'localhost';
if ($usertype == 'read') {
$user = 'root';
$pwd = '';
} elseif ($usertype == 'write') {
$user = 'root';
$pwd = '';
} else {
exit('Unrecognized connection type');
}
// Connection goes here...
if ($connectionType == 'mysqli') {
return new mysqli($host, $user, $pwd, $db);
} elseif ($mysqli->connect_error) {
die('Connect Error: ' . $mysqli->connect_error);
}
}
?>
I've been trying to follow some examples out of a book PHP Solutions: Dynamic Web Design Made Easy found HERE ...but I already had an issue with the connection.inc.php file (snippet shown above) where I had to correct "or die ('Cannot open database');" and replace it with the IF based statement you see above for the mysqli_error. So I am wondering if this book is riddled with some basic, fundamental errors - at least that when presented to novices like me leave us baffled.
Any help guys?
Thank you
I think the problem you're having is because you're combining object-oriented and non-OO calls to the MySQLi library.
The mysqli_error() function does indeed require a parameter -- it requires the connection variable; in your case, $conn.
mysqli_error($conn)
Howwever, if you'd written it in an OO manner, as you have done for most of the rest of the database calls, you would have written it like this:
$conn->error
Since all the rest of your code is written using object-oriented calls, it would make sense to use it for this call as well.
So your full line of code would look like this:
$result = $conn->query($sql) or die($conn->error);
You can see further examples in the PHP manual: http://php.net/manual/en/mysqli.error.php
Hope that helps.
With regard your question about the book you're using: I can't comment directly on the book itself as I haven't read it. But note that there are two MySQL libraries for PHP; the older mysql library, and the newer mysqli library. The older library also has a mysql_error() function, which differs from the newer one in that it does not require a connection variable. If there is an error in the book you are using, this may be the source of the confusion.