This question already has answers here:
Can I mix MySQL APIs in PHP?
(4 answers)
Closed 4 years ago.
This is my code:
$con = mysqli_connect($databasehost,$dbname,$dbpassword) or die(mysql_error());
mysqli_select_db($con, $dbname) or die(mysql_error());
mysqli_query($con,"SET CHARACTER SET utf8");
$query = file_get_contents("php://input");
$sth = mysqli_query($con,$query);
if (mysql_errno()) {
header("HTTP/1.1 500 Internal Server Error");
echo $query.'\n';
echo mysql_error();
}
else
{
$rows = array();
while($row = mysql_fetch_assoc($sth)){
$rows[] = $r;
}
print json_encode($rows);
}
What's the problem?
This is the error:
[19-Jan-2018 09:17:39 UTC] PHP Warning: mysql_fetch_assoc() expects parameter 1 to be resource on line 22
You are mixing mysqli and mysql.
The latter is deprecated as of version 5.5.0.
Try using mysqli_fetch_assoc($sth) instead of mysql_fetch_assoc($sth)
See http://php.net/manual/en/mysqli-result.fetch-assoc.php also.
Your problem is you are mixing mysqli and mysql.
Try to use mysqli_fetch_assoc($sth) instead of mysql_fetch_assoc($sth).
And, in mysqli_connect("host", "user", "password", "database"), you are missing the user name.
// (1) make sure $username is defined.
$con = mysqli_connect($databasehost,$username,$dbname,$dbpassword) or die(mysqli_error());
// check to use mysqli_, not mysql_
mysqli_select_db($con, $dbname) or die(mysqli_error());
mysqli_query($con,"SET CHARACTER SET utf8");
$query = file_get_contents("php://input");
$sth = mysqli_query($con,$query);
if (mysql_errno()) {
header("HTTP/1.1 500 Internal Server Error");
echo $query.'\n';
echo mysql_error();
}
else
{
$rows = array();
// (2) Use mysqli_fetch_assoc instead of mysql_fetch_assoc.
while($row = mysqli_fetch_assoc($sth)) {
// (3) $r is not defined. Use $row.
$rows[] = $row;
}
print json_encode($rows);
}
Related
This question already has answers here:
Deprecated: mysql_connect() [duplicate]
(12 answers)
Closed 6 years ago.
How can I make this query or script work properly with having these deprecated errors I am getting when using Wamp I am not getting the errors while using Xampp though. The errors or warning i'm getting is on this pic
this is my php script
<?php
$con = mysql_connect("localhost","root","");
if (!$con) {
die('Could not connect: ' . mysql_error());
}
mysql_select_db("db", $con);
$result = mysql_query("SELECT gender as gender_occupation, COUNT(*) as total FROM hostel_blocks GROUP BY gender");
$rows = array();
while($r = mysql_fetch_array($result)) {
$row[0] = $r[0];
$row[1] = $r[1];
array_push($rows,$row);
}
print json_encode($rows, JSON_NUMERIC_CHECK);
mysql_close($con);
?>
when i changed some connection code i got a blank page.
PHP mysql was deprecated in PHP 5.5 and removed starting PHP 7
You should move to mysqli instead.
Your code using mysqli would look like this:
<?php
$con = mysqli_connect("localhost","root","");
if (!$con) {
die('Could not connect: ' . mysqli_error($con));
}
mysqli_select_db( $con, "db");
$result = mysqli_query($con, "SELECT gender as gender_occupation, COUNT(*) as total FROM hostel_blocks GROUP BY gender");
$rows = array();
while($r = mysqli_fetch_array($result)) {
$row[0] = $r[0];
$row[1] = $r[1];
array_push($rows,$row);
}
print json_encode($rows, JSON_NUMERIC_CHECK);
mysqli_close($con);
?>
I cannot connect to my Data Base using this way:
mysql_connect("localhost","root","") or die("1");
mysql_select_db("database") or die("2");
After the connection to the data base I would like to create a while loop with the elements of the data base, this is my code:
$carlistreq = mysql_query("SELECT * FROM cars WHERE brand="'.$_GET['marq'].'"") or die("3");
$count = mysql_num_rows($carlistreq);
if($count==0)
{
echo "No elements";
}
else
{
while($row= mysql_fetch_array($query))
{
//my code
}
}
The issue comes from my SQL attributes like:
mysql_connect
mysql_fetch_array
mysql_select_db
For example.
Here is the error:
Deprecated: mysql_connect(): The mysql extension is deprecated and
will be removed in the future: use mysqli or PDO instead in C:[....]
on line 2
use the mysqli-extension like this:
$mysqli = new mysqli("localhost","user","password","database") or die("1");
$sql = "SELECT * FROM cars WHERE brand='".$_GET['marq']."'";
$result = $mysqli->query($sql);
if($result){
while($row = $result->fetch_assoc()){
//do something with your $row here
}
}
http://php.net/manual/en/book.mysqli.php
Use PDO
$db = new PDO('mysql:host=localhost;dbname=databaseName', 'User' , 'password');
$db->setAttribute(PDO::ATTR_ERRMODE,PDO::ERRMODE_EXCEPTION);
Yes now servers are going to handel mysqli and PDO. Try implementing this code:
<?php
$con = mysqli_connect("localhost","root","");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
?>
// Change database to "test"
mysqli_select_db($con,"test");
// ...some PHP code for database "test"...
mysqli_close($con);
?>
You can user http://php.net/manual/en/book.mysqli.php as reference to understand better how mysqli works.
so I am new in StackOverflow :), I have a problem to answer a question in php5,
This is the question :
Create a PHP 5.4 script to check availability of many Sites (via Echo Protocol):
* Get the list of sites (domains) from MySQL database.
so I write this script and I want if it's the good answer :
<?php
$dbname = 'mysql_dbname';
if (!mysql_connect('mysql_host', 'mysql_user', 'mysql_password')) {
echo 'Could not connect to mysql';
exit;
}
$sql = "SHOW TABLES FROM $dbname";
$result = mysql_query($sql);
if (!$result) {
echo "DB Error, could not list tables\n";
echo 'MySQL Error: ' . mysql_error();
exit;
}
while ($row = mysql_fetch_row($result)) {
echo "Table: {$row[0]}\n";
}
mysql_free_result($result);
?>
try something like this
$con = mysqli_connect(‘hostname’, ‘username’, ‘password’, ‘database’);
$result = mysqli_query($con , "select column_name from table_name");
while($data = mysqli_fetch_array($result))
{
echo $data['column_name'];
}
this will tell you the databases that have rights to be connected to
mysql_connect('localhost') or die ("Connect error");
$res = mysql_query("SHOW DATABASES");
while ($row = mysql_fetch_row($res)) {
echo $row[0], '<br/>';
}
You need to select a database before you use mysql_query.
$connection = mysql_connect('mysql_host', 'mysql_user', 'mysql_password');
//Write code to Check the connection
mysql_select_db('database_name', $connection);
//Write code to query
But as Robin mentioned in the comment, mysql* apis are deprecated. Use Mysqli or Pdo_Mysql. details here: http://in3.php.net/mysql_select_db
I got a simple query looking like this:
$con = mysqli_connect("host","username","password","default_database");
if(mysqli_connect_errno($con))
{
mysqli_connect_error();
}
else
{
$query = "SELECT * FROM `users_t`";
$result = mysqli_query($query) or die (mysql_error());
while($row = mysqli_fetch_assoc($result))
{
echo $row['email'];
}
}
Running this query gives me absolutely nothing. No errors but no result at the same time. I can't figure out whats wrong, help me please.
It looks like there may be a few things going on here:
According to the mysqli_connect_errno() Documentation, you should be checking for !$con rather than mysqli_connect_errno($con) in your if statement.
When a connection error is encountered, you're calling the error function but not printing it.
According to the mysqli_query() documentation, the first argument should be the database connection, the second being the query itself.
When a query errors out, you're calling mysql_error() when you should be calling mysqli_error(), passing it the connection. Again, according to documentation
Try this out and see if this resolves your problems:
<?php
$con = mysqli_connect("host","username","password","default_database");
if(!$con) {
print mysqli_connect_error();
} else {
$query = "SELECT * FROM `users_t`";
$result = mysqli_query($con, $query) or die (mysqli_error($con));
while($row = mysqli_fetch_assoc($result)) {
echo $row['email'];
}
}
$result = mysqli_query($query) or die (mysql_error());
Needs a mysqli resource link and also its not mysql_error()
if ($result = mysqli_query($con,$query))
{
while($row = mysqli_fetch_assoc($result))
{
echo $row['email'];
}
}
PHP is case-sensitive, so there is a very good change that your column in the database is actually named Email and then you have no result when you check $row['email'], because the E has to be upper case.
What you can do is make sure you use the right case in PHP, or select just the fields you want. MYSQL isn't case-senstive, so if you do the following it will work.
Also a good way to check if you have the right case: use var_dump($row) (I've added it to the code just to show you)
$con = mysqli_connect("host","username","password","default_database");
if(mysqli_connect_errno($con))
{
mysqli_connect_error();
}
else
{
$query = "SELECT email FROM `users_t`";
$result = mysqli_query($query) or die (mysql_error());
while($row = mysqli_fetch_assoc($result))
{
echo $row['email'];
var_dump($row);
}
}
I was wondering if there's a way in PHP to list all available databases by usage of mysqli. The following works smooth in MySQL (see php docs):
$link = mysql_connect('localhost', 'mysql_user', 'mysql_password');
$db_list = mysql_list_dbs($link);
while ($row = mysql_fetch_object($db_list)) {
echo $row->Database . "\n";
}
Can I Change:
$db_list = mysql_list_dbs($link); // mysql
Into something like:
$db_list = mysqli_list_dbs($link); // mysqli
If this is not working, would it be possible to convert a created mysqli connection into a regular mysql and continue fetching/querying on the new converted connection?
It doesn't appear as though there's a function available to do this, but you can execute a show databases; query and the rows returned will be the databases available.
EXAMPLE:
Replace this:
$db_list = mysql_list_dbs($link); //mysql
With this:
$db_list = mysqli_query($link, "SHOW DATABASES"); //mysqli
I realize this is an old thread but, searching the 'net still doesn't seem to help. Here's my solution;
$sql="SHOW DATABASES";
$link = mysqli_connect($dbhost,$dbuser,$dbpass) or die ('Error connecting to mysql: ' . mysqli_error($link).'\r\n');
if (!($result=mysqli_query($link,$sql))) {
printf("Error: %s\n", mysqli_error($link));
}
while( $row = mysqli_fetch_row( $result ) ){
if (($row[0]!="information_schema") && ($row[0]!="mysql")) {
echo $row[0]."\r\n";
}
}
Similar to Rick's answer, but this is the way to do it if you prefer to use mysqli in object-orientated fashion:
$mysqli = ... // This object is my equivalent of Rick's $link object.
$sql = "SHOW DATABASES";
$result = $mysqli->query($sql);
if ($result === false) {
throw new Exception("Could not execute query: " . $mysqli->error);
}
$db_names = array();
while($row = $result->fetch_array(MYSQLI_NUM)) { // for each row of the resultset
$db_names[] = $row[0]; // Add db name to $db_names array
}
echo "Database names: " . PHP_EOL . print_r($db_names, TRUE); // display array
Here is a complete and extended solution for the answer, there are some databases that you do not need to read because those databases are system databases and we do not want them to appear on our result set, these system databases differ by the setup you have in your SQL so this solution will help in any kind of situations.
first you have to make database connection in OOP
//error reporting Procedural way
//mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);
//error reporting OOP way
$driver = new mysqli_driver();
$driver->report_mode = MYSQLI_REPORT_ALL & MYSQLI_REPORT_STRICT;
$conn = new mysqli("localhost","root","kasun12345");
using Index array of search result
$dbtoSkip = array("information_schema","mysql","performance_schema","sys");
$result = $conn->query("show databases");
while($row = $result->fetch_array(MYSQLI_NUM)){
$print = true;
foreach($dbtoSkip as $key=>$vlue){
if($row[0] == $vlue) {
$print=false;
unset($dbtoSkip[$key]);
}
}
if($print){
echo '<br/>'.$row[0];
}
}
same with Assoc array of search result
$dbtoSkip = array("information_schema","mysql","performance_schema","sys");
$result = $conn->query("show databases");
while($row = $result->fetch_array(MYSQLI_ASSOC)){
$print = true;
foreach($dbtoSkip as $key=>$vlue){
if($row["Database"] == $vlue) {
$print=false;
unset($dbtoSkip[$key]);
}
}
if($print){
echo '<br/>'.$row["Database"];
}
}
same using object of search result
$dbtoSkip = array("information_schema","mysql","performance_schema","sys");
$result = $conn->query("show databases");
while($obj = $result->fetch_object()){
$print = true;
foreach($dbtoSkip as $key=>$vlue){
if( $obj->Database == $vlue) {
$print=false;
unset($dbtoSkip[$key]);
}
}
if($print){
echo '<br/>'. $obj->Database;
}
}