I can't run my codes.
It says:
syntax error, unexpected '$query' (T_VARIABLE).
Code
<?php
$hostname="localhost";
$username="";
$password="";
$dbname="thesis";
$usertable="product";
$yourfield="product_id";
msql_connect($hostname,$username,$password) or die ("<html><script>
language='Javascript'>alert('Unable to connect to database!.'),history.go(-1)</script></html>")
$query = "SELECT * FROM $usertable";
$result = mysql_query($query);
if($result)
{
while ($row = mysql_fetch_array($result))
{
$name = $row["$yourfield"];
echo "Name: ".$name."</br>";
}
}
?>
msql_connect($hostname,$username,$password) or die ("<html><script>
language='Javascript'>alert('Unable to connect to database!.'),history.go(-1)</script></html>")
should have a semicolon.
Replace it with this:
msql_connect($hostname,$username,$password) or die ("<html><script>
language='Javascript'>alert('Unable to connect to database!.'),history.go(-1)</script></html>");
<?php
$hostname = "localhost";
$username = "";
$password = "";
$dbname = "thesis";
$usertable = "product";
$yourfield = "product_id";
$mysqli = new mysqli($hostname, $username, $password, $dbname);
/* check connection */
if ($mysqli->connect_errno) {
printf("Connect failed: %s\n", $mysqli->connect_error);
exit();
}
/* Select queries return a resultset */
if ($result = $mysqli->query("SELECT * FROM $usertable")) {
while ($row = mysql_fetch_array($result))
{
$name = $row["$yourfield"];
echo "Name: ".$name."</br>";
}
/* free result set */
$result->close();
}
$mysqli->close();
http://php.net/manual/en/mysqli.query.php
Related
I have written following code to connect mysql in php but I am not getting output.
<?php
$servername = "localhost";
$username = "root";
$password = "pravin";
$mysql_conn = new mysqli($servername, $username, $password);
if ($mysql_conn->connect_error) {
die("Connection failed: ". $mysql_conn->connect_error);
}
echo "Connected successfully";
$name = $_POST["microorganism"];
echo $name;
$db_selected = mysql_select_db('yieldofvanillin', $mysql_conn);
if (!$db_selected){
die ('Can\'t use : ' . mysql_error());
}
$query = "SELECT * FROM vanillin WHERE Microorganism = '$name' ";
$result = $mysql_query($query);
while ($line = myql_fetch_array($result, MYSQL_ASSOC)) {
echo $line["Substrate"];
echo $line["products"];
echo $line["Microorganism"];
echo $line["yield"];
echo $line["Reference"];
}
mysql_close($mysql_conn);
?>
The database name is "yieldofvanillin" and it has five column. I an getting output Connected successfully. After that no output. Please let me know the bug in code.
i have remove errors. which i mention in comments. Code Reference PHP Manual. you should read this manual (strongly recommended)
<?php
$mysqli = new mysqli("localhost", "root", "pravin", "yieldofvanillin");
/* check connection */
if ($mysqli->connect_errno) {
printf("Connect failed: %s\n", $mysqli->connect_error);
exit();
}
$query = "SELECT * FROM vanillin WHERE Microorganism = '$name' ";
if ($result = $mysqli->query($query)) {
/* fetch associative array */
while ($row = $result->fetch_assoc()) {
echo $row["Substrate"];
echo $row["products"];
echo $row["Microorganism"];
echo $row["yield"];
echo $row["Reference"];
}
/* free result set */
$result->free();
}
you're mixing mysqli and mysql libraries.
the code should be:
<?php
$servername = "localhost";
$username = "root";
$password = "pravin";
$mysql_conn = new mysqli($servername, $username, $password);
if (mysqli_connect_errno()) {
die("Connection failed: ". mysqli_connect_error());
}
echo "Connected successfully";
$name = $_POST["microorganism"];
echo $name;
$db_selected = mysqli_select_db($mysql_conn,'yieldofvanillin');
if (!$db_selected){
die ('Can\'t use : ' . mysqli_error($mysql_conn));
}
$query = "SELECT * FROM vanillin WHERE Microorganism = '$name' ";
$result = mysqli_query( $mysql_conn,$query);
while ($line = mysqli_fetch_assoc($result)) {
echo $line["Substrate"];
echo $line["products"];
echo $line["Microorganism"];
echo $line["yield"];
echo $line["Reference"];
}
mysqli_close($mysql_conn);
?>
Remove error in your code.Read carefully php manual.
<?php
$servername = "localhost";
$username = "root";
$password = "pravin";
$db = "yieldofvanillin";
// Create connection
$mysqli = new mysqli($servername, $username, $password, $db);
/* connection string*/
if ($mysqli->connect_errno) {
die("Connection failed: " . $mysqli->connect_error);
exit();
}
$query = "SELECT * FROM vanillin WHERE Microorganism = '$name' ";
if ($result = $mysqli->query($query)) {
while ($row = $result->fetch_assoc()) {
echo $row["Substrate"];
}
$result->free();
}
$mysqli->close();
?>
Your output not showing because mysql_fetch_array is not correct.Because you are mixing mysql_ and mysqli_ functions and you called myql_fetch_array that doesn't exist in mysqli. MySQL and MySQLi are two different PHP extensions and they cannot be mixed. Because the former is deprecated in mysqli
I would need some help with showing data that I have on my database but I can't seen to be able to.`
$servername = "servername";
$username = "username";
$password = "password";
$dbname = "dbname";
$connect = mysqli_connect($servername, $username, $password, $dbname) or die ("connection failed");
//Query
$query = "SELECT * FROM 'Students'";
mysqli_master_query($dbname, $query) or die ("Error while Query");
$result = mysqli_master_query($dbname, $query);
$row = mysql_fetch_array($result);
while ($row = mysql_fetch_array($result)) {
echo "<p>".$row['Name']."</p>";
};
mysql_close($connect);
?>`
I am pretty new to this so I could have missed something simple. Any help appreciated.
Below is a sample code of the normal procedure to connect to a database and to select data from it. Please follow this type of coding since MySQL is now deprecated and MySQLi is used.
<?php
$servername = "localhost";
$username = "username";
$password = "password";
$dbname = "myDB";
// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
$sql = "SELECT id, firstname, lastname FROM MyGuests";
$result = mysqli_query($conn, $sql);
if (mysqli_num_rows($result) > 0) {
// output data of each row
while($row = mysqli_fetch_assoc($result)) {
echo "id: " . $row["id"]. " - Name: " . $row["firstname"]. " " . $row["lastname"]. "<br>";
}
} else {
echo "0 results";
}
mysqli_close($conn);
?>
For further reference check out http://php.net/manual/en/book.mysqli.php and also https://www.w3schools.com/php/php_mysql_insert.asp
I want to make a dropdown list and i use this codes:
<div class="selector">
<?php
include ("connect.php");
$db = new mysqli('localhost', $dbuser, $dbpass, $dbnam);
?>
<div class="label">Select Name:</div>
<select name="names">
<option value = "">---Select---</option>
$stmt = $db->prepare("SELECT `name` FROM `monitoare`");
$stmt->execute();
$stmt->bind_result($name);
while ($stmt->fetch()){
echo "<option value='$name'></option>";
}
$stmt->close();
?>
</select>
as index.php
and this:
<?php
$dbname = 'mydabase';
$dbuser = 'myuser';
$dbpass = 'mypass';
?>
as connect.php and after i launch this the drop stays just with ---select--- as an option
I reckon that you should do all of this using either MySQLi procedural or object oriented rather trying to do with prepared statements.
<?php
$servername = "localhost";
$username = "username";
$password = "password";
$dbname = "myDB";
// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
?>
You can save this in a 'connect.php' file and include it in the relevant file.
The solution to getting dropdown instead of --select-- without any errors is shown below
echo "<select name='names'>"
$sql = "SELECT `name` FROM monitoare";
$result =mysqli_query($db, $sql);
if (mysqli_num_rows($result) > 0) {
// output data of each row
while($row = mysqli_fetch_assoc($result)) {
echo "<option value=".$row['name'].">".$row['name']."</option>";
}
} else {
echo "No results found";
}
?>
For further reference check out http://php.net/manual/en/book.mysqli.php and also https://www.w3schools.com/php/php_mysql_insert.asp
Here's a (corrected) example, adapt to your code :
<?php
error_reporting(E_ALL);
ini_set('display_errors', 1);
$host = ""; /* your credentials here */
$user = ""; /* your credentials here */
$pwd = ""; /* your credentials here */
$db = ""; /* your credentials here */
/* connect to DBd */
$mysqli = mysqli_connect("$host", "$user", "$pwd", "$db");
if (mysqli_connect_errno()) { echo "Error connecting to DB : " . mysqli_connect_error($mysqli); }
$query = " SELECT `name` FROM `monitoare` ";
$stmt = $mysqli->prepare($query);
$stmt->execute();
$stmt->bind_result($name);
$stmt->store_result();
if ($stmt->num_rows > 0) { /* we have results */
echo"<select name=\"names\"><option value=\"\">---Select---</option>";
while($stmt->fetch()){
echo "<option value=\"$name\">$name</option>";
}
echo"</select>";
}
else
{ echo"[ no data ]"; }
?>
I have problem with 2 select Databases from the same connection simultaneously, the code is:
#$dbmssSQLGestasa_conn = mssql_connect($servidor, $usuario, $contra);
if(!$dbmssSQLGestasa_conn){
header("Location:nobase.php");
exit();
}
mssql_select_db('GESTASA', $dbmssSQLGestasa_conn);
the code of the other connection:
#$dbmssSQLTasa_conn = mssql_connect($servidor, $usuario, $contra);
if(!$dbmssSQLTasa_conn){
header("Location:nobase.php");
exit();
}
//Apertura de la base de datos
mssql_select_db('TASA', $dbmssSQLTasa_conn);
it dont work and give me error "(severity 16)".
is It possible do 2 or more selections databases en la same connection mssql?
Dont use mysql_select_db();
In Mysqli :
$servername = "localhost";
$username = "username";
$password = "password";
// Create connection
$conn = new mysqli($servername, $username, $password);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT * FROM test123.posts"; // databse.tablename
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
$mack[] = $row;
}
}
$sql1 = "SELECT * FROM portal.abouts"; // databse.tablename
$result1 = $conn->query($sql1);
if ($result1->num_rows > 0) {
// output data of each row
while($row1 = $result1->fetch_assoc()) {
$mack1[] = $row1;
}
}
echo "<pre>";
print_r($mack);
print_r($mack1);
echo "</pre>";
$conn->close();
You can easily use 2 databases in same time with Below Codes
I'm Using variables as Capital word But you can use your own variable words.
<?php
define('HOST', "YOURHOSTNAME");
define('USER', "YOURHOSTNAME");
define('PASS', "YOURHOSTNAME");
define('DATABASE1', "NAMEOFDATABASE1");
define('DATABASE2', "NAMEOFDATABASE2");
$DATABASE1 = mysqli_connect(HOST, USER, PASS, DATABASE1);
$DATABASE2 = mysqli_connect(HOST, USER, PASS, DATABASE2);
if(!$DATABASE1){
die("DATABASE1 CONNECTION ERROR: ".mysqli_connect_error());
}
if(!$DATABASE2){
die("DATABASE2 CONNECTION ERROR: ".mysqli_connect_error());
}
$sql = "SELECT * FROM TABLE"; /* You can use your own query */
$DATABASE1_QUERY = mysqli_query($DATABASE1, $sql);
$DATABASE2_QUERY = mysqli_query($DATABASE2, $sql);
$DATABASE1_RESULT = mysqli_fetch_assoc($DATABASE1_QUERY);
$DATABASE2_RESULT = mysqli_fetch_assoc($DATABASE2_QUERY);
/* SHOW YOUR RESULT HERE WHICH DATABASE YOU WANT FROM */
echo $DATABASE1_RESULT['id'];
echo $DATABASE2_RESULT['id'];
/*After complete your all work don't forgot about close database connections*/
mysqli_close($DATABASE1);
mysqli_close($DATABASE2);
?>
I want to get all the records from the while loop. I'm unable to get all the rows from the query. It shows only the first row.
Is there anything I was going wrong in my code.
function Connect($DB_HOST = 'localhost', $DB_USER = 'root', $DB_PASS = '', $DB_NAME = 'bodhilms')
{
$mysqli = new mysqli($DB_HOST, $DB_USER, $DB_PASS, $DB_NAME);
if ($mysqli->connect_errno) {
printf("Connect failed: %s\n", $mysqli->connect_error);
exit();
}
return $mysqli;
}
function GetCoeficient($coeficient = false, $con)
{
if(!$con)
return 0;
$result = array();
$sql[] = "SELECT * FROM users ";
if($coeficient != false)
$sql[] = "WHERE username = '".$coeficient."' ORDER BY u.id";
//print_r($coeficient);
$query = $con->query(implode(" ",$sql));
//print_r($query);
while($row = $query->fetch_assoc())
{
$result[] = $row;
}
return (!empty($result))? $result : 0;
}
$con = Connect();
$result = GetCoeficient($coeficient,$con);
$username = $result[0]['username'];
$firstname = $result[0]['firstname'];
$lastname = $result[0]['lastname'];
$email = $result[0]['email'];
First of all,to make sure the infomation of mysql is right,like port.
and I wonder the code of you $result = Getcourse($coeficient,$con);, how the var coeficient come from.Then
You can try the code below:
$mysqli=new mysqli("localhost","root","root","123");
$query="select * from test";
$result=$mysqli->query($query);
if ($result) {
if($result->num_rows>0){
while($row =$result->fetch_array() ){
echo ($row[0])."<br>";
echo ($row[1])."<br>";
echo ($row[2])."<br>";
echo ($row[3])."<br>";
echo "<hr>";
}
}
}else {
echo 'failure';
}
$result->free();
$mysqli->close();