function inseartinto()
{
$DEL_LOG_REP = $connection->prepare("DELETE FROM test WHERE itemname='111'")$DEL_LOG_REP->execute()$DEL_LOG_REP->close()return $DEL_LOG_REP
}
This is short tutorails on connection in insert query into php...
<?php
$servername = "localhost";
$username = "username";
$password = "password";
$dbname = "myDB";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "INSERT INTO MyGuests (firstname, lastname, email)
VALUES ('John', 'Doe', 'john#example.com')";
if ($conn->query($sql) === TRUE) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
$conn->close();
?>
For more better understand..go to http://www.w3schools.com/php/default.asp
What code are you trying to writing man..
This is not a valid code for just not insert but also for connetion..
ya go onto W3School.com as mention by #Sasa.D... it provide you with knowledge from basic to expertise..
Related
This is the code, that is not working -
<form method="post">
<input type="submit" name="zapis" id="zapisk" value="Запис" /><br/>
</form>
<?php
function zapisk()
{$servername = "localhost";
$username = "test";
$password = "";
$dbname = "myDB";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "INSERT INTO MyGuests (firstname, lastname, email)
VALUES ('John', 'Doe', 'john#example.com')";
if ($conn->query($sql) === TRUE) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
$conn->close();
}
//{echo "Your test function on button click is working";}
//if(array_key_exists('zapis',$_POST)){zapisk();}
?>
If I activate the last two rows, at the moment as comment, the function work.
Is it possible, that the function can work, without the use of the last two rows.
You must call zapisk function when form submitted.
<form method="post">
<input type="submit" name="zapis" id="zapisk" value="Запис" /><br/>
</form>
<?php
function zapisk(){
$servername = "localhost";
$username = "test";
$password = "";
$dbname = "myDB";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "INSERT INTO MyGuests (firstname, lastname, email) VALUES ('John', 'Doe', 'john#example.com')";
if ($conn->query($sql) === TRUE) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
$conn->close();
}
// Check form submitted and post has zapis value.
if(!empty($_POST['zapis'])){
zapisk();
}
?>
Or you don't use function and write your code in if statement:
<form method="post">
<input type="submit" name="zapis" id="zapisk" value="Запис" /><br/>
</form>
<?php
// Check form submitted and post has zapis value.
if(!empty($_POST['zapis'])){
$servername = "localhost";
$username = "test";
$password = "";
$dbname = "myDB";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "INSERT INTO MyGuests (firstname, lastname, email) VALUES ('John', 'Doe', 'john#example.com')";
if ($conn->query($sql) === TRUE) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
$conn->close();
}
?>
I've been trying to figure out how to insert data into mySQL database for a long time. When I try to insert data, it returns "no database selected". I'm not too sure what's wrong with the code, could someone check it out?
<?php
$servername = "localhost";
$database= "learnsc2_ts";
$username = "learnsc2_admin";
$password = "Ts#123";
$conn = new mysqli($servername, $username, $password, $database);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
else {
echo "Connection successful";
}
$query = "INSERT INTO users(fname, lname) VALUES ('Owen',
'Feng')";
mysqli_query($conn, $query);
if (mysqli_query($conn, $query)) {
echo "New record created successfully";
} else {
echo "Error: " . $query . "<br>" . mysqli_error($conn);?>
Make sure your database name is correct.
i tested it in my local, It's work Just fine.
$servername = "localhost";
$database= "test";
$username = "root";
$password = "";
$conn = new mysqli($servername, $username, $password, $database);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$query = "INSERT INTO users(fname, lname) VALUES ('Owen',
'Feng')";
$query = mysqli_query($conn, $query);
if ($query) {
echo "New record created successfully";
} else {
echo "Error: " . $query . "<br>" . mysqli_error($conn);
}
You forgot to add database name
$conn = new mysqli($servername, $username, $password, $database);
I figured it out. Something was wrong with the old username I was using. After changing to a new username and database, it worked out!
I have very strange problem. I want to run mysql query as it is shown down below, but it's not working. Connection to database is successful, INSERT query is ok too, because when I run it directly in phpmyadmin Console it works, but it's not working here in PHP code.
Could you tell me what I'm missing?
$servername = "localhost";
$username = "admin";
$password = "admin123";
$dbname = "database1";
$conn = mysqli_connect($servername, $username, $password, $dbname);
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
$sql = "INSERT INTO last_visit (ip, lastvisit) VALUES ('123', '123')";
You need to run your $sql, because now your $sql is only a string, it does nothing.
Add this :
if ($conn->query($sql) === TRUE) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
I know this question has been answered many times. I tried a few of the solutions from here but nothing worked. I'm not using any PHP framework.
I have an insert operation taking place and I want to get the id of the inserted row.
Here's my code:
$qry="INSERT INTO tablename(content) VALUES('".$content."')";
setData($qry);
setData() is a user defined function which does insert operation.
//for data submit
function setData($qry)
{
$obj=new DBCon();
$res=$obj->submitQuery($qry);
return $res;
}
//fetches result
function getData($qry)
{
$obj=new DBCon();
$res=$obj->selectQuery($qry);
return $res;
}
This is the class I made for establishing database connectivity:
class DBCon
{
private function getConnection()
{
// hostname, username, password, database
$con=mysqli_connect("localhost","root","","dbname")OR die('Could not Connect the Database');
return $con;
}
public function closeCon()
{
mysqli_close($con);
}
public function submitQuery($qry)
{
$result=0;
$con=$this->getConnection();
$result=mysqli_query($con,$qry);
return $result;
}
public function selectQuery($qry)
{
$con=$this->getConnection();
$res=mysqli_query($con,$qry);
return $res;
}
}
To obtain the last inserted id, I wrote the following query but it did not yield any result:
SELECT LAST_INSERT_ID() FROM tablename
Is there any other method to get this?
LAST_INSERT_ID() only returns values that have been auto-generated by the MySQL server for an AUTO_INCREMENT column; when you insert a specific value, no auto-generation takes place and no value will be returned by LAST_INSERT_ID().
Try : $con->insert_id;
Try This way.
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "INSERT INTO MyGuests (firstname, lastname, email)
VALUES ('First Name', 'Last Name', 'first#example.com')";
if ($conn->query($sql) === TRUE) {
$last_id = $conn->insert_id;
echo "New record created successfully. Last inserted ID is: " . $last_id;
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
MySQLi Object-oriented
<?php
$servername = "localhost";
$username = "username";
$password = "password";
$dbname = "myDB";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "INSERT INTO MyGuests (firstname, lastname, email)
VALUES ('John', 'Doe', 'john#example.com')";
if ($conn->query($sql) === TRUE) {
$last_id = $conn->insert_id;
echo "New record created successfully. Last inserted ID is: " . $last_id;
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
$conn->close();
?>
MySQLi Procedural
<?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());
}
$sql = "INSERT INTO MyGuests (firstname, lastname, email)
VALUES ('John', 'Doe', 'john#example.com')";
if (mysqli_query($conn, $sql)) {
$last_id = mysqli_insert_id($conn);
echo "New record created successfully. Last inserted ID is: " . $last_id;
} else {
echo "Error: " . $sql . "<br>" . mysqli_error($conn);
}
mysqli_close($conn);
?>
PDO
<?php
$servername = "localhost";
$username = "username";
$password = "password";
$dbname = "myDBPDO";
try {
$conn = new PDO("mysql:host=$servername;dbname=$dbname", $username, $password);
// set the PDO error mode to exception
$conn->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
$sql = "INSERT INTO MyGuests (firstname, lastname, email)
VALUES ('John', 'Doe', 'john#example.com')";
// use exec() because no results are returned
$conn->exec($sql);
$last_id = $conn->lastInsertId();
echo "New record created successfully. Last inserted ID is: " . $last_id;
}
catch(PDOException $e)
{
echo $sql . "<br>" . $e->getMessage();
}
$conn = null;
?>
$last_id = $conn->insert_id;
I'm not exactly sure what happened but this database and the php effecting it were working just fine until it hit the fourth row and now it won't insert new records at all.
if($_POST)
{
$servername = ******;
$username = ******;
$password = ******;
$db = ******;
$conn = mysqli_connect($servername, $username, $password, $db);
mysqli_select_db($conn,$db);
$uuid = $_POST['uuid'];
$sql = "INSERT INTO uuid VALUES ('$uuid');";
mysqli_query($conn,$sql);
mysqli_close($conn);
}
I'm not sure what happened but this is the relevant code for the mysqli query.
try this
<?php
if(isset($_POST['uuid']))
{
$servername = yourServerName;
$username = username;
$password = password;
$dbname = databaseName;
// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
$uuid = $_POST['uuid'];
$sql = "INSERT INTO tableName (columnName) VALUES ('$uuid')";
if (mysqli_query($conn, $sql)) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . mysqli_error($conn);
}
mysqli_close($conn);
}
?>
Also, I recommend using prepared statements.