I have faced another problem please help.
when i query to my db 1 time, 2 rows entered into the db!
Here is the code:
<?php
$servername = "localhost";
$username = "***";
$password = "***";
$dbname = "***";
$conn = mysqli_connect($servername, $username, $password, $dbname);
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
$name = $_POST['name'];
$famil = $_POST['famil'];
$email = $_POST['email'];
$amount = $_POST['amount'];
$name = mysqli_real_escape_string($conn, $name);
$famil = mysqli_real_escape_string($conn, $famil);
$email = mysqli_real_escape_string($conn, $email);
$amount = mysqli_real_escape_string($conn, $amount);
$sql = "INSERT INTO users".
"(name, famil, email, amount)".
"VALUES ('$name','$famil','$email','$amount')";
mysqli_query($conn,$sql);
if (mysqli_query($conn, $sql)) {
echo "Data entered successfully.";
} else {
echo "Error entering data: " . mysqli_error($conn);
}
mysqli_close($conn);
?>
and the result after 2 times run this code:
3 mohsen gholi ***#yahoo.com 235354346
4 mohsen gholi ***#yahoo.com 235354346
5 mohsen gholi ***#yahoo.com 235354346
6 mohsen gholi ***#yahoo.com 235354346
Remove this line as said by Fred
mysqli_query($conn,$sql);
and just use:
if (mysqli_query($conn, $sql)) {
echo "Data entered successfully.";
} else {
echo "Error entering data: " . mysqli_error($conn);
}
Related
I am trying to POST my data in my database but unable to do... Which mistake I did here? I have simply try to print Query but it prints without any value....
If I did any mistake then let me know..
And suggest me what can I do now...
<?php
if(isset($_POST['firstname'])){
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "sms";
$conn = mysqli_connect($servername, $username, $password, $dbname);
if(!$conn){
die("Could not connect to the database due to the following error --> ".mysqli_connect_error());
}
//echo "success";
$firstname = $_POST['firstname'];
$lastname = $_POST['lastname'];
$email = $_POST['email'];
$mobno = $_POST['mobno'];
$dob = $_POST['dob'];
$sql1="INSERT INTO `register`(`firstname`, `lastname`, `email`, `mobno`, `dob`) VALUES ('$firstname','$lastname','$email','$mobno','$dob')";
echo $sql1;
if($conn->query($sql1) == true){
// echo "Successfully inserted";
// Flag for successful insertion
$insert = true;
}
else{
echo "ERROR: $sql1 <br> $conn->error";
}
// Close the database connection
$conn->close();
}
?>
I want to set up a website with a form in it. The form will transfer the data to the DB, but I think it is not safe to let the personal data in the DB which is external reachable.
So I thought I should transfer the data via PHP from the DB1(server1 - external reachable) to DB2(server2 - only internal reachable).
The following picture should help to know what I am searching for.
Is there any names/methods to google for?
From php you just have to create a new connection for DB2.
<?php
$servername = "localhost";
$username = "database1";
$password = "xxxxxxxx";
$dbname = "database1";
$servername2 = "localhost";
$username2 = "database2";
$password2 = "xxxxxxxx";
$dbname2 = "database2";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
$conn2 = new mysqli($servername2, $username2, $password2, $dbname2);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
if ($conn2->connect_error) {
die("Connection failed: " . $conn2->connect_error);
}
//escape variables for security
$fname = mysqli_real_escape_string($conn, $_POST['fname']);
$lname = mysqli_real_escape_string($conn, $_POST['lname']);
$sql = "INSERT INTO mytable (fname,lname)
VALUES ('$fname','$lname')";
if ($conn->query($sql) === TRUE) {
echo "Successfully Saved";
} else {
echo "Error: Go back and Try Again ! " . $sql . "<br>" . $conn->error;
}
if ($conn2->query($sql) === TRUE) {
echo "Successfully Saved";
} else {
echo "Error: Go back and Try Again ! " . $sql . "<br>" . $conn2->error;
}
$conn->close();
$conn2->close();
?>
Good Day developers outthere! 😊😊
I just wanna ask what is the problem with my code, I'm trying to make a webpage using html,css,php and database. Now I already created a php in my html form and my database is already connected, but everytime I submit the information in the html form I created, nothing appeared in my database.
<?php
if(isset($_POST['save'])){
$FName = $_POST['FName'];
$MName = $_POST['MName'];
echo "Successfully Added";
$sql= "INSERT INTO 'tbstudinfo' (Transaction_Number, First_Name, `Middle_Name') VALUES ('000',$FName,$MName)";
} else{
echo "<p>Insertion Failed.</p>";
}
?>
Just as #executable mentioned, you are defining query in your code but not executing it.
Define Connection Object (Mysqli, PDO..)
Prepare Query and Bind Variables
Execute your query
Here's an example using prepared statements
<?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: " . $conn->connect_error);
}
if( isset($_POST['save']) ){
// prepare and bind
$stmt = $conn->prepare("INSERT INTO 'tbstudinfo' (Transaction_Number, First_Name, Middle_Name) VALUES (?, ?, ?)");
$stmt->bind_param("sss", $transaction_number, $FName, $MName);
// set parameters and execute
$transaction_number = '000';
$FName= $_POST['FName'];
$MName= $_POST['MName'];
$stmt->execute();
echo "Successfully Added";
}else{
echo "<p>Nothing Posted</p>";
}
W3Schools and PHP.Net both have pretty good examples about how to use prepared statements to make your SQL Query more secure from SQL Injections.
You simply don't execute your query. Using MySQLi :
<?php
$servername = "localhost";
$username = "root";
$password = "";
$db = "dbthesis";
$conn = new mysqli($servername, $username, $password, $db);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
if(isset($_POST['save'])){
$FName = $_POST['FName'];
$MName = $_POST['MName'];
$sql = "INSERT INTO tbstudinfo (Transaction_Number, First_Name, Middle_Name) VALUES ('000', '$FName', '$MName')";
if ($conn->query($sql) === TRUE) {
echo "Successfully Added";
} else {
echo "<p>Insertion Failed.</p>";
}
}
$conn->close();
You only making a query, not running query. This this code
$FName = $_POST['FName'];
$MName = $_POST['MName'];
$sql = "INSERT INTO tbstudioinfo (Transaction_Number, First_Name, Middle_Name) VALUES ('000','$FName','$MName')";
// code below runs your query
if (mysqli_query($conn, $sql)) {
echo "Successfully Added";
} else {
echo "Error: " . $sql . "<br>" . mysqli_error($conn);
}
Form data is not getting inserted in MYSQL . Form successfully posts data (I have checked with var dump). Please help me with this.
This is my action PHP.
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname= "test";
$conn = new mysqli($servername,$username,$password,$dbname);
if($conn->connect_error){
die("connection failed:" .$conn->connect_error);
}
else{
echo "Connected successfully";
}
$selected = mysqli_select_db($conn,$dbname);
$first_name = $_POST['first_name'];
$last_name = $_POST['last_name'];
$email = $_POST['email'];
$mobile = $_POST['mobile'];
$address = $_POST['address'];
//$gender = $_POST['male'];
//$gender = $_POST['female'];
$sql = "INSERT INTO test1 (first_name,last_name,email,mob,home_address) VALUES ('$first_name','$last_name','$email','$mobile','$address')";
var_dump($sql);
?>
You are missing the last part which executes the query to save the data:
if ($conn->query($sql) === TRUE) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
I have a user registration system but its empty. This is the script I use in forum.modxpertz.tk. It worked at first but it shows nothing now. Here is the code.
<?php
$servername = "localhost";
$username = "root";
$password = "";
// Create connection
$conn = mysqli_connect($servername, $username, $password);
mysqli_select_db($conn,'login');
// Check connection
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
$sql = "SELECT userid FROM login";
$result = mysqli_query($conn, $sql);
if (mysqli_num_rows($result) > 0) {
// output data of each row
while($row = mysqli_fetch_assoc($result)) {
$reguserid=$row["userid"];
}
$userid = mysqli_real_escape_string($conn, $_POST['userid']);
$pswrd = mysqli_real_escape_string($conn, $_POST['pswrd']);
$fname = mysqli_real_escape_string($conn, $_POST['fname']);
$lname = mysqli_real_escape_string($conn, $_POST['lname']);
$gender = mysqli_real_escape_string($conn, $_POST['gender']);
$token = rand('122332344','922332344');
$url = array('forum.modzexpertz.tk/verify.php#',$token);
$post= join($url);
if($userid!=$reguserid){
$sql = "INSERT INTO login(fname, lname, userid, pswrd, gender)VALUES('$fname', '$lname', '$userid', '$pswrd', '$gender')";
if ($conn->query($sql) === TRUE) {
echo "New record created successfully";
}else {
echo "Failed to Register.";
}} else {
echo "A user with the email youve provided has already been registered.";
}}
$conn->close();
?>
I know only little about PHP and jQuery.
Please try below code :
<?php
$servername = "localhost";
$username = "root";
$pswrd = "";
$db = "login";
$conn = mysqli_connect($servername,$username,$pswrd, $db);
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$table = 'login';
if(#mysqli_num_rows(mysqli_query($conn, "SELECT NULL FROM `$table` WHERE userid='".$_POST['userid']."'")) > 0){
$error = "1";
echo "user with same userid is already exist";
}
if(isset($_POST['fname']) && isset($_POST['lname']) && isset($_POST['gender']) && isset($_POST['userid']) && isset($_POST['pswrd']) && $_POST['fname']!="" && $_POST['lname']!="" && $_POST['gender']!="" && $_POST['userid']!="" && $_POST['pswrd']!="")
{
if($error==''){
$ins['fname'] = mysqli_real_escape_string($conn, $_POST['fname']);
$ins['lname'] = mysqli_real_escape_string($conn, $_POST['lname']);
$ins['gender'] = mysqli_real_escape_string($conn, $_POST['gender']);
$ins['userid'] = mysqli_real_escape_string($conn, $_POST['userid']);
$ins['pswrd'] = mysqli_real_escape_string($conn, $_POST['pswrd']);
$insertsql = "INSERT INTO `$table` (fname, lname, gender, userid, pswrd) VALUES ('".$ins['fname']."','".$ins['lname']."','".$ins['gender']."','".$ins['userid']."','".$ins['pswrd']."')";
#mysqli_query($conn, $insertsql);
//echo $insertsql; exit;
echo "Success";
}
}else{
echo "Please enter required parameters";
}
mysqli_close($conn);
?>