I´m trying to create a form connected to a database but when I fill out the form and I refer to the table in phpMyAdmin I see that it have entered a blank record instead of form data. I´m using PhpStorm.
I think all this code is correct...
That is the form of the .html:
<form id="form1" name="form1" method="post" action="index.php">
<label for="userSignUp">Email</label>
<input type="text" name="userSign" id="userSignUp" />
<label for="passwordSignUp">Password</label>
<input type="password" name="passwordSign" id="passwordSignUp" />
<input type="submit" name="Submit" id="Submit" value="Submit" />
</form>
I have the following .php:
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "test";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$db_selected = mysqli_select_db($conn, $dbname);
$userSignUp = ""; // If I substitute "" with characters at this time the table is well updated
$passwordSignUp = ""; // Same as before
if(isset($_POST['userSign'])){
$userSignUp = $_POST['userSign'];
}
if (isset($_POST['passwordSign'])) {
$passwordSignUp = $_POST['passwordSign'];
}
$sql = "INSERT INTO test.person (FirstName, Password) VALUES ('$userSignUp', '$passwordSignUp')";
if ($conn->query($sql) === TRUE) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
$conn->close();
Related
"I have read a lot of problem been solved in stackoverflow similar to my problem, and have seen a lot of example, yet still my code is not inserting in to mysql. however if i hard feed the php it would insert. my info is coming as submit from html post.I have good server connection and also connection to the database, can any one help me if i miss any thing. here is my code below."
<?php
$servername = "localhost";
$username = "root";
$password = "";
$db="image";
// Create connection
$connection = mysqli_connect($servername, $username, $password, $db); // Establishing Connection with Server
if (!$connection) {
die("Connection failed: " . mysqli_connect_error());
}
else{
echo "Connected successfully";
}
if(isset($_POST['submit'])){ // Fetching variables of the form which travels in URL
$name = $_POST['name'];
$image = $_POST['image'];
echo $name;
echo $image;
if($name !=''||$image !=''){
//Insert Query of SQL
$query = mysqli_query("INSERT INTO image (id, name, imagename) VALUES ('NULL', '$name', '$image')");
echo "Data Inserted successfully...!!";
}
else{
echo "Insertion Failed <br/> Some Fields are Blank....!!";
}
}
mysqli_close($connection); // Closing Connection with Server
?>
<form action = "test2.php" method="POST" enctype="multipart/form-data">
<label>name: </label><input type="text" name="name" />
<label>File: </label><input type="text" name="image" />
<input type="submit" />
</form>
</body>
</html>
" i expect output of 5/2 to be 2.5"
Write the name for submit button
<input type="submit" name="submit" />
then in php file
if(isset($_POST['submit'])){ // Fetching variables of the form which travels in URL
}
this if statement will run
you not given name attribute to button so give name="submit" and if you want to upload file then change type="file"
<?php
$servername = "localhost";
$username = "root";
$password = "";
$db="image";
// Create connection
$connection = mysqli_connect($servername, $username, $password, $db); // Establishing Connection with Server
if (!$connection) {
die("Connection failed: " . mysqli_connect_error());
}
else{
echo "Connected successfully";
}
if(isset($_POST['submit'])){ // Fetching variables of the form which travels in URL
$name = $_POST['name'];
$image = $_POST['image'];
echo $name;
echo $image;
if($name !=''||$image !=''){
//Insert Query of SQL
$query = mysqli_query("INSERT INTO image (id, name, imagename) VALUES ('NULL', '$name', '$image')");
echo "Data Inserted successfully...!!";
}
else{
echo "Insertion Failed <br/> Some Fields are Blank....!!";
}
}
mysqli_close($connection); // Closing Connection with Server
?>
<form action = "test2.php" method="POST" enctype="multipart/form-data">
<label>name: </label><input type="text" name="name" />
<label>File: </label><input type="file" name="image" />
<input type="submit" name="submit" />
</form>
</body>
</html>
You're checking isset($_POST['submit']) but there is no input field which is posted with submit name.. you need to add the name attribute in the submit button. also you're not passing the $connection in the mysqli_query.
$servername = "localhost";
$username = "root";
$password = "";
$db="image";
// Create connection
$connection = mysqli_connect($servername, $username, $password, $db); // Establishing Connection with Server
if (!$connection) {
die("Connection failed: " . mysqli_connect_error());
}
else{
echo "Connected successfully";
}
if(isset($_POST['submit'])){ // Fetching variables of the form which travels in URL
$name = $_POST['name'];
$image = $_POST['image'];
echo $name;
echo $image;
if($name !=''||$image !=''){
//Insert Query of SQL
$query = mysqli_query($connection, "INSERT INTO image (id, name, imagename) VALUES ('NULL', '$name', '$image')");
if($query !== false){
echo "Data Inserted successfully...!!";
}
else{
echo "Query failed";
}
}
else{
echo "Insertion Failed <br/> Some Fields are Blank....!!";
}
}
mysqli_close($connection); // Closing Connection with Server
?>
<form action = "test2.php" method="POST" enctype="multipart/form-data">
<label>name: </label><input type="text" name="name" />
<label>File: </label><input type="text" name="image" />
<input type="submit" name = "submit" />
</form>
</body>
</html>
One more suggestion always use PDO in code to prevent SQL injection. Your code is vulnerable to sql injection.
Whenever I try to save user input to a MySQL database, it saves empty rows as soon as I load the page. Here is my index.php file:
<?php
$servername = "localhost";
$username = "username";
$password = "password";
$dbname = "dbname";
$uName = mysqli_real_escape_string(mysqli_connect($servername, $username,
$password, $dbname), $_POST['username']);
$pass = mysqli_real_escape_string(mysqli_connect($servername, $username,
$password, $dbname), $_POST['password']);
// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "INSERT INTO users (username, password)
VALUES ('$uName', '$pass')";
if ($conn->query($sql) === TRUE) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
$conn->close();
?>
<!DOCTYPE html>
<html>
<body>
<form action="index.php" method="post">
<input type="text" name="username"> <br> <br>
<input type="password" name="password"> <br> <br>
<input type="submit" name="submit">
</form>
</body>
</html>
Please tell me if there is anything I am doing wrong.
Thanks
Please see modified code below:
$servername = "localhost";
$username = "username";
$password = "password";
$dbname = "dbname";
// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
if(isset($_POST['submit'])){
$uName = mysqli_real_escape_string(mysqli_connect($servername, $username,
$password, $dbname), $_POST['username']);
$pass = mysqli_real_escape_string(mysqli_connect($servername, $username,
$password, $dbname), $_POST['password']);
$sql = "INSERT INTO users (username, password)
VALUES ('$uName', '$pass')";
if ($conn->query($sql) === TRUE) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
$conn->close();
}
?>
<!DOCTYPE html>
<html>
<body>
<form action="" method="post">
<input type="text" name="username"> <br> <br>
<input type="password" name="password"> <br> <br>
<input type="submit" name="submit">
</form>
</body>
</html>
<?php
$servername = "localhost";
$username = "username";
$password = "password";
$dbname = "dbname";
$conn = mysqli_connect($servername, $username, $password, $dbname);
if(isset($_POST['submit']))
{
$uName=$_POST['username'];
$password=$_POST['password'];
$sql = "INSERT INTO users (username, password) VALUES ('$uName', '$pass')";
if(mysqli_query($conn, $sql)){
{
echo "New record inserted successfully";
}
else
{
echo "There have some problem";
}
}
?>
Please Try this one
I have a custom search form on my website with this code:
<center>
<form action="orderstatus.php" method="POST">
<input type="text" name="OrderLineNumber" value="">
<br>
<input type="submit" value="Submit">
</form>
</center>
And although I call orderstatus.php to action . When I press the submit button I get this Url : blablabla.de/action_page.php?OrderLineNumber=+23563623
This means instead of running orderstatus.php it's running action_page.php
My orderstatus.php is trying to print One Select Row from database like this :
<!DOCTYPE html>
<html>
<body>
<?php
$servername = "*******";
$username = "*******";
$password = "*******";
$dbname = "********";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT * FROM OrderStatus";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc() & fetch_assoc() = $_POST["name"] ) {
echo "<br> Order Line Number: ". $row["OrderLineNumber"]. " - Date Started: ". $row["Date Started"]. " " . $row["Status"] . "<br>";
}
} else {
echo "0 results";
}
$conn->close();
?>
</body>
</html>
I'm trying to register the e-mail entered in my form by users in my SQL table. but I'm not receiving any errors and the data are not saved either!
<?php
echo "I was here !!!";
if(!empty($_POST['mail']))
{
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "mailing";
echo "I was here !!!";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = 'INSERT INTO contact VALUES ("","'.$_POST['mail'].'")';
if ($conn->query($sql) === TRUE) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
$conn->close();
}
?>
and my html code:
<div class="form">
<p>Rester notifié par toutes les nouveautés !</p>
<form method="post" action="index.php" class="mainform">
<div class="field"><input type="text" class="field" name="mail" /></div>
<div class="submit"><input class="submit" type="button" value="Envoyer" /></div>
</form>
</div>
can you tell me what's the problem ?
change your button type .because if you want to submit the data by form then button type should be submit like that
<input class="submit" type="submit" value="Envoyer" />
Check if there's a value for $_POST['mail']. Your condition didn't handle empty value for $_POST['mail']. If there is a value. Change your query
INSERT INTO contact(email) VALUES ("'.$_POST['mail'].'")
Try this. Since you only need to add an email. Hope it helps.
i try to creat a table with html and php
when i insert data into my db i get num 1 like a values in all column
this my code
<html dir="rtl">
<form action="" method="post">
<label for="Nom">الاسم:</label>
<center><input type="text" name="Nom"></center>
<label for="Cin">البطاقة الوطنية:</label>
<center><input type="text" name="Cin"></center>
<label for="Tel">الهاتف:</label>
<center> <input type="text" name="Tel"></center>
<label for="DATE_donation"> تاريخ التبرع:</label>
<center><input type="date" name="DATE_donation"></center>
<center><input type="submit" value="إدخال"></center>
</form>
</html>
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "ikhlas";
$con= mysqli_connect($servername, $username, $password, $dbname);
if (!$con) {
die("Connection failed: " . mysqli_connect_error());
}
$Name =isset($_POST['Nom']);
$CIN = isset($_POST['Cin']);
$TEL = isset($_POST['Tel']);
$date = isset($_POST['DATE_donation']);
$sql="INSERT INTO persone(Nom, Cin, Tel, DATE_donation) value ('$Name','$CIN','$TEL','$date')";
if (mysqli_query($con, $sql)) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . mysqli_error($con);
}
?>
and this my ressult in dbenter image description here
Sure because you define all variable with $foo = isset($bar) instead of
if(isset($bar))
$foo = $bar;
Take a look to the doc about SQL injection too: http://php.net/manual/en/security.database.sql-injection.php
Remove your isset()
http://php.net/manual/en/function.isset.php
Replace to:
if(isset($_POST['إدخال']))
{
$Name = (!empty($_POST['Nom']))?$_POST['Nom']:"";
$CIN = (!empty($_POST['Cin']))?$_POST['Cin']:"";
$TEL = (!empty($_POST['Tel']))?$_POST['Tel']:"";
$date = (!empty($_POST['DATE_donation']))?$_POST['DATE_donation']:"";
}