I am doing a to do list webpage using HTML, PHP and MySQL but I cannot submit the datetime to the database:
HTML:
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="utf-8" />
<title>Sign Controller</title>
<link rel="stylesheet" href="/css/style.css" />
<script src="http://code.jquery.com/jquery-latest.js"></script>
<script src="update.js"></script>
</head>
<body>
<!--MAIN CONTENT HERE-->
<div class="container">
<form id="controller" action="php/update.php" method="post">
<h3>Sign display controller</h3>
<h4>Event Controller</h4>
<fieldset>
<input placeholder="sign ID" type="text" name="sign_Name" tabindex="1" required autofocus/>
</fieldset>
<h5>Event Start</h5>
<fieldset>
<input type="datetime-local" name="time_eventStart" tabindex="2" required/>
</fieldset>
<h5>Event End</h5>
<fieldset>
<input type="datetime-local" name="time_eventEnd" tabindex="3" required/>
</fieldset>
<fieldset>
<input placeholder="Event source" type="text" name="event_Source" tabindex="4"/>
</fieldset>
<fieldset>
<input placeholder="Idle source" type="text" name="idle_Source" tabindex="5"/>
</fieldset>
<button type="submit">Update</button>
</form>
</div>
</body>
</html>
PHP:
<?php
$servername = "localhost";
$username = "root";
$password = "root";
$dbname = "projectSS";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sign_Name = $_POST["sign_Name"];
$event_Source = $_POST["event_Source"];
$idle_Source = $_POST["idle_Source"];
$time_eventStart = $_POST["time_eventStart"];
$time_eventEnd = $_POST["time_eventEnd"];
$sql = "INSERT INTO display (sign_Name,event_timeStart,event_timeEnd,event_Source,idle_Source)
VALUES ('$sign_Name','$time_eventStart','$time_eventEnd','$event_Source','$idle_Source')";
if ($conn->query($sql) === TRUE) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
$conn->close();
?>
I'm having a hard time to deal with the datetime-local type and I am new to programming so.
just change date time format as per mysql standard
if your column datatype is date used below
$time_eventStart = date('Y-m-d',strtotime($_POST["time_eventStart"]));
$time_eventEnd = date('Y-m-d',strtotime($_POST["time_eventEnd"]));
if your column datatype is datetime used below
$time_eventStart = date('Y-m-d H:i:s',strtotime($_POST["time_eventStart"]));
$time_eventEnd = date('Y-m-d H:i:s',strtotime($_POST["time_eventEnd"]));
Change your query. Your query is not executing.
try this:
$sql = "INSERT INTO display (sign_Name,event_timeStart,event_timeEnd,event_Source,idle_Source)
VALUES ('".$sign_Name."','".$time_eventStart."','".$time_eventEnd."','".$event_Source."','".$idle_Source."')";
Try something like this
$date = '2014-06-06 12:24:48';
echo date('d-m-Y (H:i:s)', strtotime($date));
and then save this.
datetime-local it considered 24 hour format so if you select 07/12/2017 12:00 AM it posted 2017-07-12T00:00.
In insert query just put posted value into your query. e.g.
$_POST['time_eventStart'].
Also if you print date-time to your input echo posted value. e.g.
<input type="datetime-local" name="time_eventStart" value="<?php echo $_POST['time_eventStart']; ?>" tabindex="2" required="" />
Related
There is something wrong with my code :
On the edit page, I want to show the user the previous value in the input box.
But one error I keep getting about the value is the following :
Notice: Undefined variable: gebruikers_naam in C:\xampp\htdocs\website_herkansing\edit_gebruiker.php on line 72
I think there is something wrong with the isset/submit part but I just
can not figure it out..
Here is the code I'm working with
<?php
session_start();
define('DB_NAME', 'ochtendgloren');
$servername = "localhost";
$username = "root";
$password = "";
$db = "ochtendgloren";
$tbl = "members";
// Create connection
$conn = new mysqli($servername, $username, $password, $db);
//Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
if (isset($_POST['submit']))
{
$gebruikers_naam = mysqli_real_escape_string($db, $_POST['gebruikers_naam']);
htmlentities($gebruikers_naam);
$id = $_GET['id'];
$query = "UPDATE members
SET gebruikers_naam = '$gebruikers_naam'
WHERE id = '$id' " ;
$result = $conn->query($query);
if($result){
echo ("<SCRIPT LANGUAGE='JavaScript'>
window.alert('edit succesvol!')
window.location.href='admin_members.php';
</SCRIPT>");
}
}
?>
<html>
<head>
<link rel="stylesheet" href="boekingsform.css">
<link rel="stylesheet" href="https://fonts.googleapis.com/css?family=Abel">
</head>
<div class="boeken">
<h1>Wijzig hier de gebruiker</h1>
<form action="editrij.php?id=<?= $id ?>" method="post" >
<div class="row">
<div class="col-25">
<label for="gebruikers_naam"> vul hier de nieuwe gebruikers naam in: </label>
</div></div>
<br>
<div class="row">
<div class="col-75">
<input type="text" name="voornaam" required="required" value="<?= $gebruikers_naam['gebruikers_naam'] ?>"/>
</div>
</div>
<br>
<input type="submit" value="submit" name="submit" />
</form>
</div>
</html>
I think the problem is related with your variable usage and name of input box it must be gebruikers_naam
On line :
<input type="text" name="voornaam" required="required" value="<?= $gebruikers_naam['gebruikers_naam'] ?>"/>
You may use $gebruikers_naam only to print the name.
Also you must assign a global variable before before using it in if clause.
Just assign null like $gebruikers_naam = ""; after variables declarations.
You aren't sending the variable gebruikers_naam, but voornaam. This line of code
<input type="text" name="voornaam" required="required" value="<?= $gebruikers_naam['gebruikers_naam'] ?>"/>
Should be
<input type="text" name="gebruikers_naam" required="required" value="<?= $gebruikers_naam['gebruikers_naam'] ?>"/>
Also, because you are using the variables $id and $gebruikers_naam in the form, you should assign them a value before the if clause.
Still not completely sure the 'flow' of your code makes a lot of sense, but following should help a bit:
<?php
session_start();
//define('DB_NAME', 'ochtendgloren'); // not used in your code, commented out
$servername = "localhost";
$username = "root";
$password = "";
$db = "ochtendgloren";
$tbl = "members";
// Create connection
$conn = new mysqli($servername, $username, $password, $db);
//Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
if (isset($_POST['submit']))
{
if(isset($_POST['gebruikers_naam']) && !empty($_POST['gebruikers_naam'])) { // added condition: we need a 'gebruikers_naam'
$gebruikers_naam = mysqli_real_escape_string($conn, $_POST['gebruikers_naam']); // changed $db to $conn
// the return value (encoded string) of htmlentities is not used in your code
// so commented it out
//htmlentities($gebruikers_naam);
$id = $_GET['id'];
$query = "UPDATE members
SET gebruikers_naam = '$gebruikers_naam'
WHERE id = '$id' " ;
$result = $conn->query($query);
}
if($result){
echo ("<SCRIPT LANGUAGE='JavaScript'>
window.alert('edit succesvol!')
window.location.href='admin_members.php';
</SCRIPT>");
} else { // added else {} statement
echo "<script> alert('Error: could not update the database'); </script>"; // added: error message
}
}
?>
<html>
<head>
<link rel="stylesheet" href="boekingsform.css">
<link rel="stylesheet" href="https://fonts.googleapis.com/css?family=Abel">
</head>
<div class="boeken">
<h1>Wijzig hier de gebruiker</h1>
<form action="editrij.php?id=<?php echo $id; ?>" method="post" > <!-- you need to echo $id and close with ';' - changed: echo $id; -->
<div class="row">
<div class="col-25">
<label for="gebruikers_naam"> vul hier de nieuwe gebruikers naam in: </label>
</div></div>
<br>
<div class="row">
<div class="col-75">
<?php $gebruikers_naam = (isset($gebruikers_naam)) ? $gebruikers_naam : 'N/A'; ?> <!-- added a test to see if $gebruikers_naam is available -->
<input type="text" name="voornaam" required="required" value="<?php echo $gebruikers_naam; ?>"/> <!-- variable is $gebruikers_naam, changed (echo and ';') -->
</div>
</div>
<br>
<input type="submit" value="submit" name="submit" />
</form>
</div>
</html>
I am trying to send the data put into the input fields to my database and I cant seem to make it work out properly..
The ultimate goal is to put in the input fields into the database and show the inserted data in another window.
Here's my code
<?php
$servername = "localhost";
$username = "root";
$password = "";
// Create connection
$conn = new mysqli($servername, $username, $password);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
echo "Connected successfully";
// create a variable
$naam=$_POST['namen'];
$plaats=$_POST['plaatsen'];
$land=$_POST['landen'];
//Execute the query
mysqli_query($conn,"INSERT INTO phptoets(Namen,Plaatsen,Landen)
VALUES('$naam','$plaats','$land')");
?>
<!DOCTYPE html>
<html>
<head>
<link rel="stylesheet" type="text/css" href="main.css">
<title>PHP Toets</title>
</head>
<body>
<div class="import_intel">
</div>
<form method="POST">
<div class="invulform">
<h2>Vul hier de gegevens in die naar de database moeten</h2>
<input type="text" name="naam" class="input_name" placeholder="Naam"><br>
<input type="text" name="plaats" class="input_plaats" placeholder="Plaats"><br>
<input type="text" name="land" class="input_land" placeholder="Land"><br>
<input type="submit" name="submit" class="submit_button" value="Verstuur">
</div>
</form>
<div class="overzichtform">
<h3>Data</h3>
</div>
</body>
</html>
Try this. It will help you. I've done few changes in your code.
- Add form to post the data on server
- Add database name in connection
- Add provincie field in form (because you are trying to get that in php)
- Use the same variable in query as declared at the time of connection
<?php
if (isset($_POST['submit'])) {
$servername = "localhost";
$username = "root";
$password = "";
$database = "DATABASE";
// Create connection
$conn = new mysqli($servername, $username, $password, $database);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
// create a variable
$namen=$_POST['naam'];
$plaatsen=$_POST['plaats'];
$landen=$_POST['land'];
$provincie=$_POST['provincie'];
//Execute the query
mysqli_query($conn, "INSERT INTO employees1(naam,plaats,land,provincie) VALUES('$namen','$plaatsen','$landen','$provincie')");
}
?>
<!DOCTYPE html>
<html>
<head>
<link rel="stylesheet" type="text/css" href="main.css">
<title>PHP Toets</title>
</head>
<body>
<div class="import_intel">
</div>
<form method="POST" action="">
<div class="invulform">
<h2>Vul hier de gegevens in die naar de database moeten</h2>
<input type="text" name="naam" class="input_name" placeholder="Naam"><br>
<input type="text" name="plaats" class="input_plaats" placeholder="Plaats"><br>
<input type="text" name="land" class="input_land" placeholder="Land"><br>
<input type="text" name="provience" class="input_provience" placeholder="Provience"><br>
<input type="button" name="submit" class="submit_button" value="Verstuur">
</div>
</form>
<div class="overzichtform">
<h3>Data</h3>
</div>
</body>
</html>
This is my solution:
You forgot to set database name and the names of your input fields where not equal to your $_POST names.
<?php
$servername = "localhost";
$username = "root";
$password = "root";
$database = "test";
// Create connection
$conn = new mysqli($servername, $username, $password, $database);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
echo "Connected successfully";
// create a variable
$naam=$_POST['naam'];
$plaats=$_POST['plaats'];
$land=$_POST['land'];
//Execute the query
mysqli_query($conn,"INSERT INTO phptoets(namen,plaatsen,landen)
VALUES('$naam','$plaats','$land')");
?>
<!DOCTYPE html>
<html>
<head>
<link rel="stylesheet" type="text/css" href="main.css">
<title>PHP Toets</title>
</head>
<body>
<div class="import_intel">
</div>
<form method="post">
<div class="invulform">
<h2>Vul hier de gegevens in die naar de database moeten</h2>
<input type="text" name="naam" class="input_name" placeholder="Naam"><br>
<input type="text" name="plaats" class="input_plaats" placeholder="Plaats"><br>
<input type="text" name="land" class="input_land" placeholder="Land"><br>
<input type="submit" name="submit" class="submit_button" value="Verstuur">
</div>
</form>
<div class="overzichtform">
<h3>Data</h3>
</div>
</body>
</html>
<?php
// TESTS
$servername = "localhost";
$username = "root";
$password = "";
$db_name = "test";
// Create connection
$conn = new mysqli($servername, $username, $password, $db_name);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
echo "Connected successfully <br>";
if(isset($_POST['submit']))
{
// create a variable
$a = (string)filter_input(INPUT_POST,'naam');
$b = (string)filter_input(INPUT_POST,'plaats');
$c = (string)filter_input(INPUT_POST,'land');
$d = (string)filter_input(INPUT_POST,'provincie');
echo("executing query <br>");
//Execute the query
if($a != null && $b != null && $c != null && $c != null)
{
$sql="INSERT INTO employees1 (naam,plaats,land,provincie) VALUES (?,?,?,?)";
echo("sql ".$sql. "<br>");
if($stmt = $conn->prepare($sql))
{
$stmt->bind_param("ssss",$a,$b,$c,$d);
$stmt->execute();
$stmt->close();
}
}
$sql = "SELECT naam, plaats, land, provincie FROM employees1";
if ($stmt = $conn->prepare($sql))
{
$stmt->execute();
$stmt->bind_result($naam,$plaats,$land,$provincie);
while ($stmt->fetch())
{
printf("naam : %s, plaats: %s, land: %s, provincie: %s <br>",$naam,$plaats,$land,$provincie);
}
$stmt->close();
}
}
?>
<!DOCTYPE html>
<html>
<head>
<link rel="stylesheet" type="text/css" href="main.css">
<title>PHP Toets</title>
</head>
<body>
<div class="import_intel">
</div>
<div class="invulform">
<h2>Vul hier de gegevens in die naar de database moeten</h2>
<form class="my_form" target="_self" enctype="multipart/form-data" method="post">
<input type="text" name="naam" class="input_name" placeholder="Naam"><br>
<input type="text" name="plaats" class="input_plaats" placeholder="Plaats"><br>
<input type="text" name="land" class="input_land" placeholder="Land"><br>
<input type="text" name="provincie" class="input_land" placeholder="Provincie"><br>
<input type="submit" name="submit" class="submit_button" value="Verstuur">
</form>
</div>
<div class="overzichtform">
<h3>Data</h3>
</div>
</body>
</html>
I will just name few things that I changed in your code, not mentioning syntax errors
you dont specify db_name in your sql connection
you dont use prepared statements nor any kind of input filtering
(note : my input filtering is very basic, read more about how to
filter inputs)
to address html form you need to create one and have submit input
type inside
I'm using Cloud9 to create a website. For whatever reason, the data taken from a HTML page will not get inserted into the Database.
I have tested to see if the Database is connected and it is. I would like to be able to get the data to be inserted into the Database.
The HTML code and PHP code is:
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="UTF-8">
<title>Add Record Form
</title>
</head>
<body>
<form action="../php/keithphp/address_submit.php" method="post">
<p>
<label for="address_street">Street</label>
<input type="text" name="address_street" id="address_street">
</p>
<p>
<label for="address_street2">Street 2</label>
<input type="text" name="address_street2" id="address_street2">
</p>
<p>
<label for="address_city">City</label>
<input type="text" name="address_city" id="address_city">
</p>
<p>
<label for="address_county">County</label>
<input type="text" name="address_county" id="address_county">
</p>
<p>
<label for="eircode">Eircode</label>
<input type="text" name="eircode" id="eircode">
</p>
<!-- <p>
<label for="address_geo_latitude">Latitude</label>
<input type="float" name="address_geo_latitude" id="address_geo_latitude">
</p>
<p>
<label for="address_geo_longtitude">Longitude</label>
<input type="float" name="address_geo_longtitude" id="address_geo_longtitude">
</p> -->
<input type="submit" value="Submit">
</form>
</body>
</html>
*****************
<?php
$servername = getenv('IP');
$username = getenv('C9_USER');
$password = "";
$database = "c9";
$dbport = 3306;
// Create connection
$db = new mysqli($servername, $username, $password, $database, $dbport);
// Check connection
if ($db->connect_error) {
die("Connection failed: " . $db->connect_error);
}
/*$address_id = $_POST['address_id'];*/
$address_street = $_POST['address_street'];
$address_street2 = $_POST['address_street2'];
$address_city = $_POST['address_city'];
$address_county = $_POST['address_county'];
$address_eircode = $_POST['address_eircode'];
/*$address_geo_latitude = $_POST['address_geo_latitude'];
$address_geo_longtitude = $_POST['address_geo_longtitude'];*/
$sql = "INSERT INTO Address(address_id, address_street, address_street2, address_city, address_county, address_eircode, address_geo_latitude, address_geo_longtitude) VALUES ('$address_id', '$address_street', '$address_street2', '$address_city', '$address_county', '$address_eircode', '$address_geo_latitude', '$address_geo_longtitude')";
$success = $db->query($sql);
if (!$sucess){
die("Could not enter data: ".$db->error);
}
echo "Thank you. Address submitted!"
/*my$db->close();*/
?>
And the result after typing in test data is (no error message after text), I get;
Could not enter data:
I have created a html register page which is really basic and requires the user to enter their First name, Last name, email, and password. However only the first and last names are being recorded in the database in phpmyadmin and the email and passwords are showing as blank cloumns. I have tried to drop and add the tables and columns again without any luck, i have changed variable names and no luck as well. Not too sure what to do.
Php code
<?php
$servername = "localhost";
$username = "";
$password = "";
$dbname = "";
//Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
//echo "Connected successfully";
$first_name = $_POST['first_name'];
$last_name = $_POST['last_name'];
$email = $_POST['email'];
$password_ = $_POST['password_'];
if (isset($_POST['register'])) {
register($first_name,$last_name,$_email,$password,$conn);
}
$conn->close();
function register($first_name,$last_name,$email,$password,$conn) {
// echo $first_name . " " . $last_name . " " . $student_id . " " . $email;
$sql = "INSERT INTO `register` (`first_name`, `last_name`, `email`, `password_`) VALUES ('$first_name', '$last_name', '$email', '$password_')";
if ($conn->query($sql) === TRUE) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
}
?>
html code
<!DOCTYPE html>
<html lang="en">
<head>
<title>Bootstrap Case</title>
<meta charset="utf-8">
<meta name="viewport" content="width=device-width, initial-scale=1">
<link rel="stylesheet" href="http://maxcdn.bootstrapcdn.com/bootstrap/3.3.5/css/bootstrap.min.css">
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.3/jquery.min.js"></script>
<script src="http://maxcdn.bootstrapcdn.com/bootstrap/3.3.5/js/bootstrap.min.js"></script>
</head>
<body>
<form role="form" action="register.php" method="POST">
<div class="form-group">
<label>First Name:</label>
<input type="text" class="form-control" id="first_name" name="first_name"required>
</div>
<div class="form-group">
<label>Last Name:</label>
<input type="text" class="form-control" id="last_name" name="last_name"required>
</div>
<div class="form-group">
<label>Email address:</label>
<input type="varchar" class="form-control" name="e_mail" id="email"required>
</div>
<div class="form-group">
<label>Password:</label>
<input type="password" class="form-control" id="password" name="password"required>
</div>
<div class="form-group">
<label>Confirm Password:</label>
<input type="password" class="form-control" id="confirm_password"required >
<script>
var password = document.getElementById("password")
, confirm_password = document.getElementById("confirm_password");
function validatePassword(){
if(password.value != confirm_password.value) {
confirm_password.setCustomValidity("Passwords Don't Match");
} else {
confirm_password.setCustomValidity('');
}
}
password.onchange = validatePassword;
confirm_password.onkeyup = validatePassword;
</script>
</div>
<button type="submit" class="btn btn-default" name="register">Register</button>
</form>
</body>
</html>
In order to grab the _POST variables, your input forms must have a name attribute. For your Email form, you have only specified an ID and no name. Go back and add in the name='email' attribute and it should work. Same for password.
It looks like i was missing an underscore in the php register code where it states the function. i have added it and now my code seems to be working, thanks for all the feedback!
The below code is a simple form that is sending the data that is inputted to my local database.
<html>
<head>
<title>!!!!!!!!!!!!!!</title>
<link rel="stylesheet" type="text/css" href="style.css">
</head>
<body>
<div id="main">
<h1>Insert data into database using mysqli</h1>
<div id="login">
<h2>Student's Form</h2>
<hr/>
<form action="" method="post">
<label>Student Name :</label>
<input type="text" name="stu_name" id="name" required="required" placeholder="Please Enter Name"/><br /><br />
<label>Student Email :</label>
<input type="email" name="stu_email" id="email" required="required" placeholder="john123#gmail.com"/><br/><br />
<label>Student City :</label>
<input type="text" name="stu_city" id="school" required="required" placeholder="Please Enter Your City"/><br/><br />
<input type="submit" value=" Submit " name="submit"/><br />
</form>
</div>
<!-- Right side div -->
</div>
<?php
if(isset($_POST["submit"])){
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "Student";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "INSERT INTO students (student_name, student_email, student_school)
VALUES ('".$_POST["stu_name"]."','".$_POST["stu_email"]."','".$_POST["stu_city"]."')";
if ($conn->query($sql) === TRUE) {
echo "<script type= 'text/javascript'>alert('New Record Inserted Successfully');</script>";
} else {
echo "<script type= 'text/javascript'>alert('Error: " . $sql . "<br>" . $conn->error."');</script>";
}
$conn->close();
}
?>
</body>
</html>
The issue that I am finding is that I am trying to change the Student City input field into a drop down where the values is retrieve from the database and put into a drop down list for a new user to select.
Could someone advise on what needs to be done please.
i am trying to use the below code and send the below list to my database.
<option value="US">United States</option>
<option value="UK">United Kingdom</option>
<option value="France">France</option>
<option value="Mexico">Mexico</option>
but i am finding it hard to send the these values to the database with my above code as well as where to place this code.
As a rough example of how you could build the dropdown menu using data from your db this should give you the general idea perhaps.
/* store formatted menu options in temp array */
$html=array();
/* query db to find schools/cities */
$sql='select distinct `student_school` from `students` order by `student_school`';
$res=$mysqli_query( $conn, $sql );
/* process recordset and store options */
if( $res ){
while( $rs=mysqli_fetch_object( $res ) ){
$html[]="<option value='{$rs->student_school}'>{$rs->student_school}";
}
}
/* render menu */
echo "<select name='stu_city'>", implode( PHP_EOL, $html ), "</select>";
You need to refactor your code by moving the if (isset($_POST)) above the html:
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "Student";
// Create connection
$conn = new mysqli ( $servername, $username, $password, $dbname );
// Check connection
if ($conn->connect_error) {
die ( "Connection failed: " . $conn->connect_error );
}
$sql = "SELECT city_name FROM cities" ;
if ($conn->query ( $sql ) === TRUE) {
$cities = ... // build the cities from the query result
} else {
$cities = '<option value="none">No cities found</option>' ;
}
if (isset ( $_POST ["submit"] )) {
$sql = "INSERT INTO students (student_name, student_email, student_school)
VALUES ('" . $_POST ["stu_name"] . "','" . $_POST ["stu_email"] . "','" . $_POST ["stu_city"] . "')";
if ($conn->query ( $sql ) === TRUE) {
echo "<script type= 'text/javascript'>alert('New Record Inserted Successfully');</script>";
} else {
echo "<script type= 'text/javascript'>alert('Error: " . $sql . "<br>" . $conn->error . "');</script>";
}
$conn->close ();
}
?>
<html>
<head>
<title>!!!!!!!!!!!!!!</title>
<link rel="stylesheet" type="text/css" href="style.css">
</head>
<body>
<div id="main">
<h1>Insert data into database using mysqli</h1>
<div id="login">
<h2>Student's Form</h2>
<hr />
<form action="" method="post">
<label>Student Name :</label> <input type="text" name="stu_name"
id="name" required="required" placeholder="Please Enter Name" /><br />
<br /> <label>Student Email :</label> <input type="email"
name="stu_email" id="email" required="required"
placeholder="john123#gmail.com" /><br />
<br /> <label>Student City :</label> <select name="stu_city" multiple><?php echo $cities; ?>
</select>><br />
<br /> <input type="submit" value=" Submit " name="submit" /><br />
</form>
</div>
<!-- Right side div -->
</div>
</body>
</html>
Use the Select tag: Lets say you hav a column in your database with Student City, like this, lets say the database field is called city
City 1
City 2
City 3
Step 1: Query the database and fetch all the Cities
$sql = "SELECT city FROM table_name";
$result = $conn->query($sql);
Then you come to your dropdown:
<select name="stu_city" id="..." required>
<?php
while($cities = $conn->fetch_array($result){
extract($cities);
echo "<option value='...'>$city</option>";
}
?>
</select>
You need to refactor your code by moving the if (isset($_POST)) above the html:
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "Student";
// Create connection
$conn = new mysqli ( $servername, $username, $password, $dbname );
// Check connection
if ($conn->connect_error) {
die ( "Connection failed: " . $conn->connect_error );
}
$sql = "SELECT city_name FROM cities" ;
$result = $conn->query ( $sql );
if (isset ( $_POST ["submit"] )) {
$sql = "INSERT INTO students (student_name, student_email, student_school)
VALUES ('" . $_POST ["stu_name"] . "','" . $_POST ["stu_email"] . "','" . $_POST ["stu_city"] . "')";
if ($conn->query ( $sql ) === TRUE) {
echo "<script type= 'text/javascript'>alert('New Record Inserted Successfully');</script>";
} else {
echo "<script type= 'text/javascript'>alert('Error: " . $sql . "<br>" . $conn->error . "');</script>";
}
$conn->close ();
}
?>
<html>
<head>
<title>!!!!!!!!!!!!!!</title>
<link rel="stylesheet" type="text/css" href="style.css">
</head>
<body>
<div id="main">
<h1>Insert data into database using mysqli</h1>
<div id="login">
<h2>Student's Form</h2>
<hr />
<form action="" method="post">
<label>Student Name :</label> <input type="text" name="stu_name"
id="name" required="required" placeholder="Please Enter Name" /><br />
<br /> <label>Student Email :</label> <input type="email"
name="stu_email" id="email" required="required"
placeholder="john123#gmail.com" /><br />
<br /> <label>Student City :</label> <select name="stu_city" multiple>
<?php
if ($result == TRUE) {
while($cities = $conn->fetch_array($result)){
extract($cities);
echo "<option value=''>$city_name</option>";
}
}
else {
echo "<option value='none'>No cities found</option>";
}
?>
</select>><br />
<br /> <input type="submit" value=" Submit " name="submit" /><br />
</form>
</div>
<!-- Right side div -->
</div>
</body>
</html>