Store the data in mysql - php

Question

i want to store a data to the mysql database but if i press the submit button it display the php page.
form:
<div id="test">
<form action="demo.php" method="post">
please enter the number(1 to 100) : <input type="text" name="value">
<input type="submit">
</form>
</div>
demo.php
<?php
$value=$_POST['value'];
$servername = "localhost";
$username = "root";
$password = "*****";
$dbname = "clock";
$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "INSERT INTO input VALUES (".$_POST['value'].")";
if ($conn->query($input) == TRUE) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
$conn->close();
?>

Answer 1

Replace $input with $sql in the if statement:
From
if ($conn->query($input) == TRUE) {
To
if ($conn->query($sql) == TRUE) {
Note: Since you are storing $_POST['value'] in $value, you don't need to use $_POST['value'] in the query, instead make use of $value.

Answer 2

There is one correction:
Change
if ($conn->query($input) == TRUE) {
To:
if ($conn->query($sql) == TRUE) {
Along with that, following are suggestions:
$sql = "INSERT INTO input VALUES (".$_POST['value'].")";
This query is OK, is value of $_POST['value'] is integer, but, for strings, it will create SQL Syntax Error.
Please correct this to:
$sql = "INSERT INTO input VALUES ('".$_POST['value']."')";
Observe, enclosed semi colon.
2) You are inserting only one field value and not specifying fields. This means you table has only field. Normally we do not have tables with only one field.
Please specify field names:
$sql = "INSERT INTO input (field_one) VALUES (".$_POST['value'].")";

Answer 3

Get values like this
$value=$_REQUEST['value'];
change the query to
$sql = "INSERT INTO input VALUES ('$value')";

Answer 4

1st : Add name attribute to submit button name="submit"
<input name="submit" type="submit">
2nd : use isset like this if(isset($_POST['submit'])){ //rest of code here }
3rd : Try to use prepared statement or pdo to avoid sql injection
demo.php
<?php
$servername = "localhost";
$username = "root";
$password = "*****";
$dbname = "clock";
$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
if(isset($_POST['submit'])){
$stmt = $conn->prepare("INSERT INTO input(value) VALUES (?)");
$stmt->bind_param('s',$_POST['value']);
//The argument may be one of four types:
//i - integer
//d - double
//s - string
//b - BLOB
//change it by respectively
if ($stmt->execute() == TRUE && $stmt->affected_rows>0) {
echo "New record created successfully";
} else {
echo "Error: <br>" . $conn->error;
}
}
$conn->close();
?>