i am trying to insert simple two values in db but they are neither inserting nor giving errors. i do not know about what's happening. I also tried to sql query outside the if statement but it did not work.
please help
coding file
<?php
include 'includes/db_connection.php';
if (isset($_POST['submit']))
{
$name = test_input($_POST["name"]);
$fathername = test_input($_POST["fathername"]);
$sql = "INSERT INTO student (name,fathername)
VALUES ('$name','$fathername')";
mysqli_query($conn, $sql);
if(mysqli_query($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);
}
function test_input($data) {
$data = trim($data);
$data = stripslashes($data);
$data = htmlspecialchars($data);
return $data;
}
}
?>
<!DOCTYPE html>
<!--
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and open the template in the editor.
-->
<html>
<head>
<meta charset="UTF-8">
<title></title>
<script src="//ajax.googleapis.com/ajax/libs/jquery/1.10.2/jquery.min.js"></script>
<script src="//cdnjs.cloudflare.com/ajax/libs/jquery-form-validator/2.2.43/jquery.form-validator.min.js"></script>
</head>
<body>
<form action="valid_test.php" method="post" id="registration-form">
<div>
Name
<input name="name" data-validation="length alphanumeric"
data-validation-length="6-15"
data-validation-error-msg="User name has to be an alphanumeric value (6-15 chars)">
</div>
<div>
Father Name
<input name="fathername" data-validation="length alphanumeric"
data-validation-length="6-20"
data-validation-error-msg="User name has to be an alphanumeric value (6-20 chars)">
</div>
<div>
<input type="submit" value="Validate">
<input type="reset" value="Reset form">
</div>
</form>
When we post HTML form, elements get posted with name attribute.
You are missing name attribute for submit button.
Change:
<input type="submit" value="Validate">
To:
<input type="submit" value="Validate" name="submit">
As your submit button does not have a name attribute, the condition
if (isset($_POST['submit'])) is not getting fulfilled and hence code not working.
Also, you are repeating mysqli_query($conn, $sql);. Remove first instance of it.
First, in your PHP, where is $conn: you are trying to insert something that doesn't exist.Second, your SQL isn't quite right, instead of that you should have something like this:
$sql = "INSERT INTO student (name, fathername) VALUES (" . $name . ", " . $fathername . ");";
What you did in your question, is you put in '$name' and '$fathername' as individual characters, what you want is the value of the variable, this is how you do that.
Regards