till where i could understand its a problem in variable please help in solving this issue
what am i developing? i am making a gui interface to edit a database for my project this code is just php and a bit of html hopefully this problem will be resolved main form:
<section class="signup form">
<h2>rigister yourself please</h2>
<form action="includes/signup.inc.php" method="post">
<input type="text" name="name" placeholder="your name">
<input type="text" name="gender" placeholder="your gender " >
<input type="text" name="admission_number" placeholder="addmisopn number">
<input type="text" name="fathers_name" placeholder="father's name">
<input type="text" name="contact" placeholder="your phone no.">
<input type="text" name="email_id" placeholder="your email">
<button type="submit" name="submit">submit </button>
</form>
</section>
the link to the database:
<?php
$SERVERName ="localhost";
$dBUsername ="root";
$dBpassword="";
$dbname="library";
$conn = mysqli_connect(,$SERVERName,$dBUsername,$dBpassword,$dbname);
if (!conn) {
die("connection failed" .mysqli_connect_error())
}
error in this file
<?php
if (isset($_POST["submit"])) {
echo" The student has beeen registered now you may Continue your work!"
$name = $_POST["name"];
$gender = $_POST["gender"],
$admission_number = $_POST["admission_number"],
$fathers_name = $_POST["fathers_name"],
$contact = $_POST["contact"],
$email id = $_POST["email id"],
}
else{
header("location: ../signup.php")
}
till where i could understand its a problem in variable please help in solving this issue
what am i developing? i am making a gui interface to edit a database for my project this code is just php and a bit of html hopefully this problem will be resolved
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