how to post contact no. to sql database using php - php

Question

I want to store user's contact no. from an android registration form in SQL database but it shows 'failure' instead of 'success' when i try to add $_post['contact'] code to my PHP file.
This is my PHP file.
register.php
<?php
define('HOST','mysql8.000webhost.com');
define('USER','a6293046_******');
define('PASS','*********');
define('DB','a6293046_insti');
$con = mysqli_connect(HOST,USER,PASS,DB);
$name = $_POST['name'];
$email = $_POST['email'];
$address = $_POST['address'];
$contact =(int)$_POST['contact'];
$institute = $_POST['institute'];
$sql = "insert into Persons (name,email,address,contact-no,institute) values ('$name','$email','$address','contact-no','$institute')";
if(mysqli_query($con,$sql)){
echo 'success';
}
else{
echo 'failure';
}
mysqli_close($con);
?>
Please tell me what i am doing wrong in it. thank you

Answer 1

<?php
define('HOST','mysql8.000webhost.com');
define('USER','a6293046_******');
define('PASS','*********');
define('DB','a6293046_insti');
$con = mysqli_connect(HOST,USER,PASS,DB);
$name = $_POST['name'];
$email = $_POST['email'];
$address = $_POST['address'];
$contact =$_POST['contact'];
$institute = $_POST['institute'];
$sql = "insert into Persons (name,email,address,contact-no,institute) values ('$name','$email','$address',$contact,'$institute')";
$result=$con->query($sql);
if($result)
{
echo 'success';
}
else{
echo 'failure';
}
$con->close();
?>
remove the int for $POST_['contact'] and dnt use single quotes('') for $contact while inserting since it is a integer for string value you should give single quotes

Answer 2

Try this
<?php
define('HOST','mysql8.000webhost.com');
define('USER','a6293046_******');
define('PASS','*********');
define('DB','a6293046_insti');
$con = mysqli_connect(HOST,USER,PASS,DB);
//change 'somename' with the name of submit button!
if(isset($_POST['somename'])){
$error=0;
$name = $_POST['name'];
$email = $_POST['email'];
$address = $_POST['address'];
$contact = $_POST['contact'];
$institute = $_POST['institute'];
if(is_numeric($contact)){
$error=1;
}
if($error==1){
$sql = "insert into Persons (name,email,address,contact-no,institute) values ('$name','$email','$address','$contact','$institute')";
$insert= mysqli_query($con,$sql);
echo 'success';
}
else{
echo 'failure';
}
}
?>
this code will insert data in database only if $contact is numeric otherwise there will not be insert

Answer 3

'contact-no' is not variable.
change 'contact-no' to $contact
try this code
$sql = "insert into Persons (name,email,address,contact-no,institute) values ('$name','$email','$address','$contact','$institute')";

Answer 4

Offcoarse it will: you have an mistake in your SQL, you lost dollar sign and you should use $contact variable not $contact-no
$sql = "insert into Persons (name,email,address,contact-no,institute) values ('$name','$email','$address','$contact','$institute')";