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rated 0 times [  134] [ 5]  / answers: 1 / hits: 51554  / 8 Years ago, tue, august 2, 2016, 12:00:00

I am trying to insert value using ajax in php, but data is not inserted in database. I have taken this code from the questions answered in other question from this site. Can anyone suggest where am I making mistake..?



<script>
$(#submit).click(function() {
var name= $(#name).val();
var password= $(#password).val();

$.ajax({
type: POST,
url: insert.php,
data: name= + name+ &password= + password,
success: function(data) {
alert(sucess);
}
});


});

<script src=https://ajax.googleapis.com/ajax/libs/jquery/1.10.0/jquery.min.js></script>

<?php
//------insert.php------
$servername = localhost;
$username = username;
$password = password;
$dbname = dbname;

// Create connection
$conn = new mysqli($servername, $username, $password,$dbname);

// Check connection
if ($conn->connect_error) {
die(Connection failed: . $conn->connect_error);
}


$name=$_POST['name'];
$pass=$_POST['password'];
$sql= mysqli_query($conn,INSERT INTO insert_tbl(name,pass) VALUES('.$name.','.$pass.'));

?>

More From » php

 Answers
7
<script>
$(#FORM_ID).submit(function() {
var name= $(#name).val();
var password= $(#password).val();

$.ajax({
type: POST,
url: insert.php,
data: name= + name+ &password= + password,
success: function(data) {
alert(sucess);
}
});


});
</script>


and also either load



<script src=https://ajax.googleapis.com/ajax/libs/jquery/1.10.0/jquery.min.js></script>


before your script tag or use



 <script>
$(document).ready(function(){
$(#FORM_ID).submit(function() {
var name= $(#name).val();
var password= $(#password).val();

$.ajax({
type: POST,
url: insert.php,
data: name= + name+ &password= + password,
success: function(data) {
alert(sucess);
}
});


});
});
</script>

[#61170] Sunday, July 31, 2016, 8 Years  [reply] [flag answer]
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raymondd

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