mySQL insert statement NOT working? - php

Question

I've been looking around for an answer. I've been cross-referencing my code with others. And I don't seem to see any blatant errors...
My database does not update, my database does nothing. Oh and the page also does nothing too...... It's frustrating because, I'm just using notepad ++ and can't pinpoint the error. I'm using XAmpp as well, and, all the values match the ones I have made.
The .post statement (Using jQuery 1.7.1):
//Make sure DOM is loaded before running jQuery based code: [STATIC CODE]
$(document).ready(function(){
$('#uploadbtn').click(function() {
//Parse name and song.
var name = $('#songname').val();
var song = $('#songupload').val();
$.ajax(
{
type: 'POST',
data: 'db/upload.php?name=' + name + 'song=' + song,
success: function(res) {
$('#nav-playlist').html(res);
}
}
)
});
});
Now here is my php file:
<?php
/*Connection to database.
First with mysql_connect (to log in). Then selecting the master database.
*/
echo "Upload.php accessed...";
$connection = mysql_connect("localhost", "root", "root") or die ( mysql_error() );
$database = mysql_select_db("betadb") or die( mysql_error() );
//Properties (to be inserted into database).
$name = mysql_real_escape_string($_POST["name"]);
$song = mysql_real_escape_string($_POST["song"]);
//Insertion formula for mySQL
$query = "INSERT INTO songs SET name= '$name' song='$song' ";
if (mysql_query($query)){
echo "Success";
}
else {
}
?>
ADDITIONAL NOTES:
the song table consists of id, name, song (in that order).
Song is the BLOB datatype, as it is used to store .mp3s

Answer 1

The problem is with the following line
data : 'db/upload.php?name='+name+'song='+song,
data should be an array containing the values, such as
var data
data["name"] = name
data["song"] = song
The $.ajax call is also missing the url parameter which is needed to carry out the request
url: 'db/upload.php'

Answer 2

Try to specify your column.
$query = "INSERT INTO songs (youcoulmn1,youcolumn2) VALUES ('$name', '$song')";
See also:
PHP MySQL Insert Into
Regards

Answer 3

$name = mysql_real_escape_string($_POST["name"]); //Assign to name & song variables.
$song = mysql_real_escape_string($_POST["song"]);
//Insertion formula for mySQL
$query = "INSERT INTO songs VALUES ('$name', '$song')";
$result = mysql_query($query, $connection) or die ("Unsucessful");

Answer 4

had better using SET, is more easier for some conditions, and for Jquery Command use $.ajax, you can try this one
for javascript / JQuery
$(function(){
$('#uploadbtn').click(function(){
var name = $('#songname').val();
var song = $('#songupload').val();
$.ajax({
type :'POST',
data : 'db/upload.php?name='+name+'song='+song,
success :function(res){
$('#nav-playlist').html(res);
}
});
});
});
and for Insert command in the php
$name = mysql_real_escape_string($_POST["name"]);
$song = mysql_real_escape_string($_POST["song"]);
$query = "INSERT INTO songs SET name='$name' song='$song'";
if(mysql_query($query)){
// some process if success
}else{
// some proses if not
}
use mysql_real_escape_string to filtering data before insert to database

Answer 5

in order to debug it
1) do print_r($_POST); to check do you have anything to insert
2) then instead of
$result = mysql_query($query, $connection) or die ("Unsucessful");
do
$result = mysql_query($query, $connection) or die (mysql_error());
to get the exact error and search for the error fix

Answer 6

The single quotes around variables might be causing the problem.. check it..

Answer 7

As far as I remember, the line
$query = "INSERT INTO songs SET name= '$name' song='$song' ";
should be
$query = "INSERT INTO songs SET name= '$name', song='$song' ";
Pay attention to commas!
And also:
data: 'db/upload.php?name=' + name + 'song=' + song,
should be at least:
data: 'db/upload.php?name=' + name + '&song=' + song,
because there is no delimiter between fields right now.