mySQL insert statement NOT working?

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

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 4

$name = mysql_real_escape_string($_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 5

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 6

变量周围的单引号可能导致问题..检查它..

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.