PHP MYSQL Access denied for user Error [duplicate] - php

Question

This question already has answers here:
Unable to connect to database: Access denied for user ''#'localhost' to database 'socialdb'
(6 answers)
Closed 9 years ago.
I'm trying to get the 'Content' value from a specific row ID in mySql, I echoed the result in PHP, But i got "Access denied for user ''#'localhost' (using password: NO)".
I tryed to check my query in phpMyAdmin, and i got the exact result I expected for.
Here is my code:
$con = mysql_connect("localhost", "XXX", "XXX") or die (mysql_error());
$db = mysql_select_db("XXX", $con) or die (mysql_error());
function head($d) {
$query = mysql_query("SELECT Content FROM list WHERE ID = '$d'") or die(mysql_error());
$post = mysql_fetch_assoc($query);
echo $post["Content"];
}
<h1> <?php head(1); ?> </h1>
Thank you so much.

Answer 1

The error message you quote sounds like the database isn't connected yet when you're making the call. If no connection exists, mysql_query() will try to establish a connection with default values (which usually are empty).
You should check your program flow and make sure you are actually connected to the database before making any queries.

Answer 2

move your first two lines to your function:
function head($d) {
$con = mysql_connect("localhost", "XXX", "XXX") or die (mysql_error());
$db = mysql_select_db("XXX", $con) or die (mysql_error());
$query = mysql_query("SELECT Content FROM list WHERE ID = '$d'") or die(mysql_error());
$post = mysql_fetch_assoc($query);
echo $post["Content"];
}
<h1> <?php head(1); ?> </h1>

Answer 3

issue is with
mysql_connect("localhost", "ncode_hana", "NOAM7778")
chech that sqn user and pass are correct and this user has right over your selected db...
also, never share your orignial user ID and password with anyone..
also, as Pekka told you need something like this
<h1> <?php include('yourdatabaseconnectionfile.php'); ?> <?php head(1); ?> </h1>
ALTERNATIVELY
<?php
$con = mysql_connect("localhost", "XXX", "XXX") or die (mysql_error());
$db = mysql_select_db("XXX", $con) or die (mysql_error());
<h1> <?php head(1); ?> </h1>
function head($d) {
$query = mysql_query("SELECT Content FROM list WHERE ID = '$d'") or die(mysql_error());
$post = mysql_fetch_assoc($query);
echo $post["Content"];
}
ONE MORE OPTION
add connection string in function (not recommended though)
function head($d) {
mysql_connect("localhost", "ncode_hana", "NOAM7778");
$query = mysql_query("SELECT Content FROM list WHERE ID = '$d'") or die(mysql_error());
$post = mysql_fetch_assoc($query);
echo $post["Content"];
}