Access denied for user 'testuser'@'localhost' to database 'testdb'

Question

Trying to connect to a db with

$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_error) {
  die("Connection failed: " . $conn->connect_error);
} else {
  echo '.';
}

and I get

Connection failed: Access denied for user '*'@'localhost' to database '*'

but if I use

$conn = mysqli_connect($servername, $username, $password, $dbname);
if ($conn->connect_error) {
  die("Connection failed: " . $conn->connect_error);
} else {
  echo '.';
}

it works! What gives?

I've used new mysqli before in and it worked fine. I know my username/password are right. Its GoDaddy and I clicked all of the permissions. What is going on?

Edit:

Running this:

<?php 
ini_set('display_errors',1);
error_reporting(E_ALL);

$servername = "localhost";
$username = "testuser";
$password = "testpassword";
$dbname = "testdb";

mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);
$conn1 = mysqli_connect($servername, $username, $password, $dbname);
$conn2 = new mysqli($servername, $username, $password, $dbname);
var_dump($conn1->stat, $conn2->stat);

Output:

Fatal error: Uncaught exception 'mysqli_sql_exception' with message 'Access denied for user 'testuser'@'localhost' to database 'testdb'' in /home/public_html/wj-test.php:11 
Stack trace: 
    #0 /home/public_html/wj-test.php(11): mysqli_connect('localhost', 'testuser', 'testpassword', 'testdb') 
    #1 {main} thrown in /home/public_html/wj-test.php on line 11

I know you said the variables are right, but check them again. It looks like you have `$username = '*'` and `$dbname = '*'` in the first code. Maybe it's a variable scope problem. — Barmar, Mar 09 '17 at 04:29
There has to be some other difference in the two codes. `mysqli_connect()` is simply a function that calls `new mysqli`. That's how all the procedural vs. OOP variants of `mysqli` work. — Barmar, Mar 09 '17 at 04:36
Turn on full error reporting in the script, to make sure you're not getting any warnings about undefined variables. — Barmar, Mar 09 '17 at 04:37
its on ~E_ALL but I tried E_ALL as well and no other errors. I swear I can just swap in mysqli_connect() and it works. Ugh. — Matthew Ediger, Mar 09 '17 at 05:02
As much as I hate leaving a mystery like this unsolved, maybe you should just give in an use `mysqli_connect`, it's not that big a deal. You can still use OOP everywhere else with `$conn`. — Barmar, Mar 09 '17 at 07:08
What php version are you using? Note that the PHP manual page for `mysqli::$connect_error` states that it "works as of PHP 5.2.9 and 5.3.0." specifically with regard to using it with the OOP constructor. It seems unlikely that you would be using such an old php version, but if you are, then this is probably the answer. (and if you are, then please please upgrade immediately!) — Spudley, Oct 18 '18 at 20:53

score -2 · Answer 1 · answered Mar 09 '17 at 05:10

-2

Here ,May be two issues :

The database password may be wrong or empty password.
User permission , this particular user may not having permission for this this DB.

Login to root user (super user) and check the mysql table for user privileges

answered Mar 09 '17 at 05:10

Prithabai R

58
2

Your Common Sense · Answer 2 · 2017-03-09T08:11:49.927

-2

Use your common sense.

As you probably know, there is no magic in this world. Therefore, if you face some strange, impossible behavior, it means there is a human error to blame.

Obviously, some of your premises are wrong. Most likely, procedural version doesn't work either. So you have to test it. And, if it comes to asking a question, to provide a trustable code that makes an evident test. While from the way you are wording it, you didn't make a side by side comparison.

Why don't you write your variables definition, then procedural, and then OOP variants and then run it at once?

This is what you should be doing, instead of producing some speculations or conjectures.

A robust error reporting instead of blunt die("Connection failed: also helps.

<?php 
ini_set('display_errors',1);
error_reporting(E_ALL);

$servername = "";
$username = "";
$password = "";
$dbname = "";

mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);
$conn1 = mysqli_connect($servername, $username, $password, $dbname);
$conn2 = new mysqli($servername, $username, $password, $dbname);
var_dump($conn1->stat, $conn2->stat);

This is what you had to post here as a sensible question. You see, telling us verbatim what you did something and swearing what you seen something doesn't work in programming. Only a code does.

So please run this exact code (setting your credentials of course) and copy and paste its output here.

edited Mar 09 '17 at 08:11

answered Mar 09 '17 at 06:13

Your Common Sense

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I've tested it both ways already bra. Can you tell me what premise is wrong? – Matthew Ediger Mar 09 '17 at 07:09
**Most likely, procedural version doesn't work either.** The question specifically says it does. Who's not using their common sense? – Barmar Mar 09 '17 at 07:09
@Barmar if many years on this site has been taught me anything, its a motto that reads "Never trust to what the OP *says*" :) – Your Common Sense Mar 09 '17 at 08:15
@MatthewEdiger Please post the code you run in the question as well – Your Common Sense Mar 09 '17 at 08:28
Edited. First time I copied it into this comment box and it messed up the code. – Matthew Ediger Mar 09 '17 at 08:37
@MatthewEdiger well line 8 is `$dbname = "testdb";`. Please check which particular code you are running. – Your Common Sense Mar 09 '17 at 08:39
@YourCommonSense you caught me. I was including a file with my info in it, trying not to broadcast it across the interwebs. But It is actually line 11. – Matthew Ediger Mar 09 '17 at 08:52
so 11 is mysqli_connect. **THE PROBLEM SOLVED** – Your Common Sense Mar 09 '17 at 08:54
@Barmar so, what about common sense now? – Your Common Sense Mar 09 '17 at 09:00
Running the code throws an error with either $conn1 or $conn2 on line 11 actually. – Matthew Ediger Mar 09 '17 at 09:03
@MatthewEdiger So now you're getting an error from `mysqli_connect`, not just `new mysqli`? – Barmar Mar 09 '17 at 09:03
It still works as far as my connect function. 'if ($conn->connect_error){}else{}' – Matthew Ediger Mar 09 '17 at 09:06
both are throwing errors but one actually works. i don't know why it is throwing an error but i'm just going to surrender to use the mysqli_connect way and corresponding functions I guess.. – Matthew Ediger Mar 09 '17 at 09:10
1

@MatthewEdiger there are no "corresponding functions" whatsoever. all functions are just aliases to methods and fully interchangeable. **including connect**. – Your Common Sense Mar 09 '17 at 09:27

score -2 · Answer 3 · answered Oct 18 '18 at 20:44

I experienced the exact opposite, i.e. new mysqli() did work for me and mysqli_connect() failed. I solved this by defining the port of the SQL server and changing "localhost" to "127.0.0.1".

So in the above case, try:

$dbport = 3307;
$conn = new mysqli($servername, $username, $password, $dbname, $dbport);

and/or

$servername = "127.0.0.1";

Maybe this issue is related to running MySQL and PHP on a Synology NAS.

score -4 · Answer 4 · answered Mar 14 '17 at 06:02

What helped me was changing vars into single quotes...

    $servername = 'localhost';
    $username = 'testuser';
    $password = 'testpassword';
    $dbname = 'testdb';

// Create connection
$mysqli = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($mysqli->connect_error) {
    die("Connection failed: " . $mysqli->connect_error);
}

else {
   echo "Connected.";
}

$conn->close();

Access denied for user 'testuser'@'localhost' to database 'testdb'

4 Answers4