Failed to query database [duplicate] - php

Question

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How do I get PHP errors to display?
(27 answers)
Closed 5 years ago.
<?php
//Get Value
$username = $_POST['user'];
$password = $_POST['pass'];
//Connet To The Server And Select Database
mysqli_connect("192.168.xxx.xxx", "xxx", "xxxxxxxxxxxxx");
mysqli_select_db("xxxxx");
//Query The Database For User
$result = mysqli_query("select * from user where username = '$username' and password = '$password'")
or die("Failed to query database ".mysqli_connect_error());
$row = mysqli_fetch_array($result);
if (empty($username)) {
header('Location: fa.html');
} elseif (empty($password)) {
header('Location: fa.html');
} elseif ($row['username'] == $username && $row['password'] == $password){
header('Location: su.html');
} else{
header('Location: fa.html');
}
?>
I have no experience to code PHP so i have no idea what's wrong is my code.
I have replace "mysql" into "mysqli" but it is still not working correctly.
It's work fine when running "mysql_*" and using my local Window web server.
But when i put it into Linux server it occur error message "Failed to query database ".

Answer 1

Add the fourth parameter here as database name like
mysqli_connect("192.168.xxx.xxx", "xxx", "xxxxxxxxxxxxx", "database_name");
and remove
mysqli_select_db("xxxxx");
A sample way of writing php code would be
$link = mysqli_connect("server_name","username","password","database_name");
if(mysqli_connect_error()) {
die("There was an error connecting to the database");
}
$query = 'yourQuery';
$result=mysqli_query($link,$query);
if ( false==$result ) {
printf("error: %s\n", mysqli_error($link));
}
This will work
Hope it helps

Answer 2

You need to trap for more errors to see where it's failing. Right now all you know is that it's failing at-least by the mysqli_query() line where you do trap for errors.
First you have to ensure that you can actually connect to the database server.
Change:
mysqli_connect("192.168.xxx.xxx", "xxx", "xxxxxxxxxxxxx");
to
if(!mysqli_connect("192.168.xxx.xxx", "xxx", "xxxxxxxxxx"))
{
die("Could not connect");
}
then you should also trap for errors on selecting the database. Change:
mysqli_select_db("xxxxx");
to
if(!mysqli_select_db("xxxx"))
{
die("Could not select database");
}
Edit: Also you really didn't need to blank-out your IP. 192.168.1.* is a class C private address - meaning it is not accessible from outside your local network.

Answer 3

You need to learn more, because these are really just the basics. The reason why your code works locally and doesn't on the remote server is probably caused by your DB access - it obviously isn't the same. You need to change your credentials and your db IP to match your remote server if you want to deploy your code there.