[RESOLVED] Help with fatal error

Docie

Hello guys i need some help with this code.
I have to build a simple bookstore ( for a university class ) but i stuck in here for many hours... The code is similar to a code from a book but still doesn't work...
The error says:

Fatal error: Call to undefined function query() in C:\Program Files\EasyPHP

2.0b1\www\eshop\results.php on line 32

So line 32 is part of this code:

@ $db = mysql('localhost','eshop');

if (mysqli_connect_errno())
{
	echo "Error:No Access to Database. Try again later.";
	exit;
}

$query = "select * from books where ".$searchtype." like '%".$searchterm."%'";
$result = $db->query($query);

$num_results = $result->num_rows;

Please help!

The complete code is and i call this php document from a html form etc:

<html>
<head>Roads Bookstore</head>
<title>Roads Bookstore</head>

<body>
<h1>Roads E-search Results</h1>
<?php
$searchtype=$_POST['searchtype'];
$searchterm=$_POST['searchterm'];
$searchterm=trim($searchterm);
if (!$searchtype || !$searchterm)
{
	echo "Not Enough Details. Please Try again.";
	exit;
}

if (!get_magic_quotes_gpc())
{
	$searchtype = addslashes($searchtype);
	$searchterm = addslashes($searchterm);
}

@ $db = mysql('localhost','eshop');

if (mysqli_connect_errno())
{
	echo "Error:No Access to Database. Try again later.";
	exit;
}

$query = "select * from books where ".$searchtype." like '%".$searchterm."%'";
$result = $db->query($query);

$num_results = $result->num_rows;

echo "<p>Number of items found: ".$num_results."</p>";

for ($i=0; $i < $searchresults; $i++)
{
	$row = $result->fetch_assoc();
	echo "<p><strong>".($i+1)."Title: ";
	echo htmlspecialchars(stripslashes($row["title"]));
	echo "<strong><br /> Author: ";
	echo stripslashes($row["author"]);
	echo "<br />ISBN: ";
	echo stripslashes($row["isbn"]);
	echo "<br />Price: ";
	echo stripslashes($row["price"]);
	echo "</p>";
}
	$result->free();
	$db->close();

?>
</body>
</html>

bradgrafelman

If you'll remove the '@' error suppressor from the $db declaration line, you'll probably see that PHP is trying to tell you that there is no such function as mysql().

Perhaps you're missing the 'new' keyword and instead wanted to use the [man]mysqli[/man] extension instead?

Also note that neither [man]addslashes/man nor magic_quotes_gpc should not be used to sanitized data for a SQL query. Instead you should be using prepared statements OR a DBMS-specific sanitizing fuction, such as [man]mysqli_real_escape_string/man. In addition, you shouldn't ever have to use [man]stripslashes/man on data coming from your DB; if you do, it likely points to a problem when the data was being inserted.

Docie

I used the mysqli() function, but the result was a new error...

Warning: mysqli::mysqli() [function.mysqli-mysqli]: (28000/1045): Access denied for user 'eshop'@'localhost' (using password: NO) in C:\Program Files\EasyPHP 2.0b1\www\eshop\results.php on line 23
Error:No Access to Database. Try again later.

The new code is:

<html>
<head>Roads Bookstore</head>
<title>Roads Bookstore</head>

<body>
<h1>Roads E-search Results</h1>
<?php
$searchtype=$_POST['searchtype'];
$searchterm=$_POST['searchterm'];
$searchterm=trim($searchterm);
if (!$searchtype || !$searchterm)
{
	echo "Not Enough Details. Please Try again.";
	exit;
}

if (!get_magic_quotes_gpc())
{
	$searchtype = addslashes($searchtype);
	$searchterm = addslashes($searchterm);
}

$db = new mysqli('localhost','eshop');

if (mysqli_connect_errno())
{
	echo "Error:No Access to Database. Try again later.";
	exit;
}

$query = "select * from books where ".$searchtype." like '%".$searchterm."%'";
$result = $db->query($query);

$num_results = $result->num_rows;

echo "<p>Number of items found: ".$num_results."</p>";

for ($i=0; $i < $searchresults; $i++)
{
	$row = $result->fetch_assoc();
	echo "<p><strong>".($i+1)."Title: ";
	echo htmlspecialchars(stripslashes($row["title"]));
	echo "<strong><br /> Author: ";
	echo stripslashes($row["author"]);
	echo "<br />ISBN: ";
	echo stripslashes($row["isbn"]);
	echo "<br />Price: ";
	echo stripslashes($row["price"]);
	echo "</p>";
}
	$result->free();
	$db->close();

?>
</body>
</html>

am desperate...

bradgrafelman

Well, there's nothing we can do to help you... the error message pretty much tells you what's wrong:

Access denied for user 'eshop'@'localhost' (using password: NO)

halojoy

new MySQLi()
takes normally 4 arguments, if you select the database name, too.
http://php.net/manual/en/mysqli.connect.php

dbhost, often 'localhost'
dbuser, defaults to 'root', but can be changed later
dbpass, is empty '' (no password) by default when mysql installs, but should be changed
dbname, probably in this case: 'eshop'

One typical mysqli db connection could be:

<?php
$dbhost = 'localhost';
$dbuser = 'root';
$dbpass = 'xxxxxx'; // your secret db password
$dbname = 'eshop';

// connect & login to database with name 'eshop'
$db = new mysqli ( $dbhost, $dbuser, $dbpass, $dbname );
if ($mysqli->connect_error) {
    exit('Connect Error: '.$mysqli->connect_error);
}

Docie

soooooooo much guys!!!
It eventually worked!
I was so stupid. I wrote it like this:

$db = new mysqli('localhost','root','','eshop');

[/B]

and worked!!!after 6 hours of trial and error!!!you saved me!

halojoy

You solved it, with some help.
Great 🙂