Have a website with some PHP scripting in it. I get a MySQL error in it. I've used this code before on other websites, the exact same code as in copy-paste. For some reason it does not work.

Warning:  mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given in /homepages/31/d231284211/htdocs/home/editor/editor.php on line 22

Try not using mysqli another words your mysql server may not support that...Use mysql (without the i ) and that may work.

Not all servers will work the same, you need to check the configuration of the server before making your decision on what server you want to use. mySQL 2 mySQL doesn't mean it's all going to be the same.

Good luck

Each time I've used this script it has been on the same server. I've been doing this long enough to know that you need to check which version of MySQL and PHP you are running and to make sure that the mysqli extension has been enabled.

The client's host is the same as mine, I've double checked the .htaccess on the site to make sure that the latest version of PHP and MySQL are being run.

All local mind readers are fired. Show the code.

<?php
$DBConnect = mysqli_connect('db18.perfora.net','<removed>','<removed>','<removed>');

$SQL = "SELECT * FROM Content WHERE Page = 'Staff'";
$AboutUsResult = mysqli_query($DBConnect, $SQL);

$About_Content = mysqli_fetch_array($AboutUsResult);
$Staff = $About_Content[0];


$SQL = "SELECT * FROM Content WHERE Page = 'Staff'";
$StaffResult = mysqli_query($DBConnect, $SQL);

$Staff_Content = mysqli_fetch_array($StaffResult);
$Staff = $Staff_Content[0];

$SQL = "SELECT * FROM counter";
$QueryResult = mysqli_query($DBConnect, $SQL);

$PageVist = mysqli_fetch_array($QueryResult);
$PageVists = $PageVist[0];

?>

Note: the $PageVisit variable works fine. the mysqli_fetch_array() works perfectly for it as the number is actually displayed. However for the rest of them it does not work.

It's not usually considered a good idea to leave usernames and passwords to your database visible in a public forum.

It's been there for almost 11 hours now, I would suggest changing them pretty quickly!

And which of these lines causes an error?

Also add some debug code after each mysqli_query() like the following:

PHP Code:

$res = mysqli_query($connection, $sql);
if (!$res) {
   echo mysqli_error();
}
else {
   // fetch_data
} 


This will show you what exactly is wrong with your query.

Originally Posted by chrishirst

It's not usually considered a good idea to leave usernames and passwords to your database visible in a public forum.

It's been there for almost 11 hours now, I would suggest changing them pretty quickly!

Thank you for catching that!! I thought I had removed all of the sensitive information from the code. I appreciate you removing it!!

22: $About_Content = mysqli_fetch_array($AboutUsResult);
$Staff = $About_Content[0];


$SQL = "SELECT * FROM Content WHERE Page = 'Staff'";
$StaffResult = mysqli_query($DBConnect, $SQL);

29: $Staff_Content = mysqli_fetch_array($StaffResult);
$Staff = $Staff_Content[0];

The Error indicates that the problems are located on lines 22 and 29. I have placed the line numbers in the code to indicate the lines in question.

Originally Posted by [b]lackwater

Warning:  mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given in /homepages/31/d231284211/htdocs/home/editor/editor.php on line 22

You have the error stated right in front of you, you're passing a boolean (true|false) parameter to the fetch_array instead of an SQL resultset from the db.

Try something like this

PHP Code:

$DBConnect = mysqli_connect("host", "user", "password", "db");
/* first check the connection, do this always! */
if (mysqli_connect_errno()) {
    printf("Connect failed: %s\n", mysqli_connect_error());
    exit();
}

$sql = "SELECT * FROM Content WHERE Page = 'Staff'";
/* try it like this */
if ($AboutUsResult = mysqli_query($DBConnect, $sql)) {

/* return the numeric array */
$row = mysqli_fetch_array($AboutUsResult, MYSQLI_NUM);
printf ("%s (%s)\n", $row[0]);  

} else {
printf("Error: %s\n", mysqli_error($DBConnect));
}
 
/* when you're done free the resultset */
mysqli_free_result($AboutUsResult);

/* close connection */
mysqli_close($DBConnect); 


The other possible thing could be that your query doesn't return a resultset, to think of it, but to test it, you should do like follows:

PHP Code:

if(mysqli_query($DBConnect, $sql) === TRUE) {
 printf("successfully queried.\n");
} else {
printf("query failed.\n");
}