Warning: mysql_result() expects parameter 1 to be resource, boolean given [duplicate]

The problem is that mysql_query() is returning a boolean instead of a result resource. There are two reasons this can happen:

1. You performed query that returns success/fail instead of a result set (e.g. UPDATE)
2. Your query failed

In your case the query failed. The reason it failed is because you have escaped the back ticks in the PHP string where you did not need to.

Your lines look like this:

$siteTitle = mysql_result(mysql_query("SELECT \`siteTitle\` FROM siteSettings"), 0);

When they should simply be this:

$siteTitle = mysql_result(mysql_query("SELECT `siteTitle` FROM siteSettings"), 0);

Now, some side notes:

• Don't write new code that uses the mysql_* functions. They are deprecated and will eventually be removed from PHP. Use MySQLi or PDO instead (I personally recommend PDO, YMMV)
• Nesting database functions in this way is not a particularly good way to write your code. It is much better to check the errors explicitly after every function call.

For example:

$result = mysql_query("SELECT somecol FROM sometable");
if (!$result) {
  // Handle error here
}
// Now process the result

• You should quote either all identifiers, or none, in your queries (preferably all). Quoting only some makes it harder to read.

E.g.

SELECT `siteTitle` FROM `siteSettings`