Solved

PHP Error

Posted on 2010-11-30
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Last Modified: 2012-05-10
I am not sure what is wrong with the code below, but when ever I attempt to execute it I get the following error:

PHP Warning:  mysql_fetch_assoc() expects parameter 1 to be resource, boolean given in mysql.main.php on line 17

According to the PHP manual, I have mysql_fetch_assoc() written correctly. Not sure why this error is being thrown.
<?php

	// require the configurations file
	require(dirname(__FILE__) . "/configurations/config.inc.php");
	
	// open connection to mysql - source and destination
	$mysql_source_link_id = mysql_connect($argv[1], mysql_user, mysql_user_password);
	$mysql_destination_link_id = mysql_connect(destination_mysql_host, mysql_user, mysql_user_password);
	
	// open the databases for both - source and destination mysql instances
	$mysql_database_source = mysql_select_db($argv[2], $mysql_source_link_id);
	$mysql_database_destination = mysql_select_db(destination_database_name, $mysql_destination_link_id);
	
	// execute the select statement on the source
	$query_select = mysql_query($sql_select, $mysql_source_link_id); 
	
	for ($i=0; $i < ($row = mysql_fetch_assoc($query_select)); $i++) {
		
		echo $row["world_id"] . "\n";
		echo $row["room_id"] . "\n";
		echo $row["player_id"] . "\n";
		echo $row["timestamp"] . "\n";
		echo $row["is_blocked"] . "\n";
		echo $row["message"] . "\n";
		
	}
	
	mysql_free_result($query_select);
	
	// close connection to mysql - source and destination
	mysql_close($mysql_destination_link_id);
	mysql_close($mysql_source_link_id);

?>

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Question by:plecostomus
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2 Comments
 
LVL 83

Expert Comment

by:Dave Baldwin
ID: 34241768
I don't see '$sql_select' defined anywhere in that code.  Your 'mysql_query()' won't work without it.
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Accepted Solution

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jebpotly earned 500 total points
ID: 34241771
http://us2.php.net/manual/en/function.mysql-query.php
"For SELECT, SHOW, DESCRIBE, EXPLAIN and other statements returning resultset, mysql_query() returns a resource on success, or FALSE on error.

"For other type of SQL statements, INSERT, UPDATE, DELETE, DROP, etc, mysql_query() returns TRUE on success or FALSE on error."

I would guess that your "mysql_query($sql_select, $mysql_source_link_id); " statement on line 15 is returning a boolean which is why your warning is complaining about a boolean. What is the value of your $sql_select variable and what what do you get if you do a print_r($query_select)  on line 16 right after you do the query?
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