Anakin Posted March 12, 2006 Share Posted March 12, 2006 Hi,Can someone please assist with the following code.(1) I get the following error when I run the script.(2) I also need to have the results output in alphabetical order (ORDER BY).Much appreciated for assistance!!!RegardsAnakin=============================================================<?xml version="1.0" encoding="UTF-8" ?> - <menu>- <menu-title label="menu"> <menu-item label="Select City ..." /> <br /> [b]<b>Warning</b> : mysql_fetch_array(): supplied argument is not a valid MySQL result resource on line <b>24</b>[/b] <br /> </menu-title> </menu>=============================================================<?php// This line connects to the database.include_once("config.php");global $connection;$country_id = $_GET[country_id];$query = "SELECT city_id, City FROM city WHERE country_id = ".$country_id;$result = mysql_query($query,$connection); // And now we need to output an XML document.// We use the names of columns as <row> properties.echo '<?xml version="1.0" encoding="UTF-8"?>'; echo '<menu>'; echo '<menu-title label="menu">';echo '<menu-item label="Select City ..." />';while($row=mysql_fetch_array($result)){ $line = '<menu-item data="'.$row[city_id].'" label="'.$row[City].'"/>'; echo $line;}echo '</menu-title>';echo '</menu>';?> Quote Link to comment Share on other sites More sharing options...
azuka Posted March 12, 2006 Share Posted March 12, 2006 The most obvious problem seems to be your naming structure. You have a table called city with a column city. Try changing your query to:[code]$query = "SELECT city_id, city.City FROM city WHERE country_id = '". $country_id . "'";[/code] Quote Link to comment Share on other sites More sharing options...
keeB Posted March 12, 2006 Share Posted March 12, 2006 [code]// This line connects to the database.include_once("config.php");global $connection;$country_id = $_GET[country_id]; [i].....[/i]$result = mysql_query($query,$connection);[/code]the problem is there. $connection wont have a value because it seem's you're (re)declaring it to have no value.if inside of config.php you have [i]$connection = mysql_connect();[/i] then $connection will be a usable variable WITHOUT the global declaration...I hope this helps. Quote Link to comment Share on other sites More sharing options...
kenrbnsn Posted March 12, 2006 Share Posted March 12, 2006 No, the statement[code]<?php global connection; ?>[/code]doesn't assign a value, it just tells PHP that the variable was used (created) outside the function and to use that variable instead of a local variable. In this instance, it is meanless and should be remoed anyway.Change this line[code]<?php $result = mysql_query($query,$connection); ?>[/code]to[code]<?php $result = mysql_query($query) or die('Problem with the query: ' . $query . '<br />' . mysql_error()); ?>[/code]And change your query to[code]<?php $query = "SELECT city_id, City FROM city WHERE country_id = '" . $country_id . "'"; ?.[/code]This should help you debug your problem.Ken Quote Link to comment Share on other sites More sharing options...
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