如何一次查询两个表以查找具有相同名称但数据不同的字段？

Question

I have two tables, both containing a field named "id" that has different data in each table. I need to retrieve both ids to construct an URL. I'm using the following code:

<?php // no direct access
mysql_connect("localhost", "user", "pass") or die(mysql_error());
mysql_select_db("db") or die(mysql_error());

$value = $_POST['password'];


$content = mysql_query("SELECT * FROM jos_content WHERE pass='$value'") or die(mysql_error());  
$menu = mysql_query("SELECT * FROM jos_menu WHERE menutype='mymenutype' AND alias='$value'") or die(mysql_error());  

$id = mysql_fetch_array( $content );
$itemid = mysql_fetch_array( $menu );
// Print out the contents of each row into a table 

if(isset($_POST['password'])){
    header('Location: ' . 'index.php?option=com_content&view=article&id=' . $id['id'] . '&Itemid=' . $itemid['id']);
    exit;
}

?>

However, I get data only from the first query, while the second returns nothing. Am I doing anything wrong? Perhaps my SELECT query is not spelled correctly? Please, help.

Thanks in advance, S.

Answer 1

First thing that I see is that you use the password in your second query also, which my intuition tells me is not quite what you want.

Anyway, in order to debug the query, first, have a look in the error log and see if or die(mysql_error()) part is throwing something.

If not, echo the $menu string (query), run that exactly as it will be shown into mysql and see if returns some records. When you make it return something, change the string, and there you go - it should work

Answer 2

You can use var_dump($menu) to see the results for debuging. if $itemid is a multidimensional array u have to use index properly like $itemid[0]['id'] and you can try echo the query and and use same query in mysql as (Tudor Constantin) suggested. try to use mysql joins like this

select jc.id as id, jm.itemid as itemid FROM jos_content as jc left join jos_menu as jm on jm.alias=jc.pass WHERE jm.menutype='mymenutype' and jc.pass='$value'