Hi,
You can use LEFT OUTER JOIN
Table1 : software list
Table2 : installed ones
$res=mysql_query("select table1.software_name, table2.software_name from
table1 left outer join table2 on
table1.software_name=table2.software_name");
Check if there is not a matching record on table 2
while($row=mysql_fetch_array($res)){
echo $row[0]; //print software name
//check if there is a matching record in table 2 display tick
if (!empty($row[1])) display_tick();
}
Hope this helps
Mustafa
----- Original Message -----
From: "Benjamin Jeeves" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Friday, March 05, 2004 1:07 PM
Subject: [PHP-DB] Help please
hi all,
I my querying two tables in mysql the first table I query returns a list of
software that I have.
Then I query a second table that show me what is installed on a computer.
I would then like to be able to display the first list but with a tick next
to software installed on a computer.
I have treid with making a function that reuten the full list of software
but can not get it to to work function below.
function software_list(&$software) # Software is a word or it can be empty.
{
$db_search = "SELECT software_name
FROM software_list
WHERE software_name
LIKE '".$software."%'";
$db_result = mysql_query($db_search);
$db_count = mysql_num_rows($db_result);
while($sw_result = mysql_fetch_array($db_result)){
$temp = $sw_result[software_name];
for($i=0;$i<count($temp); $i++){
$sw_list = $temp;
}
return $sw_list;
}
}
But when I go to print the return values in $sw_list all I get is my first
value
so can I get a full list to be return. I have been trying to print it by
using the following
for($i=0; $i<count(software_list($sw_list));$i++){
print software_list($sw_list);
}
If any one can help with the comparing of the two list and then display a
tick next too any match that would be a great help too.
Thank you
Ben
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