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Need to display mysql query results after clicking hyperlink

kel

By kel
in PHP Coding Help

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kel Newbie

kel

Posted
Posted

Hi

 

I'm trying to create a system that amongst other things displays data (in this case printer information) from a mysql database. I have created a page that displays some stuff in a table with the number of records displayed and hyperlinked next to it. Each count depends on the status of each record (e.g. by date). Anyway, what I am trying to do is as follows:

 

When the respective count/total hyperlink is selected, then I want the data that this total represents to be pulled from the database and displayed (maybe in a table) below the table and hyperlinks (see above).

 

something like:

 

Total printers on site - some number (hyperlinked)

Printers that require maintenance - some number (h/linked)

blah

blah

 

and then when whichever link is selected, the respective data should be displayed here (bearing in mind that the data returned will be hyperlinked too - but I think I can do that bit) - I can send it to a different page (using POST) but what I really want is for it to display on the same page as the totals.

 

Here is my code so far: As you can see i want to use a function (see top of code) to display this stuff, preferably want a reusable function that will display depending on the hyperlink selected.

 

<?php

function listAllPrinters()

{

        print "hi";  //function to get data displayed under table

}

?>

<Html>

<Head><B><font size = "18">Printer Maintenance</font></Head>

<Body>

<?php $db_link = mysql_connect('localhost', 'deeb', 'deeb');?>  <! db_link is just a variable name, could've called it foo, this bit connection to mysql server -->

<?php if (!$db_link)

{

die('Could not connect: ' . mysql_error());

}

?>

<?php $db_selected = mysql_select_db('maint_db', $db_link); ?> <! selects the relevant database -->

<?php $db_link = mysql_connect('localhost', 'deeb', 'deeb');?>

<?php if (!$db_link)

{

die('Could not connect: ' . mysql_error());

}

?>

<?php $db_selected = mysql_select_db('maint_db', $db_link); ?> <! selects the relevant database -->

<?php

 

//All printers

$prt_count = mysql_query("SELECT asset_tag, serial, prt_name, sched_date FROM item");

$prt_num = mysql_num_rows($prt_count);

 

//printers that need maintaining

$maint_count = mysql_query("SELECT * FROM item WHERE sched_date < Date_sub(current_date(), interval 3 month)");

$maint_num = mysql_num_rows($maint_count);

 

//Printers that are overdue for maintenance

$over_count = mysql_query("SELECT * FROM item WHERE sched_date < Date_sub(current_date(), interval 4 month)");

$over_num = mysql_num_rows($over_count);

?>

<TABLE BORDER=1><TR BGCOLOR="RED"></TR>

<TD><B>PRINTERS (HP AND ZEBRA)</TD>

<TD>

        <?php echo ("<A HREF=\"maint_index.php?action=printall\">$prt_num</A>\n"); ?> <!displays the number of printers and hyperlinks the result -->

</TD>

</TR>

<TR>

<TD><B>PRINTERS REQUIRING MAINTENANCE</TD>

<TD>

        <?php echo ("<A HREF=\"maint_index.php?action=printall\">$maint_num</A>\n"); ?> <!displays the number of printers that are due for maintenance and hyperlinks the result -->

</TD>

</TR>

<TR>

<TD><B>MAINTENANCE OVERDUE</TD>

<TD>

        <?php echo ("<A HREF=\"maint_index.php?action=printall\">$over_num</A>\n"); ?> <!displays the numer of printers that are overdue for maintenance and hyperlinks the result -->

</TD>

</TR>

</TABLE>

<BR>

<?php

if ( $_GET['action'] == "printall" )

{

    listAllPrinters();

}

?>

</Body>

</Html>

 

Suggestions welcome - I'm fairly new to all of this stuff

 

Thanks

 

 

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