Jump to content

previous next links


Recommended Posts

I have the following code allowing me to navigate to the previous or next record in the db.  In the code below, I would like to swap 'new_date' (which is in the 000-00-00) format for the existing 'record_id' in the script.  I need to navigate via the new_date because it is possible with my program to delete a record and reenter it later for that passed date causing the record_id to be out of sinc with the reord_id.


Example:  My record_id is auto-increment so if I find a mistake in a past record and go to fix it for that date in the past, my auto-increment would be 235, 236, 237, 568, 239, 240 etc., so I  can't use record_id to navigate next and previous. 


When I use new_date instead of record_id, my error says either that the query was empty, or that Variable id not defined. Script terminating.


I would appreciate any help you can give, Thanks!




This is also on a web page, so here's my code:




$current_id = $_GET['record_id'];
$record_id = $current_id;

$prevquery = "SELECT * FROM daily_sales WHERE record_id < $current_id ORDER BY record_id DESC LIMIT 1";

echo $current_id;
$prevresult = mysql_query($prevquery) or die(mysql_error());
while($prevrow = mysql_fetch_array($prevresult))
$previd = $prevrow['record_id'];

$nextquery = "SELECT * FROM daily_sales WHERE record_id > $current_id ORDER BY record_id ASC  LIMIT 1";
$nextresult = mysql_query($nextquery) or die(mysql_error());
while($nextrow = mysql_fetch_array($nextresult))
$nextid  = $nextrow['record_id'];






<a href="http://www.mywebsite.com/view.php?record_id=<?php echo $previd; ?>">Previous</a>

<a href="http://www.mywebsite.com/view.php?record_id=<?php echo $nextid; ?>">next</a>



Link to comment
Share on other sites

You are making this more difficult than it needs to be! Your prev and next links don't need to have the IDs of those prev/next records in them. In fact, it is a waste to run those two queries. Your prev/next links can instead have a current ID and then a second parameter to define whether it is prev/next. then the receiving page will use that information to get the prev or next link



EDIT: Here is some sample code to illustrate

$rec_id = $_GET['record_id'];
$COMPARE_OP = '=';
    if($_GET['offset']==1) {
        $COMPARE_OP = '>';
    } else {
        $COMPARE_OP = '<';

//Get the current record
$query = "SELECT *
          FROM daily_sales
          WHERE record_id {$COMPARE_OP} {$rec_id}
          ORDER BY record_id DESC
          LIMIT 1";
$result = mysql_query($query) or die(mysql_error());
$current_record = mysql_fetch_assoc($prevresult);
$current_id = $current_record['record_id'];
//Insert code to display the current record

//Create links to prev and next records
echo "<a href=\"http://www.mywebsite.com/view.php?record_id={$current_id}&offset=-1\">Prev</a>";
echo "<a href=\"http://www.mywebsite.com/view.php?record_id={$current_id}&offset=-\">Next</a>";
Edited by Psycho
Link to comment
Share on other sites

Thanks for your input.  It looks like the code should work, but when I run it I get the following error:


Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in /www/htdocs/applealleycafe.com/books/view.php on line 322

Link to comment
Share on other sites

I said the code was for illustrative purposes - not that it was rady to copy/paste into your application. Besides, how the heck am I supposed to know where line 322 is? But, the query is apparenlty failing. Echo it to the page to verify the content

Edited by Psycho
Link to comment
Share on other sites

I really do appreciate your quick response and expertise. 


Line 322 is:   $current_record = mysql_fetch_assoc($prevresult);




The error message says it is not a valid argument, but I'm not experienced enough to know how to find out what is making it not work.


Any suggestions?

Link to comment
Share on other sites

I looked over the code more carefully and changed this:


$result = mysql_query($query) or die(mysql_error());


to this:


$prevresult = mysql_query($query) or die(mysql_error());


This got rid of the error, but the links don't work.  I will look at them next.  The code is good if I can make it work.  It is a much better approach to the problem.


Thanks for giving me something better to work with!

Link to comment
Share on other sites

Does anyone know how to associate the date with the file?  I have two column names with dates: new_date in the 01/01/2013 format and record_date with the 000-00-00 format.  when I run the original code with the record_id, everything works, but I can't seem to get the others to work.


Any help would be appreciated.  Thanks!

Link to comment
Share on other sites

This thread is more than a year old. Are you sure you have something important to add to it?

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

Reply to this topic...

×   Pasted as rich text.   Restore formatting

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.