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MySQL - get next record id from results

Posted on 2010-11-22
3
670 Views
Last Modified: 2012-05-10
Hi,
I have the following query which returns a result set and loops through to display the different records. Is there a way to get the id of the record coming next, perhaps using mysql_data_seek?


mysql_select_db($database_conn_data, $conn_data);
$query_view_items = "SELECT * FROM items ORDER BY sortid";
$view_items = mysql_query($query_view_items, $conn_data) or die(mysql_error());
$row_view_items = mysql_fetch_assoc($view_items);



do {


//get current record id
$id = $row_view_items['id'];

//code here to push pointer forward by one

$next_id =  $row_view_items['id'];

//code here to move pointer back by one
......
...

} while ($row_view_items = mysql_fetch_assoc($view_items));

Thanks in Advance for your feedback.

Cheers,

Andrew
0
Comment
Question by:sabecs
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3 Comments
 
LVL 17

Expert Comment

by:sweetfa2
ID: 34193888
$numrows = mysql_num_rows($result);
for ($i = 0; $i < $numrows; $i++) {
    if (!($row = mysql_fetch_assoc($result))) {
        continue;
    }
    if (!mysql_data_seek($result, $i + 1)) {
        echo "Cannot seek to row $i: " . mysql_error() . "\n";
        continue;
    }

    if (!($nextrow = mysql_fetch_assoc($result))) {
        continue;
    }

    echo $row['last_name'] . ' ' . $row['first_name'] . "<br />\n";
}

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0
 
LVL 17

Accepted Solution

by:
sweetfa2 earned 500 total points
ID: 34193961
Will try again - forgot a step.

numrows = mysql_num_rows($result);
for ($i = 0; $i < $numrows; $i++) {
    if (!mysql_data_seek($result, $i)) {
        echo "Cannot seek to row $i: " . mysql_error() . "\n";
        continue;
    }

    if (!($row = mysql_fetch_assoc($result))) {
        continue;
    }
    if (!mysql_data_seek($result, $i + 1)) {
        echo "Cannot seek to row $i: " . mysql_error() . "\n";
        continue;
    }

    if (!($nextrow = mysql_fetch_assoc($result))) {
        continue;
    }

}

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0
 

Author Closing Comment

by:sabecs
ID: 34201258
Thanks for your help.
0

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