I recently had a script that worked where I could count the rows in a table with a variable like so:
$i = 1;
while($row = mysqli_fetch_array($query5)) {
echo "$i";
echo "<br />";
$i++;
}
However, I recently implemented this pagination script here http://papermashup.com/easy-php-pagination/ which ruins it. As I go to the second page, it counts the rows all over again and starts identifying each row as #1. However, I would like it to continue to name the same row # from the previous page. Is this possible?
You must have page number somewhere, so start counting from page number * items per page
$i = $page_number * $page_size + 1;// assuming page_number starts from 0
while($row = mysqli_fetch_array($query5))
{
echo "$i";
echo "<br />";
$i++;
}
Related
I have some code that LIMITs data to display only 4 items per page. The column I'm using has about 20-30 items, so I need to make those spread out across the pages.
On the first page, I have:
$result = mysqli_query($con,"SELECT * FROM menuitem LIMIT 4");
{
echo "<tr>";
echo "<td align='center'><img src=\"" . $row['picturepath'] . "\" /></td>";
echo "<td align='center'>" . $row['name'] . "</td> <td align='center'> <input type='button' value='More Info'; onclick=\"window.location='more_info.php?';\"> </td>";
echo "<td align='center'>" . $row['price'] . "</td> <td align='center'> <input type='button' value='Add to Order' onclick=''> </td>";
echo "</tr>";
}
echo "</table>";
mysqli_close($con);
?>
<table width="1024" align="center" >
<tr height="50"></tr>
<tr>
<td width="80%" align="right">
NEXT
</td>
<td width="20%" align="right">
MAIN MENU
</td>
</tr>
</table>
You'll notice towards the bottom of the page my anchor tag within lists the second page, "itempage2.php". In item page 2, I have the same code, except my select statement lists the offset of 4.
$result = mysqli_query($con,"SELECT * FROM menuitem LIMIT 4 offset 4");
This works, this way when there is a pre-determined number of items within my database. But it's not that good. I need to create a new page only if there are more items, not hard-coded into it like it is now.
How can I create multiple pages without having to hard-code each new page, and offset?
First off, don't have a separate server script for each page, that is just madness. Most applications implement pagination via use of a pagination parameter in the URL. Something like:
http://yoursite.com/itempage.php?page=2
You can access the requested page number via $_GET['page'].
This makes your SQL formulation really easy:
// determine page number from $_GET
$page = 1;
if(!empty($_GET['page'])) {
$page = filter_input(INPUT_GET, 'page', FILTER_VALIDATE_INT);
if(false === $page) {
$page = 1;
}
}
// set the number of items to display per page
$items_per_page = 4;
// build query
$offset = ($page - 1) * $items_per_page;
$sql = "SELECT * FROM menuitem LIMIT " . $offset . "," . $items_per_page;
So for example if input here was page=2, with 4 rows per page, your query would be:
SELECT * FROM menuitem LIMIT 4,4
So that is the basic problem of pagination. Now, you have the added requirement that you want to understand the total number of pages (so that you can determine if "NEXT PAGE" should be shown or if you wanted to allow direct access to page X via a link).
In order to do this, you must understand the number of rows in the table.
You can simply do this with a DB call before trying to return your actual limited record set (I say BEFORE since you obviously want to validate that the requested page exists).
This is actually quite simple:
$sql = "SELECT your_primary_key_field FROM menuitem";
$result = mysqli_query($con, $sql);
$row_count = mysqli_num_rows($result);
// free the result set as you don't need it anymore
mysqli_free_result($result);
$page_count = 0;
if (0 === $row_count) {
// maybe show some error since there is nothing in your table
} else {
// determine page_count
$page_count = (int)ceil($row_count / $items_per_page);
// double check that request page is in range
if($page > $page_count) {
// error to user, maybe set page to 1
$page = 1;
}
}
// make your LIMIT query here as shown above
// later when outputting page, you can simply work with $page and $page_count to output links
// for example
for ($i = 1; $i <= $page_count; $i++) {
if ($i === $page) { // this is current page
echo 'Page ' . $i . '<br>';
} else { // show link to other page
echo 'Page ' . $i . '<br>';
}
}
A dozen pages is not a big deal when using OFFSET. But when you have hundreds of pages, you will find that OFFSET is bad for performance. This is because all the skipped rows need to be read each time.
It is better to remember where you left off.
If you want to keep it simple go ahead and try this out.
$page_number = mysqli_escape_string($con, $_GET['page']);
$count_per_page = 20;
$next_offset = $page_number * $count_per_page;
$cat =mysqli_query($con, "SELECT * FROM categories LIMIT $count_per_page OFFSET $next_offset");
while ($row = mysqli_fetch_array($cat))
$count = $row[0];
The rest is up to you.
If you have result comming from two tables i suggest you try a different approach.
Use .. LIMIT :pageSize OFFSET :pageStart
Where :pageStart is bound to the_page_index (i.e. 0 for the first page) * number_of_items_per_pages (e.g. 4) and :pageSize is bound to number_of_items_per_pages.
To detect for "has more pages", either use SQL_CALC_FOUND_ROWS or use .. LIMIT :pageSize OFFSET :pageStart + 1 and detect a missing last (pageSize+1) record. Needless to say, for pages with an index > 0, there exists a previous page.
If the page index value is embedded in the URL (e.g. in "prev page" and "next page" links) then it can be obtained via the appropriate $_GET item.
I have some code that LIMITs data to display only 4 items per page. The column I'm using has about 20-30 items, so I need to make those spread out across the pages.
On the first page, I have:
$result = mysqli_query($con,"SELECT * FROM menuitem LIMIT 4");
{
echo "<tr>";
echo "<td align='center'><img src=\"" . $row['picturepath'] . "\" /></td>";
echo "<td align='center'>" . $row['name'] . "</td> <td align='center'> <input type='button' value='More Info'; onclick=\"window.location='more_info.php?';\"> </td>";
echo "<td align='center'>" . $row['price'] . "</td> <td align='center'> <input type='button' value='Add to Order' onclick=''> </td>";
echo "</tr>";
}
echo "</table>";
mysqli_close($con);
?>
<table width="1024" align="center" >
<tr height="50"></tr>
<tr>
<td width="80%" align="right">
NEXT
</td>
<td width="20%" align="right">
MAIN MENU
</td>
</tr>
</table>
You'll notice towards the bottom of the page my anchor tag within lists the second page, "itempage2.php". In item page 2, I have the same code, except my select statement lists the offset of 4.
$result = mysqli_query($con,"SELECT * FROM menuitem LIMIT 4 offset 4");
This works, this way when there is a pre-determined number of items within my database. But it's not that good. I need to create a new page only if there are more items, not hard-coded into it like it is now.
How can I create multiple pages without having to hard-code each new page, and offset?
First off, don't have a separate server script for each page, that is just madness. Most applications implement pagination via use of a pagination parameter in the URL. Something like:
http://yoursite.com/itempage.php?page=2
You can access the requested page number via $_GET['page'].
This makes your SQL formulation really easy:
// determine page number from $_GET
$page = 1;
if(!empty($_GET['page'])) {
$page = filter_input(INPUT_GET, 'page', FILTER_VALIDATE_INT);
if(false === $page) {
$page = 1;
}
}
// set the number of items to display per page
$items_per_page = 4;
// build query
$offset = ($page - 1) * $items_per_page;
$sql = "SELECT * FROM menuitem LIMIT " . $offset . "," . $items_per_page;
So for example if input here was page=2, with 4 rows per page, your query would be:
SELECT * FROM menuitem LIMIT 4,4
So that is the basic problem of pagination. Now, you have the added requirement that you want to understand the total number of pages (so that you can determine if "NEXT PAGE" should be shown or if you wanted to allow direct access to page X via a link).
In order to do this, you must understand the number of rows in the table.
You can simply do this with a DB call before trying to return your actual limited record set (I say BEFORE since you obviously want to validate that the requested page exists).
This is actually quite simple:
$sql = "SELECT your_primary_key_field FROM menuitem";
$result = mysqli_query($con, $sql);
$row_count = mysqli_num_rows($result);
// free the result set as you don't need it anymore
mysqli_free_result($result);
$page_count = 0;
if (0 === $row_count) {
// maybe show some error since there is nothing in your table
} else {
// determine page_count
$page_count = (int)ceil($row_count / $items_per_page);
// double check that request page is in range
if($page > $page_count) {
// error to user, maybe set page to 1
$page = 1;
}
}
// make your LIMIT query here as shown above
// later when outputting page, you can simply work with $page and $page_count to output links
// for example
for ($i = 1; $i <= $page_count; $i++) {
if ($i === $page) { // this is current page
echo 'Page ' . $i . '<br>';
} else { // show link to other page
echo 'Page ' . $i . '<br>';
}
}
A dozen pages is not a big deal when using OFFSET. But when you have hundreds of pages, you will find that OFFSET is bad for performance. This is because all the skipped rows need to be read each time.
It is better to remember where you left off.
If you want to keep it simple go ahead and try this out.
$page_number = mysqli_escape_string($con, $_GET['page']);
$count_per_page = 20;
$next_offset = $page_number * $count_per_page;
$cat =mysqli_query($con, "SELECT * FROM categories LIMIT $count_per_page OFFSET $next_offset");
while ($row = mysqli_fetch_array($cat))
$count = $row[0];
The rest is up to you.
If you have result comming from two tables i suggest you try a different approach.
Use .. LIMIT :pageSize OFFSET :pageStart
Where :pageStart is bound to the_page_index (i.e. 0 for the first page) * number_of_items_per_pages (e.g. 4) and :pageSize is bound to number_of_items_per_pages.
To detect for "has more pages", either use SQL_CALC_FOUND_ROWS or use .. LIMIT :pageSize OFFSET :pageStart + 1 and detect a missing last (pageSize+1) record. Needless to say, for pages with an index > 0, there exists a previous page.
If the page index value is embedded in the URL (e.g. in "prev page" and "next page" links) then it can be obtained via the appropriate $_GET item.
I have some code that LIMITs data to display only 4 items per page. The column I'm using has about 20-30 items, so I need to make those spread out across the pages.
On the first page, I have:
$result = mysqli_query($con,"SELECT * FROM menuitem LIMIT 4");
{
echo "<tr>";
echo "<td align='center'><img src=\"" . $row['picturepath'] . "\" /></td>";
echo "<td align='center'>" . $row['name'] . "</td> <td align='center'> <input type='button' value='More Info'; onclick=\"window.location='more_info.php?';\"> </td>";
echo "<td align='center'>" . $row['price'] . "</td> <td align='center'> <input type='button' value='Add to Order' onclick=''> </td>";
echo "</tr>";
}
echo "</table>";
mysqli_close($con);
?>
<table width="1024" align="center" >
<tr height="50"></tr>
<tr>
<td width="80%" align="right">
NEXT
</td>
<td width="20%" align="right">
MAIN MENU
</td>
</tr>
</table>
You'll notice towards the bottom of the page my anchor tag within lists the second page, "itempage2.php". In item page 2, I have the same code, except my select statement lists the offset of 4.
$result = mysqli_query($con,"SELECT * FROM menuitem LIMIT 4 offset 4");
This works, this way when there is a pre-determined number of items within my database. But it's not that good. I need to create a new page only if there are more items, not hard-coded into it like it is now.
How can I create multiple pages without having to hard-code each new page, and offset?
First off, don't have a separate server script for each page, that is just madness. Most applications implement pagination via use of a pagination parameter in the URL. Something like:
http://yoursite.com/itempage.php?page=2
You can access the requested page number via $_GET['page'].
This makes your SQL formulation really easy:
// determine page number from $_GET
$page = 1;
if(!empty($_GET['page'])) {
$page = filter_input(INPUT_GET, 'page', FILTER_VALIDATE_INT);
if(false === $page) {
$page = 1;
}
}
// set the number of items to display per page
$items_per_page = 4;
// build query
$offset = ($page - 1) * $items_per_page;
$sql = "SELECT * FROM menuitem LIMIT " . $offset . "," . $items_per_page;
So for example if input here was page=2, with 4 rows per page, your query would be:
SELECT * FROM menuitem LIMIT 4,4
So that is the basic problem of pagination. Now, you have the added requirement that you want to understand the total number of pages (so that you can determine if "NEXT PAGE" should be shown or if you wanted to allow direct access to page X via a link).
In order to do this, you must understand the number of rows in the table.
You can simply do this with a DB call before trying to return your actual limited record set (I say BEFORE since you obviously want to validate that the requested page exists).
This is actually quite simple:
$sql = "SELECT your_primary_key_field FROM menuitem";
$result = mysqli_query($con, $sql);
$row_count = mysqli_num_rows($result);
// free the result set as you don't need it anymore
mysqli_free_result($result);
$page_count = 0;
if (0 === $row_count) {
// maybe show some error since there is nothing in your table
} else {
// determine page_count
$page_count = (int)ceil($row_count / $items_per_page);
// double check that request page is in range
if($page > $page_count) {
// error to user, maybe set page to 1
$page = 1;
}
}
// make your LIMIT query here as shown above
// later when outputting page, you can simply work with $page and $page_count to output links
// for example
for ($i = 1; $i <= $page_count; $i++) {
if ($i === $page) { // this is current page
echo 'Page ' . $i . '<br>';
} else { // show link to other page
echo 'Page ' . $i . '<br>';
}
}
A dozen pages is not a big deal when using OFFSET. But when you have hundreds of pages, you will find that OFFSET is bad for performance. This is because all the skipped rows need to be read each time.
It is better to remember where you left off.
If you want to keep it simple go ahead and try this out.
$page_number = mysqli_escape_string($con, $_GET['page']);
$count_per_page = 20;
$next_offset = $page_number * $count_per_page;
$cat =mysqli_query($con, "SELECT * FROM categories LIMIT $count_per_page OFFSET $next_offset");
while ($row = mysqli_fetch_array($cat))
$count = $row[0];
The rest is up to you.
If you have result comming from two tables i suggest you try a different approach.
Use .. LIMIT :pageSize OFFSET :pageStart
Where :pageStart is bound to the_page_index (i.e. 0 for the first page) * number_of_items_per_pages (e.g. 4) and :pageSize is bound to number_of_items_per_pages.
To detect for "has more pages", either use SQL_CALC_FOUND_ROWS or use .. LIMIT :pageSize OFFSET :pageStart + 1 and detect a missing last (pageSize+1) record. Needless to say, for pages with an index > 0, there exists a previous page.
If the page index value is embedded in the URL (e.g. in "prev page" and "next page" links) then it can be obtained via the appropriate $_GET item.
The code belows gives me only 18 records
<?php
$result4 = mysql_query("select Distinct Country from trendsmtable where WHORegionAC='Europe - all countries' GROUP BY Country ");
echo "<table width=880 align=center>";
echo "<tr><td colspan=4 style='font-family:Arial;'><b>European Region</b></td></tr>";
$num_columns4 = 4;
$num_rows4 = mysql_num_rows($result4);
$i=0;
while($row4 = mysql_fetch_assoc($result4)){
$results[$i] = $row4['Country'];
$i++;
}
unset($i);
$k=0;
for ($i=0;$i<=($num_rows4/($num_columns4+1));$i++){
echo '<tr>';
for($j=1;$j<=$num_columns4;$j++){
echo "<td width=220 style='font-family:Arial; font-size:12px'>".$results[$k].'</td>';
$k++;
}
echo '</tr>';
$k++;
}
echo "</table>";
?>
while the select statement
select Distinct Country from trendsmtable where WHORegionAC='Europe -
all countries'
returns 22 rows while I execute it in mysql which is correct!
Please help me to found the error.
Well, you have an extra $k++ in there that you don't need:
$k++; // keep this one
}
echo '</tr>';
$k++; // remove this one
}
Edit
I've gone through your code and I've made some edits. Hopefully they'll clean your issue along the way:
// got rid of group by. With a select distinct, it isn't necessary
$result4 = mysql_query("select Distinct Country from trendsmtable where ".
"WHORegionAC='Europe - all countries'");
// a good practice is to always double-quote your HTML attributes.
echo "<table width=\"880\" align=\"center\">";
echo "<tr><td colspan=4 style=\"font-family:Arial;\">".
"<b>European Region</b></td></tr>";
$num_columns4 = 4; // where does 4 come from.
// $results was never initialized before.
$results = array();
while($row4 = mysql_fetch_assoc($result4)){
// This is generally thought of as "cleaner" than using an index.
// And since there cannot be more than, say, 220 rows, cleanliness
// should be a high-priority standard.
$results[] = $row4['Country'];
}
// mysql_num_rows rarely provides any major benefit at all.
$num_rows4 = count($results);
foreach( $results as $key => $val )
{
// This happens every time we fill all columns and need a new row.
if( !( $key % $num_columns4 ) )
{
// makes it so that after the first time this closes the prev
// row before starting a new one.
if( $key ) echo '</tr>';
echo '<tr>';
}
// insert nag about CSS
echo '<td width="220" style="font-family:Arial; font-size:12px">'.
$val.
'</td>';
}
echo '</tr></table>';
I think the PHP looks a little odd, it looks like you are looping from 0 to (22 / 5)
then in the inner loop 4 times.
The group by should not make any difference as you are doing a select distinct
Your SQL code in the PHP file contains a GROUP BY clause, which can reduce the number of rows returned, if you have a country more than once in the db.
I'm having some problems with the php script below that I'm currently working on. What I am trying to do is make a list with 5 events that are being shown ordered by date.
In my database I have a table with events. Each event has a date (DATETIME), an id and a name. What the php needs to do is check the table with events and filter the ones that have already passed. If an event has already passed, it's not shown. If it still has to happen, it's shown.
Now the problem is that in the do while loop, the script doesn't seem to go to a next row when it has had a run.
For example: if the database table has 10 events in it, it will show 10 times the event that's on the first row of the table when testing.
I need to know what I'm doing wrong, or if there is a way to make the row increase after each run of the loop.
<?php
$test_query_kalender = "SELECT * FROM kalender ORDER BY datum ASC";
$test_result_kalender = mysql_query($test_query_kalender);
$rij_kalender = mysql_fetch_assoc($test_result_kalender);
$vandaag_unix = time();
$datum_unix = strtotime($rij_kalender['datum']);
$i = 0; //this variable is used to insure that only 5 items are being shown on the page
do{
if($datum_unix >= $vandaag_unix) //checks if the date of the event has already passed
{
//if the date has not passed, the event will be shown
echo "<p>" . date("d-m-Y", $datum_unix) . " " . $rij_kalender['naam'] . "</p>";
$i++;
}
else
{ //if it has already passed then it should put nothing, but for testing I put a line in it
echo "<p>" . $rij_kalender['naam'] . "</p>";
}
} while(($i <= 4) && ($rij_kalender = mysql_fetch_assoc($test_result_kalender)));
echo "<p>While loop finished</p>"; //just some checking
?>
Your code loads the date once and then compares it to today each time. Move
$datum_unix = strtotime($rij_kalender['datum']);
into the loop, before the date check.
Try this:
<?php
$test_query_kalender = "SELECT * FROM kalender ORDER BY datum ASC";
$test_result_kalender = mysql_query($test_query_kalender);
$rij_kalender = mysql_fetch_assoc($test_result_kalender);
$vandaag_unix = time();
$datum_unix = strtotime($rij_kalender['datum']);
$i = 0; //this variable is used to insure that only 5 items are being shown on the page
while($rij_kalender = mysql_fetch_assoc($test_result_kalender))
{
$datum_unix = strtotime($rij_kalender['datum']);
if($datum_unix >= $vandaag_unix) //checks if the date of the event has already passed
{
//if the date has not passed, the event will be shown
echo "<p>" . date("d-m-Y", $datum_unix) . " " . $rij_kalender['naam'] . "</p>";
$i++;
}
else
{
//if it has already passed then it should put nothing, but for testing I put a line in it
echo "<p>" . $rij_kalender['naam'] . "</p>";
}
if ($i == 5) break;
}
echo "<p>While loop finished</p>"; //just some checking
?>