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MySQLi equivalent of mysql_result()?
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Closed 1 year ago.
I tried to create pagination with database iteration. This is my code so far.
$per_page = 5;
$result = mysqli_query($connection,"SELECT * FROM inquiries");
$total_records = mysqli_num_rows($result);
$total_pages = ceil($total_records / $per_page);
if (isset($_GET['page'])) {
$show_page = $_GET['page'];
if ($show_page > 0 && $show_page <= $total_pages) {
$start = ($show_page - 1) * $per_page;
$end = $start + $per_page;
} else {
$start = 0;
$end = $per_page;
}
} else {
$start = 0;
$end = $per_page;
}
$page = intval($_GET['page']);
$tpages=$total_pages;
if ($page <= 0)
$page = 1;
for ($i = $start; $i < $end; $i++) {
if ($i == $total_records) {
break;
}
echo mysqli_fetch_array($result,$i,'message');
that. Since it makes following error.
Warning: mysqli_fetch_array() expects at most 2 parameters, 3 given in..
Can someone help me to solve this error.
You can try SQL_CALC_FOUND_ROWS with limit at first query then SELECT FOUND_ROWS(); as total count.
SELECT SQL_CALC_FOUND_ROWS, Id, Name FROM my_table WHERE <give your condition> LIMIT 0, 10;
# Find total rows
SELECT FOUND_ROWS();
Related
I have done this pagination using PHP and SQL. Id 11 and 12 is missing in the page 2 module. I can't find out where my mistake is.
Here my source code:
<?php
$per_page = 5;
$rec_count = $row['id'];
$total_pages = ceil($rec_count / $per_page);
if (isset($_GET['page'])) {
$show_page = $_GET['page']; //current page
if ($show_page > 0 && $show_page <= $total_pages) {
$start = ($show_page - 1) * $per_page;
$end = $start + $per_page;
} else {
$start = 0;
$end = $per_page;
}
} else {
$start = 0;
$end = $per_page;
}
$page = intval($_GET['page']);
$tpages=$total_pages;
if ($page <= 0)
$page = 1;
$sql = "SELECT COUNT(Id) AS id FROM [Register].[dbo].[RegisData] WHERE DeleteStatus = 1";
$result = sqlsrv_query($conn,$sql);
$row = sqlsrv_fetch_array($result, SQLSRV_FETCH_ASSOC);
$rec_count = $row['id'];
$total_pages = ceil($rec_count / $per_page);
$sql1 = "SELECT [Id],[FirstName],[LastName],[ProfilePic],[Gender]
,[Email],[MobileNo],[Company],[Designation],[Country],[State],[Address],[City]
,[Pincode],[Hobbies],[DietaryHabits],[DeleteStatus]
FROM (SELECT *, ROW_NUMBER() OVER (ORDER BY RegId) as row
FROM [Register].[dbo].[RegisData])
a WHERE row between $start and $end
AND DeleteStatus = 1
ORDER BY RegId";
$result = sqlsrv_query($conn,$sql1);
$z = 1;
for ($x=1; $x<=$total_pages; $x++) {
echo $x;
}
?>
the way you did the pagination combined with DeleteStatus = 1 will always create broken and incomplete pages (if you take all the rows between 1 and 5 and row number 4 is DeleteStatus = 0 you will only get 4 rows)
you should use SQL's LIMIT functionality that allows you to select the result row number to start and how many rows to show after the WHERE clause
so try
$page = intval($_GET['page']);
if($page == 0) $page = 1;
$per_page = 5;
$offset = ($page -1) * $per_page;
$sql1 = "SELECT [Id],[FirstName],[LastName],[ProfilePic],[Gender]
,[Email],[MobileNo],[Company],[Designation],[Country],[State],[Address],[City]
,[Pincode],[Hobbies],[DietaryHabits],[DeleteStatus]
FROM (SELECT *, ROW_NUMBER() OVER (ORDER BY RegId) as row
FROM [Register].[dbo].[RegisData])
a WHERE DeleteStatus = 1
ORDER BY RegId
LIMIT {$per_page} OFFSET {$offset}";
for more on LIMIT in SQL look at this short explanation: https://www.techonthenet.com/sql/select_limit.php it's really what you're looking for
---EDIT----
since you don't want to use my suggested solution (for unknown reasons) you can try to make this change to your code:
$sql1 = "SELECT [Id],[FirstName],[LastName],[ProfilePic],[Gender]
,[Email],[MobileNo],[Company],[Designation],[Country],[State],[Address],[City]
,[Pincode],[Hobbies],[DietaryHabits],[DeleteStatus]
FROM (SELECT *, ROW_NUMBER() OVER (ORDER BY RegId) as row
FROM [Register].[dbo].[RegisData] WHERE DeleteStatus = 1)
a WHERE row between $start and $end
AND DeleteStatus = 1
ORDER BY RegId";
what I did is add the WHERE DeleteStatus = 1 condition to the internal query that numbers the lines, this way your filter should work as expected - although you should use LIMIT
To prevent duplicates (5 appearing both for page 1 and page 2) you should change this if:
if ($show_page > 0 && $show_page <= $total_pages) {
$start = ($show_page - 1) * $per_page;
$end = $start + $per_page;
}
to this:
if ($show_page > 1 && $show_page <= $total_pages) {
$start = ($show_page - 1) * $per_page + 1;
$end = $start + $per_page;
}
again, if you would have used LIMIT this would not be an issue
I created php script following the tutorial, but it has a mistake. It displays in the last page information which is in the previous page - it's because $perpage. How can I display only data which wasn't display yet.
EXAMPLE - If I set $perpage to 3 and I have 7 records (named 1,2,3,4,5,6,7) on page one is 1,2,3 on page two is 4,5,6 and on the last page is 5,6,7 (I want to display only record 7)
$query = mysql_query("SELECT ID FROM clanek");
$count = mysql_num_rows($query);
$perpage = 3; // řádků na stránku
$pages_count = ceil($count / $perpage); //celkem stránek zaokrohleno
$page = isset($_GET['page']) ? $_GET['page'] : 1;
$is_first = $page == 1; //první stránka
$is_last = $page == $pages_count;
$prev = max(1, $page - 1);
$next = min($pages_count , $page + 1); /
$data = mysql_query("SELECT DATE_FORMAT(datum_pridani_c,'%d.%m.%Y %H:%i:%s')as datumcas,nazev,kratky_popis,ID FROM clanek ORDER BY datum_pridani_c DESC LIMIT ".($page - 1).", ".$perpage); /
while ($zaznam = mysql_fetch_array($data)) {
//some info her
}
if($pages_count>0) {
if(!$is_first) {
echo '<a class="predchozistranka" href="index.php?page='.$prev.'">Předchozí</a>';
}
echo '<span class="stranka">Stránka '.$page.' / '.$pages_count.'</span>';
if(!$is_last) {
echo '<a class="dalsistranka" href="index.php?page='.$next.'">Další</a>';
}
}
$data = mysql_query("SELECT DATE_FORMAT(datum_pridani_c,'%d.%m.%Y %H:%i:%s')as datumcas,nazev,kratky_popis,ID FROM clanek ORDER BY datum_pridani_c DESC LIMIT ".($page - 1).", ".$perpage);
It has been a while since I have used MySQL but I fired up my Workbench just now and I see that the syntax for LIMIT is LIMIT offset,rowcount. The doc says so too http://dev.mysql.com/doc/refman/5.0/en/select.html
For your query to work then, instead of ($page - 1), $perpage, it should be ($page - 1)*$perpage, $perpage
$query = mysql_query("SELECT ID FROM clanek");
$count = mysql_num_rows($query);
Irrelevant but the above code is highly inefficient for getting the total number of rows. It would be better if you use SELECT count(id) FROM clanek
Im trying to return all records from my database and paginate them accordingly,
I have the following I've managed to wrangle together from a tutorial only It doesnt seem to output anything to my browser so I presume I've made an error in my syntax?
// No County Selected
try
{
$per_page = '3';
$res = $conn->prepare("SELECT * FROM directory WHERE user_active != ''");
$rows = $res ->fetchAll();
$total_records = count($rows);
$pages = ceil($total_records / $per_page);
$page = (isset ($_GET['page'])) ? (int) $_GET['page'] : 1 ;
$start = ($page - 1) * $per_page;
$query = $db->query("SELECT * FROM directory LIMIT $start , $per_page");
while($row = $query->fetch(PDO::FETCH_ASSOC)){
echo '<br>' . $row['id'];
} ?>
<br><br>
<?php
if ($pages >= 1 && $page <= $pages){
//if ($page>1 && $page <= $pages){$previous=($page -1); echo 'Previous';}
for($x=1; $x<=$pages; $x++){ echo ($x == $page) ? ' <strong>'.$x.'</strong> ' : ' '.$x.' ';}
//if ($page>1 && $page <= $pages){ $next=($page+1) ; echo 'Next';}
}
You have a logical flaw in your queries. The condition mismatches between the query you use to obtain the total count:
SELECT * FROM directory WHERE user_active != ''
and the query you use to fetch the limited (paginated) rows:
SELECT * FROM directory LIMIT $start , $per_page
Both queries should share the same conditions. Additionally, the first query should only COUNT the rows, not return them all:
SELECT COUNT(1) FROM directory WHERE user_active != ''
You will then find the number of rows (count) in the first column of the first result-row.
I am trying to display 10 records per page by using ROW_NUMBER function of SQL SERVER 2008.
I think there is something wrong with my query because I only define where to start and where to end displaying records (from 1 to 10), but not the amount of records displayed per page.
As I go to the next page I get no results displayed at all because I don't know how to add $per_page variable to my SQL query properly.
All I want is be able to display the first 10 product IDs on page 1, and once I click on the "next" button, the next 10 product IDs will be displayed, etc.
This is the code that I have right now:
$per_page = 10;
if(!isset($_GET['page']))
{
$page = 1;
}
else
{
$page = $_GET['page'];
}
if($page<=1)
{
$start = 0;
}
else
{
$start = $page * $per_page - $per_page;
}
$tsql = " SELECT *
FROM (SELECT ROW_NUMBER() OVER(ORDER BY productID) AS
rownum, productID FROM products) AS products1
WHERE rownum >= $start AND rownum <= $per_page";
$num_rows = sqlsrv_num_rows(sqlsrv_query($conn,$tsql));
$num_pages = ceil($num_rows / $per_page);
$stmt = sqlsrv_query($conn,$tsql);
while($row = sqlsrv_fetch_array($stmt)){
echo $row['productID']. "<br/>";
}
$prev = $page - 1;
$next = $page + 1;
echo "<hr>";
//prev
if($prev > 0)
{
echo "<a href='?page=$prev'>prev</a> ";
}
//numbers
$number = 1;
for($number; $number <= $num_pages; $number +=1)
{
if($page==$number)
{
echo " <b>[$number]</b> ";
}
else
{
echo "<a href='?page=$number'>$number</a> ";
}
}
//next
echo "<a href='?page=$next'>next</a>";
I'm not familiar with sql server, but it seems to me you just need the end point:
$tsql = " SELECT *
FROM (SELECT ROW_NUMBER() OVER(ORDER BY productID) AS
rownum, productID FROM products) AS products1
WHERE rownum >= $start AND rownum < ($start + $per_page)";
^ changed ^
And if $page is supposed to be an integer, it's always best to make sure that it is:
$page = (int) $_GET['page'];
This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result
I have the following mysql query which is generating the above error and i can not work out why, any help would be appreciated.
$tableName="livetrack";
$targetpage = "visitors.php";
$limit = "$pagination";
$query = "SELECT COUNT(*) as num FROM $tableName where member_id = '$site_id' and display = 'yes'";
$total_pages = mysql_fetch_array(mysql_query($query));
$total_pages = $total_pages[num];
$stages = 3;
$page = mysql_escape_string($_GET['page']);
if($page){
$start = ($page - 1) * $limit;
}else{
$start = 0;
}
// Get page data
$query1 = "SELECT * FROM $tableName WHERE date >= ( NOW() - '$fdate' ) and ip ='$ip' and member_id = '$site_id' and display = 'yes' and category= '$categories' and type ='$vtype' order by id DESC LIMIT $start, $limit";
$result = mysql_query($query1);
// Initial page num setup
if ($page == 0){$page = 1;}
$prev = $page - 1;
$next = $page + 1;
$lastpage = ceil($total_pages/$limit);
$LastPagem1 = $lastpage - 1;
$paginate = '';
if($lastpage > 1)
{
The code below is where the error message is generating from.
while($row = mysql_fetch_array($result))
{
You should print out the raw query and then run it manually in mysql (either through mysql tools or something like phpMyAdmin).
Do you get an error? (probably)
If yes, fix the query.
If not, you should always do a check for mysql errors whenever you run a query and before you try to access the results.
For instance:
$query = 'SELECT * FROM users;';
$result = mysql_query($query);
if(mysql_error() != '') { echo 'error on mysql query: #'.mysql_errno().' - '.mysql_error();
else { $data = mysql_fetch_array($result); }
A second likely possibility is that you don't have a valid connection to a database, or you are connected to the wrong database.