How to show 5 post each page with next & previous button feature? - php

I have a page named 'job.php', currently this page is showing all posted job. But now I want to show only 5 latest posts. And if anyone want to check the previous posts, they can click next button thus more 5 posts will be seen. There should be a previous button too.
Following is my code:
$result1 = mysql_query("SELECT * FROM job ORDER BY ID DESC");
$num_row = mysql_num_rows($result1);
while($row1 = mysql_fetch_array($result1)){
$cat=$row1['Category'];
$title=$row1['Title'];
echo "Job field: $cat<br/> Title: $title<br/>";
}
N:B: It's not pagination. I don't want to show page numbers, just want to show next & previous button.

There are 100s of articles available on the Internet
Create Awesome PHP/MYSQL Pagination
PHP / MySQL select data and split on pages
If you want to do on your own:
Use LIMIT keywords in your query.
Pass the page and the multiplier to the LIMIT.
Some code
<?php
$limit = 5;
$start = (int)(($page - 1 ) * $limit);
$page = mysql_real_escape_string($_GET["page"]);
$query = "SELECT * FROM `table` LIMIT $start, {(int)$page + $limit}"
?>

There are two ways to achieve this.
1) In the query itself by using LIMIT
$result1 = mysql_query("SELECT * FROM job ORDER BY ID DESC LIMIT 1, 5");
2 ) By using loop
$i = 0;
while($row1 = mysql_fetch_array($result1)){
if($i < 5) {
$cat=$row1['Category'];
$title=$row1['Title'];
echo "Job field: $cat<br/> Title: $title<br/>";
$i++;
}
}
You can pass the current value to URL and get it back by using $_GET..

You can use this query:
Select * from table_name ORDER BY ID DESC LIMIT 5;

I feel its better to use pagination. If you want dont want to show page nos, better hide it or just modify the code. If you planning to do it manually its a bit messy

You can use LIMIT in you mysql query:
$result1 = mysql_query("SELECT * FROM job ORDER BY ID DESC LIMIT 0, 5");

Related

Next and previous buttons

I know there are libraries etc that I could use to get this sorted but Im almost there with my code.
A little about the code and what it's trying to do. I have a mysql table where there are various news articles and grouped in categories of news.
I have managed to get a forward button working. So it looks for the next news article that is in the same category. This works and the code is below.
//Gets the next story from the same story type in line.
$query= "SELECT * FROM news WHERE storytype2 = '$storytype2' AND id > '$currentid'";
$result = mysql_query($query) or die ("Error in query: $query " . mysql_error());
$row = mysql_fetch_array($result);
$num_results = mysql_num_rows($result);
if ($num_results > 0){
echo "<td width=\"20%\"align=\"right\"><img title=\"Next News\" src=\"webImg/forwardarrow.png\"/></td></tr>";
}else{
echo "<td width=\"20%\"align=\"right\"></td></tr>";
}
//End of the next button
However, when I try do the same for the previous button. All I ever seem to get back is the first id of that category regardless of where my iteration is. For example, if I am on news article 10 and try to go to previous one which say has an id of 7 it will automatically show the first news article within that category, say id 4.
Below is the code.
//Gets the next story from the same story type in line.
$query= "SELECT * FROM news WHERE storytype2 = '$storytype2' AND id < '$currentid'";
$result = mysql_query($query) or die ("Error in query: $query " . mysql_error());
$row = mysql_fetch_array($result);
$num_results = mysql_num_rows($result);
if ($num_results > 0){
echo "<td width=\"20%\"align=\"left\"><img title=\"Previous News\" src=\"webImg/backarrow.png\"/></td>";
}else{
echo "<td width=\"20%\"align=\"left\"></td>";
}
//End of the next button
What have I done wrong?
Thanks
Neither of your queries is correct. Your "Next" code selects any row whose ID is higher than the current, not necessarily the next one; if you get the next one, it's just by accident.
You should use ORDER BY and LIMIT to control which row is selected:
Next:
SELECT *
FROM news
WHERE storytype2 = '$storytype2' AND id > '$currentid'
ORDER BY id
LIMIT 1
Previous:
SELECT *
FROM news
WHERE storytype2 = '$storytype2' AND id < '$currentid'
ORDER BY id DESC
LIMIT 1
Without any further information, I don't think you can assume that the first row of your queries will be the ID you're looking for. Ordering by ID first will probably solve your problem; you can also limit your query to one row, since it's the only one you're looking at. Something like the following would probably solve your problem (where x is $storytype2 and y is $currentid:
SELECT * FROM news
WHERE storytype2 = x
AND id < y
ORDER BY id DESC /* <-- THIS */
LIMIT 1
Use ORDER BY id ASC for the other case.
Note that the MySQL family of PHP is deprecated and support thereof will disappear, if it hasn't yet. Please look into PDO or MySQLi.
Note also that you are inserting a variable into SQL code directly, which is never a good idea. I hope you have some good input checks on your variables.
Let's look at the PDO way to get the previous article ID:
$dbh = new PDO(..);
// Use ? where dynamic input will come
$sql = $dbh->prepare('SELECT * FROM news
WHERE storytype2 = ?
AND id < ?
ORDER BY id DESC
LIMIT 1');
// Fill the ? safely with PDO's execute function
$sql->execute(array($storytype2, $currentid));
$result = $sql->fetch(PDO::FETCH_ASSOC);
if($result && isset($result['id'])) {
// Process previous ID
}

PHP postgresql select, getting id of the one that follows the current one

I am getting 10 rows with the highest ID from a table ...
$result = pg_query($dbconn, "SELECT w_news_id, name, w_newsnachricht, w_newsdatum FROM adempiere.w_news ORDER BY w_news_id DESC LIMIT 10");
... then I build 10 divs in a while loop:
while ($row = pg_fetch_row($result)) {
// building divs here
}
BUT I want to also include the name that belongs to the next w_news_id in that same div (as a "teaser" for the "next"-arrow). So I was thinking I have to run a second query with the ID that would be next in the loop.
How is the SQL syntax for that? Or is there maybe a better way to do this?
You can use the lead window function:
SELECT w_news_id, name, w_newsnachricht, w_newsdatum,
LEAD (name) OVER (ORDER BY w_news_id) AS next_name
FROM adempiere.w_news
ORDER BY w_news_id DESC
LIMIT 10
You can select one more record - LIMIT 11, instead of 10, and process it differently in PHP.
$result = pg_query($dbconn, "SELECT w_news_id, name, w_newsnachricht, w_newsdatum FROM adempiere.w_news ORDER BY w_news_id DESC LIMIT 11");
$i = 0;
while ($row = pg_fetch_row($result) and $i < 10) {
$i++;
// building divs here
}
// then process the arrow and teaser separately (if present)
if ($row = pg_fetch_row($result)) {
// show teaser div
}
UPDATE
The window function solution provided by Mureinik is better.

Paging in php displaying data over multiple pages

Hi I am trying to display my users data over pages
Here is my code:
//Run a query to select all the data from the users table
$perpage = 2;
$result = mysql_query("SELECT * FROM users LIMIT $perpage");
It does display this the only two per page but I was wondering how you get page numbers at the bottom that link to your data
here is my updated code
$result = mysql_query("SELECT * FROM users"); // Let's get the query
$nrResults=mysql_num_rows($result); // Count the results
if (($nrResults%$limit)<>0) {
$pmax=floor($nrResults/$limit)+1; // Divide to total result by the number of query
// to display per page($limit) and create a Max page
} else {
$pmax=floor($nrResults/$limit);
}
$result = mysql_query("SELECT * FROM users LIMIT 2 ".(($_GET["page"]-1)*$limit).", $limit");
// generate query considering limit
while($line = mysql_fetch_array( $result ))
{
?>
error
Parse error: syntax error, unexpected $end in E:\xampp\htdocs\Admin.php on line 98
In order to do this you also need to use the offset value in your SQL Statement, so it would be
SELECT * FROM users LIMIT $offset, $perpage
Example:
SELECT * FROM tbl LIMIT 5,10; # Retrieve rows 6-15
Then to get the links to put on the bottom of your page you would want to get a count of the total data, divide the total by the per page value to figure out how many pages you are going to have.
Then set your offset value based on what page the user clicked.
Hope that helps!
Update:
The unexpected end most likely means that you have an extra closing bracket } in your code which is causing the page to end and still has more code after it. Look through your code and match up the brackets to fix that. There are a few other issues in the code sample you pasted.
$result = mysql_query("SELECT * FROM users" ); //Note if you have a very large table you probably want to get the count instead of selecting all of the data...
$nrResults = mysql_num_rows( $result );
if( $_GET['page'] ) {
$page = $_GET['page']
} else {
$page = 1;
}
$per_page = 2;
$offset = ($page - 1) * $per_page; //So that page 1 starts at 0, page 2 starts at 2 etc.
$result = mysql_query("SELECT * FROM users LIMIT $offset,$per_page");
while( $line = mysql_fetch_array( $result ) )
{
//Do whatever you want with each row in here
}
Hope that helps
You can then use the nrResults number to figure out how many pages you are going to have... if you have 10 records and you are displaying 2 per page you would then have 5 pages, so you could print 5 links on the page each with the correct page # in the URL...
Use requests ! http://php.net/manual/en/reserved.variables.request.php
if (((isset($_GET['page'])) AND (is_int($_GET['page']))) {
$perpage = $_GET['page'];
}
$result = mysql_query("SELECT * FROM users LIMIT $perpage");
...
Link http://yourwebsite.com/userlistforexample.php?page=3
or
http://yourwebsite.com/userlistforexample.php?somethingishere=21&page=3

How to build next and previous links with php?

I use this code to get the informations about a certain id
$sql = dbquery("SELECT * FROM `videos` WHERE `id` = ".$local_id." ");
while($row = mysql_fetch_array($sql)){
$video_id = $row["id"];
$video_title = $row["title"];
}
Let's say the link of a page would be example.com/video.php?id=34
How can i get the next and previous $video_id and $video_title depending on the current id?
A problem is that i can't increase or decrease the value of the current id by 1 because the 35 or 33 may be deleted in the meanwhile...
How can i achieve this?
//edit
I have a very big problem: the previous link sends me to the right link but the next link always sends me to the last video added in the database.
If i go to the last or first videos added in the database i get an error because there are no more next and previous videos added.
Perhaps two more queries would work ...
select id,title from videos where id < $local_id order by id desc limit 1
select id,title from videos where id > $local_id order by id asc limit 1
You have to use select and limit to get that one row you want, i.e.,
SELECT * FROM `videos` WHERE `id` < " . $local_id . " ORDER BY `id` DESC LIMIT 1
Or use > and ORDER BYidASC for the next video, instead of the previous I showed above.
Here you go through
// entry per page
$rowsPerPage = 3;
// Set page number
if(isset($_GET['page']))
$pageNum = $_GET['page'];
else
$pageNum = 1;
Note: Following code is use to show/set next & previous page number.
Note: If the current page is homepage then default $pageNum is 1 and
if $pageNum is set as 2, it will work as
if($pageNum){
$PreviousPageNumber = $pageNum - 1;
$NextPageNumber = $pageNum + 1;
echo '<a href="?page='. $PreviousPageNumber .'>Previous Page</a>';
echo '<a href="?page='. $NextPageNumber .'>Next Page</a>";
}
Note: Following code is use to show record affected by page numbers
$GetPreviousRecord = ($pageNum - 1) * $rowsPerPage;
Note: The first, optional value of LIMIT is the start position. Note:
And the second required value is the number of rows to retrieve.
$query = "SELECT * FROM post WHERE LIMIT $GetPreviousRecord, $rowsPerPage";
$result = mysql_query($query);
And in last use the WHILE loop to get all records from database by using LIMIT.

how to check current position in mysql

how can i check current number in mysql where....
my query is
$aid = 16;
$get_prev_photo = mysql_query("SELECT * FROM photos WHERE album_id='$aid' AND pic_id<'$picid' ORDER BY pic_id LIMIT 1");
$get_next_photo = mysql_query("SELECT * FROM photos WHERE album_id='$aid' AND pic_id>'$picid' ORDER BY pic_id LIMIT 1");
while i am getting current photo with following query
$photo = mysql_query("SELECT * FROM photos WHERE pic_id='$picid' LIMIT 1");
and getting total photos in album with following query
$photos = mysql_query("SELECT * FROM photos WHERE album_id='$aid'");
$total_photos = mysql_num_rows($photos);
now i want to check where i am and show it as Showing 1 of 20, showing 6 of 20 and so on...
now i want to check where i am actually...
i think you are referring to pagination, which can be achieved using LIMIT and OFFSET sql
decide the number of results you want per page, then select that many
create links like:
View the next 10
and dynamically change those every time
queries look ~like~
$offset=$_GET['view'];
SELECT * FROM table WHERE `condition`=true LIMIT 5 OFFSET $offset
this translates roughly as
select 5 from the table, starting at the 10th record
This is bad:
$photos = mysql_query("SELECT * FROM photos WHERE album_id='$aid'");
Because it grabs all the fields for the entire album of photos when all you really want is the count. Instead, get the total number of photos in the album like this:
$result = mysql_query("SELECT count(1) as total_photos FROM photos
WHERE album_id='$aid'");
if ($result === false) {
print mysql_error();
exit(1);
}
$row = mysql_fetch_object($result);
$total_photos = $row->total_photos;
mysql_free_result($result);
Now you have the count of the total number of photos in the album so that you can set up paging. Let's say as an example that the limit is set to 20 photos per page. So that means that you can list photos 1 - 20, 21 - 40, etc. Create a $page variable (from user input, default 1) that represents the page number you are on and $limit and $offset variables to plug into your query.
$limit = 20;
$page = $_POST['page'] || 1; // <-- or however you get user input
$offset = $limit * ($page - 1);
I'll leave the part where you code the list of pages up to you. Next query for the photos based on the variables you created.
$result = mysql_query("SELECT * FROM photos WHERE album_id='$aid'
ORDER BY pic_id LIMIT $limit OFFSET $offset");
if ($result === false) {
print mysql_error();
exit(1);
}
$photo_num = $offset;
while ($row = mysql_fetch_object($result)) {
$photo_num++;
$pic_id = $row->pic_id;
// get the rest of the variables and do stuff here
// like print the photo number for example
print "Showing photo $photo_num of $total_photos\n";
}
mysql_free_result($result);
I'll leave better error handing, doing something with the data, and the rest of the details up to you. But that is the basics. Also I did not check my code for errors so there might be some syntax problems above. To make a single photo per page just make $limit = 1.

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