0

wats roang in here ..

  1. <?php
  2. ini_set('display_errors', 1);
  3. //connecting to database
  4. $error = "nu am putut sa ma conectez :(";
  5. mysql_connect('localhost','root','')or die ($error);
  6. mysql_select_db('dan') or die ($error);
  7. //max display pet page
  8. $per_page = 2;
  9. //get start variabel
  10. $start = $_GET ;
  11. //cont record
  12. $record_count = mysql_num_rows( mysql_query("SELECT *FROM news"));
  13. //numara numarul maximal a paginilor
  14. $max_pages = $record_count / $per_pages;
  15. if (!$start)
  16. $start = 0 ;
  17. $get =$record_count = mysql_query("SELECT *FROM news LIMIT $start,$per_page");
  18. while($row = mysql_fetch_assoc($get))
  19. {
  20. //get data
  21. $id=$row;
  22. $title = $row;
  23. $body = $row;
  24. $date = $row;
  25. echo"
  26. <hr><b>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<a href='pagina 11,2,3.php?start='>$title</a> $date</b><hr>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;$body<br/>
  27. ";
  28. }
  29. $prev = $start - $per_page;
  30. $next = $start + $per_page;
  31. if (!($start<=0))
  32. echo"<a href='pagina 11,2,3.php?start=$prev'>Prev</a> ";
  33. //show page
  34. //set variable of first page
  35. $i = 1;
  36. for ($x=0;$x<$record_count;$x=$x+$per_page)
  37. {
  38. if ($start!= $x)
  39. echo "<a href='pagina 11,2,3.php?start=$x'>$i</a> ";
  40. else
  41. echo "<a href='pagina 11,2,3.php?start=$x'><b>$i</b></a> ";
  42. $i++;
  43. }
  44. //show buton
  45. if (!($start>=$record_count-$per_page))
  46. echo "<a href='pagina 11,2,3.php?start=$next'>next</a> ";
  47. ?>

thx 4 help

Edited by dan1992: I made mistakes

2
Contributors
12
Replies
13
Views
7 Years
Discussion Span
Last Post by dan1992
0

Hey.

There is nothing wrong with the syntax itself.

How is this supposed to be working?
And how is it actually working?

Are you getting any errors? (We like error messages! ;-])
What have you tried to do to fix this?

We can only work with what you give us.

P.S.
Try code-tags next time, rather than (list) :-)

Edited by mike_2000_17: Fixed formatting

0

wel it gives error in line 14 [Warning: Division by zero in D:\Dan\web\bundle\xampp\xampp\htdocs\test\pagina 11,2,3.php on line 17]

and this is a scrrip theat show toy your news and he is duing pages exempel [Prev 1 2 3 4 next ]
but i have this error and if i dont fix this ,the line 45 will not work
sory for my englesh , i'm learning just 3 ears
... i like this blog :)

Edited by dan1992: n/a

0

Ok, on line #14, the problem is that you use a variable named $per_pages , which doesn't exist.
You have a variable on line #8 $per_page . Perhaps you meant to use that one?

0

thx but i dont now why my page cheanger dont work i have 17 news but i 1 page i put : 2 news and i have just 2 pages ,it sean theat the lin 12 don't work .. or i dont now :9

0

[Please do not ask me to help you in a PM. Use the forums.]
i dont understand wat you mean PM

0

Ok, try this.
On line #17, you have this code:

$get =$record_count = mysql_query("SELECT *FROM news LIMIT $start,$per_page");

Remove the $record_count from there:

$get = mysql_query("SELECT *FROM news LIMIT $start,$per_page");

Also, just a heads up:
Line #12 of your code:

$record_count = mysql_num_rows( mysql_query("SELECT *FROM news"));

.
This is by far the most wasteful thing you can do.
You essentially fetch the ENTIRE table; every single piece of data in it, just so you can count how many rows there are.

A much MUCH better way to do this is to use the COUNT() function in the SQL query and read the results of that:

$countResult = mysql_query("SELECT COUNT(*) FROM news") or die(mysql_error());
$countRow = mysql_fetch_row($countResult);
$record_count = $countRow[0];

[Please do not ask me to help you in a PM. Use the forums.]
i dont understand wat you mean PM

PM = Private Message.
It's just my signature. Don't worry about that ;-)

0

im new at php and im duing this by a algorithm and i read much ,but it gives errors

Edited by dan1992: n/a

0
<?php

if($_Post['title'])
{
//get data
$title = $_POST ['title'];
$body = $_POST ['body'];

//chceck for existance

if($title&&$body)
{
mysql_connect("localhost","root","")or die(mysql_error());
mysql_select_db("dan")or die (mysql_error());

$data ("Y-m-d");
//insert data
$insert = mysql_query("INSERT INTO news VALUES('','$title','$body','$data')")or die (mysql_error());
die("Anuntul a fost adugat");
}
else
  echo "pune info<p>";
}
?>
<Form action="" method='post'>
      Title:<br>
      <input type='text' name='title' id="title"><p>
      Body:<br>
      <textarea rows="6"cols='35' name = 'body' id="body" ></textarea>
      <p>
      <input type="submit" name='submit' value='Adauga' >
      <hr>
      <input tipe='text' name='photo'>

it gives i error in [Parse error: syntax error, unexpected T_VARIABLE in D:\Dan\web\bundle\xampp\xampp\htdocs\news\post.php on line 18]
this is a script theat put info in mysql data ... is like addinfo .

Edited by dan1992: n/a

0

it gives i error in [Parse error: syntax error, unexpected T_VARIABLE in D:\Dan\web\bundle\xampp\xampp\htdocs\news\post.php on line 18]
this is a script theat put info in mysql data ... is like addinfo .

Lines #16 and #19 are missing the semi-colon to end the line. (The ; char)

Also, please use code tags around your code. Makes it so much easier to read.

Edited by happygeek: fixed formatting

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