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i got error while coding of pagination in php5?

开发者 https://www.devze.com 2023-01-17 13:27 出处:网络
i am writing code of pagination & i got an error. this is my code : <?php mysql_connect(\"localhost\", \"root\", \"\") or die(mysql_error());

i am writing code of pagination & i got an error. this is my code :

<?php 
mysql_connect("localhost", "root", "") or die(mysql_error()); 
mysql_select_db("A开发者_运维技巧dmin") or die(mysql_error()); 

if (isset($_GET["page"]))
 {
 $page  = $_GET["page"]; 
 }
 else
 {
 $page=1;
 }



$start_from = ($page-1) * 2; 
$sql = "SELECT * FROM events ORDER BY event ASC LIMIT $start_from, 2"; 
$result = mysql_query ($sql) or die(mysql_error()); 
$num=mysql_numrows($result);
$x=0;
?> 
<table>
<tr><td>Event</td><td>Types</td></tr>
<?php 
while ($x<$num) {
$row1 = mysql_result($result,$x,'event');
$row2 = mysql_result($result,$x,'types'); 
?> 
            <tr>
            <td><? echo $row1; ?></td>
            <td><? echo $row2; ?></td>
            </tr>
<?php 
$x++;
} 
?> 
</table>
<?php 
$sql = "SELECT COUNT(event) FROM events"; 
$result = mysql_query($sql) or die(mysql_error()); 
$row = mysql_fetch_row($result); 
$total_records = $row[0]; 
$total_pages = ceil($total_records / 2); 

for ($i=1; $i<=$total_pages; $i++) { 
            echo "<a href='pagination.php?page=".$i."'>".$i."</a> "; 
}

?>

now i got an error.

  object not found


Note that instead of terrible while ($x<$num) you should use way more neat while($row=mysql_fetch_array($res))
so, make it

<?
$per_page = 2;
$page = 1;
if (isset($_GET['page'])) $page = $_GET['page'];
$start_from = ($page-1) * $per_page; 
$sql = "SELECT * FROM events ORDER BY event ASC LIMIT $start_from, $per_page"; 
$res = mysql_query ($sql) or trigger_error(mysql_error().": ".$sql);
?> 
<table>
  <tr><td>Event</td><td>Types</td></tr>
<?php 
while($row=mysql_fetch_array($res)) {
?> 
  <tr>
    <td><? echo $row['event'] ?></td>
    <td><? echo $row['types'] ?></td>
  </tr>
<?php 
} 
?> 
</table>
<?php 
$sql = "SELECT COUNT(event) FROM events"; 
$res = mysql_query ($sql) or trigger_error(mysql_error().": ".$sql);
$row = mysql_fetch_row($res); 
$total_records = $row[0]; 
$total_pages = ceil($total_records / $per_page); 

for ($i=1; $i<=$total_pages; $i++) { 
  echo "<a href='pagination.php?page=".$i."'>".$i."</a> "; 
}
?>


mysql_query returns false - i.e. your query failed and you are passing a wrong argument to mysql_num_rows. I think you have a typo in the table name in the SQL query, it should be events instead. Generally, you should check if the query succeeded:

$result = mysql_query ($sql) or trigger_error(mysql_error().": ".$sql);
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