sabo86

I got the same problem.. when you click on the last page, it's blank!!! what's the solution?

Hello guys.. I am just trying the page limiting code for mysql results.. and here is the code.. # <?php $server = "localhost"; # $user = "root"; # $pass = "password"; # $databasename = "fleifel"; # $db = mysql_connect($server, $user); # mysql_select_db($databasename,$db); # # $sql = "SELECT * FROM items "; # $query = mysql_query($sql,$db); # $total_results = mysql_num_rows($query); # $limit = "3"; //limit of archived results per page. # $total_pages = ceil($total_results / $limit); //total number of pages # if (empty($page)) # { # $page = "1"; //default page if none is selected # } # $offset = ($page - 1) * $limit; //starting number for displaying results out of DB # # $query = "SELECT * FROM items LIMIT $offset, $limit"; # $result = mysql_query($query); # //This is the start of the normal results... # # while ($row = mysql_fetch_array($result)) # { # // display your results as you see fit here. # } # mysql_close(); # # # // This is the Previous/Next Navigation # echo "<font face=Verdana size=1>"; # echo "Pages:($total_pages) "; // total pages # if ($page != 1) # { # echo "<a href=$PHP_SELF?page=1><< First</a> "; // First Page Link # $prevpage = $page - 1; # echo " <a href=$PHP_SELF?page=$prevpage><<</a> "; // Previous Page Link # } # if ($page == $total_pages) # { # $to = $total_pages; # } # elseif ($page == $total_pages-1) # { # $to = $page+1; # } # elseif ($page == $total_pages-2) # { # $to = $page+2; # } # else # { # $to = $page+3; # } # if ($page == 1 || $page == 2 || $page == 3) # { # $from = 1; # } # else # { # $from = $page-3; # } # # for ($i = $from; $i <= $to; $i++) # # { # if ($i == $total_results) $to=$total_results; # if ($i != $page) # { # echo "<a href=$PHP_SELF?showold=yes&page=$i>$i</a>"; # } # else # { # echo "<b><font face=Verdana size=2>[$i]</font></b>"; # } # if ($i != $total_pages) # echo " "; # } # if ($page != $total_pages) # { # $nextpage = $page + 1; # echo " <a href=$PHP_SELF?page=$nextpage>>></a> "; // Next Page Link # echo " <a href=$PHP_SELF?page=$total_pages>Last >></a>"; // Last Page Link # } # echo "</font>"; # ?> I got the following result: # Pages:(2) [1] Notice: Undefined variable: PHP_SELF in C:\Program Files\EasyPHP 2.0b1\www\test fleifel\example.php on line 141 2 Notice: Undefined variable: PHP_SELF in C:\Program Files\EasyPHP 2.0b1\www\test fleifel\example.php on line 165 >> Notice: Undefined variable: PHP_SELF in C:\Program Files\EasyPHP 2.0b1\www\test fleifel\example.php on line 167 Last >> What is the problem?

The while loop worked... but the results are as following: 1.Model : Executive Office Arco Description: Executive offices wooden frames anthracite lacquered; worktops, doors and drawer-fronts. Category: European Office Furniture 2.Model : MOD. SERIE 2000 Description: Work Station Category: Local Office Furniture 1.Model : MOD.098 Description: Anthropornetric desk Category: School and University Furniture 2.Model : Operative Seatings T Description: I love this descript Category: Seatings In other words, the numerations are 1 & 2 only!!! how can this be solved? here is the code again: $i=0; while ($row = mysql_fetch_array($result)) { echo '<table width="800" border="0">'; if( $i = $i%2 ) { echo ' <tr bgcolor="#EFEFEF">';} else { echo ' <tr bgcolor="#D3D3D3">'; }; echo '<td width="68" align="left" valign="top"><div align="left"><a href="'.$row["Picture"].'" target="_blank"><img src="'.$row["PictureThumbnail"].'" border="0" /></a></td>'; echo " <td><strong> ".($i+1).".Model : </strong>"; echo htmlspecialchars(stripslashes($row["Model"])); echo "<br><strong> Description: </strong>" ; echo htmlspecialchars (stripslashes($row["Description"])); echo "<br><strong> Category: </strong>"; echo htmlspecialchars (stripslashes($row["Category"])); echo "<br></td></table>";$i++; } Thanks much

but the for loop was working before I put the if condition? what do you think the problem is?

but i want to use the for loop inorder to make different row background colors! Besides I need to numerate the results

This one may make it more clear $num_results = mysql_num_rows($result); echo "<p><i>Number of records found: ".$num_results."</i></p>"; $Model = $Description = $Category = $Image = null; for ($i=0; $i <$num_results; $i++){ $row = mysql_fetch_array($result); echo '<table width="800" border="0">'; if( $i = $i%2 ) { echo ' <tr bgcolor="#EFEFEF">';} else { echo ' <tr bgcolor="#C3C3C3">'; }; echo '<td width="68" align="left" valign="top"><div align="left"><a href="'.$row["Picture"].'" target="_blank"><img src="'.$row["PictureThumbnail"].'" border="0" /></a></td>'; echo " <td><strong> ".($i+1).".Model : </strong>"; echo htmlspecialchars(stripslashes($row["Model"])); echo "<br><strong> Description: </strong>" ; echo htmlspecialchars (stripslashes($row["Description"])); echo "<br><strong> Category: </strong>"; echo htmlspecialchars (stripslashes($row["Category"])); echo "<br></td></table>"; }

hello guys here part of my code: for ($i=0; $i <$num_results; $i++){ $row = mysql_fetch_array($result); echo '<table width="800" border="0">'; if( $i = $i%2 ) { echo ' <tr bgcolor="#EFEFEF">';} else { echo ' <tr bgcolor="#C3C3C3">'; } echo '<td width="68" align="left" valign="top"><div align="left"><a href="'.$row["Picture"].'" target="_blank"><img src="'.$row["PictureThumbnail"].'" border="0" /></a></td>'; echo " <td><strong> ".($i+1).".Model : </strong>"; echo htmlspecialchars(stripslashes($row["Model"])); echo "<br><strong> Description: </strong>" ; echo htmlspecialchars (stripslashes($row["Description"])); echo "<br><strong> Category: </strong>"; echo htmlspecialchars (stripslashes($row["Category"])); echo "<br></td></table>"; } The results seems to be infinite loop of rows that doesn't end... any suggestions?

any help plz?

one more question before i try the automated thing.. I created an attribute for the thumbnail... and Now i have the following result .. and i want to make it in this way: http://img382.imageshack.us/my.php?image=res2kn9.jpg How can I edit my code?

I know graphic design, but haven't heard about an automated way as a programming way!

isn't there is another way of resizing the same picture..

btw, is there anyway that makes this image a thumbnail in a way i can click it to get the actual size?

thanks wildteen88 for your help.. it really works!! I appreciate

I have this code now (I've been editing it thousands times ) for ($i=0; $i <$num_results; $i++){ $row = mysql_fetch_array($result); echo " <p><strong>".($i+1).".Model: </strong>"; echo htmlspecialchars(stripslashes($row["Model"])); echo "<br><strong> Description: </strong>" ; echo htmlspecialchars (stripslashes($row["Description"])); echo "<br><strong> Category: </strong>"; echo htmlspecialchars (stripslashes($row["Category"])); echo "<br><strong> Image: </strong>"; echo $row["Picture"];} How can I retrieve the image and display it in the correct place? I've heard that storing them in blob format will exhaust mysql..

Ok.. well.. I want simply to display the database's records in a neat table so that it would be compatible with my layout that i will design later on

Now my code is: $Model .= '<td>'.$row['Model'].'</td>'; $Description .= '<td>'.$row['Description'].'</td>'; $Category .= '<td>'.$row['Category'].'</td>'; $Image .= '<td>'.$row['Picture'].'</td>'; } for ($i=0; $i <$num_results; $i++) { ?> <table> <tr><td><?php echo "<br><p><strong>".($i+1).". Model:"; ?></td><?=$Model?></tr> <tr><td><?php echo "<strong> Description:"; ?></td><?=$Description?></tr> <tr><td><?php echo "<strong> Category:"; ?></td><?=$Category?></tr> <tr><td><?php echo "<strong> Picture:"; ?></td><?=$Image?></tr> <?php } and results are like: Number of records found: 4 1. Model: Description: Category: Picture: 2. Model: Description: Category: Picture: 3. Model: Description: Category: Picture: 4. Model: Description: Category: Picture: How can I get the records in these attributes?

ok i got Fleifel Notice: Undefined variable: field in C:\Program Files\EasyPHP 2.0b1\www\test fleifel\results.php on line 54 Number of records found: 4 1Model 1Description 1Category 1Picture 2Model 2Description 2Category 2Picture 3Model 3Description 3Category 3Picture 4Model 4Description 4Category 4Picture But I am not getting the results from the database.. just the atribute names...

OK I tried this one, the results are as following: Notice: Undefined variable: field in C:\Program Files\EasyPHP 2.0b1\www\test fleifel\results.php on line 54 Number of records found: 4 Notice: Undefined variable: Model in C:\Program Files\EasyPHP 2.0b1\www\test fleifel\results.php on line 77 Notice: Undefined variable: Description in C:\Program Files\EasyPHP 2.0b1\www\test fleifel\results.php on line 78 Notice: Undefined variable: Category in C:\Program Files\EasyPHP 2.0b1\www\test fleifel\results.php on line 79 Notice: Undefined variable: Image in C:\Program Files\EasyPHP 2.0b1\www\test fleifel\results.php on line 80 1Model 1Description 1Category 1Picture 2Model 2Description 2Category 2Picture 3Model 3Description 3Category 3Picture 4Model 4Description 4Category 4Picture

it didn't work as well.. the same error!

Didn't work error: Parse error: parse error, unexpected T_ENCAPSED_AND_WHITESPACE, expecting T_STRING or T_VARIABLE or T_NUM_STRING in C:\Program Files\EasyPHP 2.0b1\www\test fleifel\results.php on line 77

I didn't get your point? would you plz explain?

I am trying to display my results, I tried this way: echo "<p>Number of records found: ".$num_results."</p>"; while ($row = mysql_fetch_array($result)){ $Model.="<td>$row['Model']</td>"; $Description.="<td>$row['Description']</td>"; $Category.="<td>$row['Category']</td>"; $Image.="<td>$row['Picture']</td>"; } for ($i=0; $i <$num_results; $i++) { <table> <tr><td>echo "<p><strong>.($i+1).".Model</td><?=$Model</tr> <tr><td>echo "<p><strong>.($i+1).".Description</td><?=$Description</tr> <tr><td>echo "<p><strong>.($i+1).".Category</td><?=$Category</tr> <tr><td>echo "<p><strong>.($i+1).".Picture</td><?=$Image</tr> } but i got this error: Parse error: parse error, unexpected T_ENCAPSED_AND_WHITESPACE, expecting T_STRING or T_VARIABLE or T_NUM_STRING in C:\Program Files\EasyPHP 2.0b1\www\test fleifel\results.php on line 76 What's the problem?

here is part of my code if($_POST['searchtype']=='Seatings') $field='Seatings'; elseif ($_POST['searchtype']=='Accessories') $field='Accessories'; elseif ($_POST['searchtype']=="Local Office Furniture") $field='Local Office Furniture'; elseif ($_POST['searchtype']=="European Office Furniture") $field='European Office Furniture'; elseif ($_POST['searchtype']=="School and University Furniture") $field='School and University Furniture'; $query = "select * FROM items where Category = '$field' order by Model" ; $query2="select * FROM items order by Model"; if ($_POST['searchtype']=='All') {$result = mysql_query($query2)or trigger_error(mysql_error().'<br />Query was:'.$sql,E_USER_ERROR); ;} else $result = mysql_query($query)or trigger_error(mysql_error().'<br />Query was:'.$sql,E_USER_ERROR); ; $num_results = mysql_num_rows($result); echo "<p>Number of records found: ".$num_results."</p>"; When I select "All" from the drop menu, I am getting the following result: Notice: Undefined variable: field in C:\Program Files\EasyPHP 2.0b1\www\test fleifel\results.php on line 53 Number of records found: 4 1.Model:Executive Office Arco Description:Executive offices wooden frames anthracite lacquered; worktops, doors and drawer-fronts. Category: European Office Furniture Image: 2.Model:MOD. SERIE 2000 Description:Work Station Category: Local Office Furniture Image: 3.Model:MOD.098 Description:Anthropornetric desk Category: School and University Furniture Image: http://localhost/test%20fleifel/desk.JPG 4.Model:Operative Seatings T Description:I love this descript Category: Seatings Image: http://localhost/test%20fleifel/desk.JPG The results are right, but why is the NOTICE thing is displayed?

Thanks a lot

if($_POST['searchtype']="Accessories") $field='Accessories'; elseif ($_POST['searchtype']="Seatings") $field='Seatings'; elseif ($_POST['searchtype']="Local Office Furniture") $field='Local Office Furniture'; elseif ($_POST['searchtype']="European Office Furniture") $field='European Office Furniture'; elseif ($_POST['searchtype']="School and University Furniture") $field='School and University Furniture'; $query = "select * FROM items where Category = '$field'"; Just wondering why $field is always taking the string 'Accessories'? knowing that searchtype is the drop menu form..??