Details Problem

[Vur]

I am very new to php and learning how to use php and mysql recently. I am trying to constract a simple booksite which shows books which are published by my company. I desing it (or I tried to desing it) to show list of books at index (index.php) page and to show the details of selected book at another page (detay.php)

 

At the end of index page I used:

<td><div align='right'><a href='../detay.php?id=$ara_row[12]'>Detaylar</a></div><br><br><br></td>

 

because row 12 is the auto_increment, primary, id column (which is also named as id) and when I click on details it passas to another page which is addressed as ...detay.php?id?=1.

 

On details (detay) page I used

 

$ara = mysql_query("select * from Kitap where $id=id");

$ara_sayi = mysql_num_rows($ara);

if ($ara_sayi == 0)

  {

    echo '<br><br><br><br><br><br>';

    echo "<center><b>Aradığınız kriterlere uyan bir sonuç bulunamadı.</b></center>";

  }

 

to show the book if page id = id , but else show "there is no such thing"

but it allways shows "Aradığınız kriterlere uyan bir sonuç bulunamadı" which means "tehere is no such thing"

I couldn't find where is the problem, it seems "all right" to me,

 

I will be very glad if anybody gives me a small solution. Thank you

 

help?help? :'(

[DarkWater]

You mixed up the column and variable in the query. =P

 

$ara = mysql_query("select * from Kitap where id=i$id") OR die(mysql_error());

 

Try that.

 

You had:

$ara = mysql_query("select * from Kitap where $id=id");  See how you have "$id = id"?

[Minase]

this isnt going to work lol =))

$ara = mysql_query("select * from Kitap where id=i$id") OR die(mysql_error());

maybe

$ara = mysql_query("select * from Kitap where id=$id") OR die(mysql_error());

[Vur]

You mixed up the column and variable in the query. =P

 

$ara = mysql_query("select * from Kitap where id=i$id") OR die(mysql_error());

 

Try that.

 

You had:

$ara = mysql_query("select * from Kitap where $id=id");  See how you have "$id = id"?

 

Dear friend,

 

I changed one of the variables to not to mix again like

 

<td><div align='right'><a href='../detay.php?kimlik=$ara_row[12]'>Detaylar[/url]</div>

 

and I made parameters as

 

$ara = mysql_query("select * from Kitap where $kimlik=id"); didnt work

$ara = mysql_query("select * from Kitap where id=$kimlik"); crashed

$ara = mysql_query("select * from Kitap where $id=kimlik"); didnt work

$ara = mysql_query("select * from Kitap where kimlik=$id"); crashed

 

:'( :'( :'(

[Minase]

try this it should work,but first be sure that variable $id has content.

$ara = mysql_query("select * from Kitap where id=$id") OR die(mysql_error());

[Vur]

try this it should work,but first be sure that variable $id has content.

$ara = mysql_query("select * from Kitap where id=$id") OR die(mysql_error());

 

Sorry my friend it didn't work even with this way. id colum has a number content with auto_increment. it is olsa primary column

 

I made mistake somewhere but where?

[Minase]

ah i think i know what you mean

$ara = mysql_query("select * from Kitap where id=$_GET['ID']") OR die(mysql_error());

 

try that

[Vur]

ah i think i know what you mean

$ara = mysql_query("select * from Kitap where id=$_GET['ID']") OR die(mysql_error());

 

try that

 

Nope, sorry, didn't work

Really thank you

but I think problem is somwhere else

 

[Ken2k7]

Try this:

$ara = mysql_query("select * from Kitap where id='%d'") OR die(mysql_error());
$ara = sprintf($ara, $_GET['id']);

 

There is another way to do that, but I prefer this way.

 

You just need to have the quote there.

[Vur]

I have tried many things since morning and I think problem is at first code. Somehow,

 

<td><div align='right'><a href='../detay.php?id=$ara_row[12]'>Detaylar[/url]</div>

 

code is not sending to next page data of $id

 

so $id stays blank (or empty) so it never gets equal with id column

 

any idea?

[Ken2k7]

From the code, I don't see the declaration or initiation of the variable $ara_row. Can you post the entire code?

[revraz]

we would have to see the sql before that along with the rest of the code on that page.

[Vur]

ok thanks guys.

so this is index page

 

<?php
					if ($k<>0)
						{
						}
						else
						{
						$k=0;
						}
					$ara = mysql_query("select * from Kitap where bolum=1 order by adi limit $k,6");
					$ara_sayi = mysql_num_rows($ara);//kaç satır kayıt var onu öğrenirsin, yazdırmak için gerek yok sadece bilgi
					if ($ara_sayi == 0)
						{
						echo '<br><br><br><br><br><br>';
						echo "<center><b>Aradığınız kriterlere uyan bir sonuç bulunamadı.</b></center>";
						} 
					while($ara_row = mysql_fetch_row($ara)){ 
						echo "<table width='498' border='0' cellspacing='0' cellpadding='0'>
  <tr>
    <td colspan='2' height='20' bgcolor='#F9F3DF'><b> $ara_row[0]</b></td>
  </tr>
  <tr>
    <td width='21%' height='120'><div align='center'><img height=107 
                  src='../images/kitap/$ara_row[7].jpg' 
                width=74></div></td>
    <td width='79%'><table width='96%' border='0' align='center' cellpadding='0' cellspacing='0'>
      
      <tr>
        <td><p align=justify>$ara_row[4]</p></td>
      </tr>
      <tr>
        <td><div align='right'><a href='/detay.php?id=$ara_row[12]'>Detaylar</a></div><br><br><br></td>
      </tr>
  
  <tr><br></tr>
    </table></td>
  </tr>
  <tr>
    <td height='20'> </td>
    <td height='20'> </td>
  </tr>
</table>";	
				}


	?>

 

 

 

and this is details page

 

 

 

<?php 		
						 $ara = mysql_query("select * from Kitap $id=id");
					$ara_sayi = mysql_num_rows($ara);
					if ($ara_sayi == 0)
						{
						echo '<br><br><br><br><br><br>';
						echo "<center><b>Aradığınız kriterlere uyan bir sonuç bulunamadı.</b></center>";
						} 

					while($ara_row = mysql_fetch_row($ara)){ 
									if ($ara_row[4]==0)
										{
											$ara_row[4]='-  ';
										}

									........
?>

[Ken2k7]

Please at least

 or [php] your code.

[Vur]

any idea?

[Vur]

I found a new problem which is connected directly with first one. I tried to do so paging wich is limited with 6 books. somehow it doesnt go to next page. When I click on "next" it shows same page with same books not with next ones..

 

<?
						//echo $k;
						$sayfa = $k/6;
						$sayfa = $sayfa+1;
						$sayfa_sayisi = mysql_query("select * from Kitap where bolum=1");
						$sayfa_sayisi2 = mysql_num_rows($sayfa_sayisi);
						$toplam_sayfa = intval($sayfa_sayisi2/6);
						$toplam_sayfa = $toplam_sayfa + 1;
						echo "Sayfa ";
						echo $sayfa;
						echo " / ";
						echo $toplam_sayfa;
						?>
                                    |</div></td>
                            <td width="40%" height="20" valign="bottom">
						<? 
							$ileri = mysql_query("select * from Kitap where bolum=1");
							$ileri_sayi = mysql_num_rows($ileri);
							if ($ileri_sayi>6)
								{
									$z = $k + 6;
									if ($z > $ileri_sayi)
									{
									}
									else
									{
										echo "<a href='index.php?k=$z'>İleri</a>";
									}
								} 

						?>

 

 

I think both problem is related with this code

 

<a href='/detay.php?id=$ara_row[12]'>Detaylar
echo "<a href='index.php?k=$q'>Geri</a>"
echo "<a href='index.php?k=$z'>İleri</a>"

 

I th'nk they are not sending data to next page..... how can I sove it? Any idea?

 

[Vur]

pls? ??? :'(

[Ken2k7]

echo "<a href='/detay.php?id={$ara_row[12]}'>Detaylar</a>";
echo "<a href='index.php?k=$q'>Geri</a>";
echo "<a href='index.php?k=$z'>İleri</a>";