just don't get it

pcwizzzz

Hi,
I have solved a problem but still dont get it. Maybe someone would make me understand.
Im going to write a sort of pseudo-code to be shorter.

query tab1
while
{
echo $row[name]
...........
query tab2,tab3
while
{
echo $row[name]
.......
}
}

the $row[name] on the second while it doesn't work, it's empty. Why?

eoghain

Who knows your psudo code doesn't have enough detail to show what is happening. Maybe you forgot to query for a field called name, or you reset the $row variable somehow.

There is no way to answer your question correctly given the data you supplied.

pcwizzzz

OK, HERE IS THE CODE

while($row=mysql_fetch_assoc($rez))
{
echo "<table border=2 cellspacing=4 width=\"100%\">";
echo "<tr><td bgcolor=#FFE4C4 class=hotelh>".$row[nume]."&nbsp;&nbsp;".$row[stele];
     if($row[zona]!="")
     {
      echo "<div class=zona>Zona:".$row[zona]."</div>";
      }
echo"</td></tr>";
echo "</table>";


$idhh=$row[id_hotel];
$sql1="SELECT * FROM descriere,poze WHERE descriere.id_hotel=$idhh AND poze.id_hotel=$idhh";
$rez1=mysql_query($sql1);

while($row=mysql_fetch_assoc($rez1))
{
echo "<table border=0 width=\"100%\">";
echo "<tr><td bgcolor=#FFE4C4><img src=\"/turcia/pics/$row[poza_ext]\" alt=\"$row[nume]\" border=1><br><br>
........................

The $row[name] inside ALT doesn't work unless i use
$var=$row[nume] but before the last query.

bubblenut

Try changing

$rez1=mysql_query($sql1);

$rez1=mysql_query($sql1) or die($sql."\n".mysql_error());

To view any errors arising from the second query, although I don't think this is the problem. Also, you could try this code just to see the structure of the array being returned by mysql_fetch_assoc().

$sql1="SELECT * FROM descriere,poze WHERE descriere.id_hotel=$idhh AND poze.id_hotel=$idhh";
$rez1=mysql_query($sql1); 
$row=mysql_fetch_assoc($rez1);
print_r($row); exit();

Finally, if this doesn't show up anything unexpected could you provide the first sql statement and the structure for your tables please?

pcwizzzz

Na, i tried what you had sugessted but no error. The query is working fine. I'll put the whole code.

<?php
include "../classes/conectare.php";
$obj=new conectare("kusadasi_turcia");
$sql="SELECT * FROM orase WHERE id_oras=$idh";
$rez=mysql_query($sql);
$row=mysql_fetch_assoc($rez);

echo "<title>HOTELURI IN ".$row[oras]."</title>";
echo "<style>
      BODY {
  font-family: arial,'sans-serif';
  font-size:7pt;
  border-color:#C0C0C0;
  background:#FFFFFF;
  padding:2;
  border-style:none;
  scrollbar-face-color: #F9F9F9;
  scrollbar-shadow-color: #BAD8E9;
  scrollbar-highlight-color: #BAD8E9;
  scrollbar-3dlight-color: #F9F9F9;
  scrollbar-darkshadow-color: #F9F9F9;
  scrollbar-track-color: #F9F9F9;
  scrollbar-arrow-color: #BAD8E9;
  }
h2 { color:#FF7F50;
  }


</style>";

echo "<META NAME=\"Generator\" CONTENT=\"Stone's WebWriter 3\">";
echo "<META HTTP-EQUIV=\"Author\" CONTENT=\"Ioana Filipescu\">";

echo "<LINK href=\"layout.css\" rel=stylesheet type=text/css>";
echo "</head>";
echo "<body BGCOLOR=\"#FFFFFF\" leftMargin=3 topMargin=15>";
echo "<center>";
switch ($nrst)
{
case "CLUBURI":
    echo "<h2>Cluburi in ".$row[oras]."</h2>";
    $sql="SELECT * FROM hoteluri WHERE id_oras=$idh AND stele IN ('H.V.1','H.V.2')";
break;

case "CATEGORIE SPECIALA":
      echo "<h2>Hoteluri categorie speciala din ".$row[oras]."</h2>";
      $sql="SELECT * FROM hoteluri WHERE id_oras=$idh AND stele LIKE 'S'";
break;

case "VILE":
     echo "<h2>Vile in ".$row[oras]."<h2>";
     $sql="SELECT * FROM hoteluri WHERE id_oras=$idh AND stele LIKE 'vile'";
break;

case "APART":
     echo "<h2>Apartamente in ".$row[oras]."<h2>";
     $sql="SELECT * FROM hoteluri WHERE id_oras=$idh AND stele LIKE 'apart'";
break;

default:
  echo "<h2>Hoteluri de ".$nrst." stele in ".$row[oras]."<h2>";
  $sql="SELECT * FROM hoteluri WHERE id_oras=$idh AND stele=$nrst";
break;
  }
echo "<a href=\"prezentare_oras.php?id=$idh\" class=linkback>~ Inapoi la $row[oras] ~</a>";
echo "<br><BR>";
//selectare hoteluri
$rez=mysql_query($sql);
while($row=mysql_fetch_assoc($rez))
{
$nume=$row[nume];
echo "<table border=2 cellspacing=4 width=\"100%\">";
echo "<tr><td bgcolor=#FFE4C4 class=hotelh>".$row[nume]."&nbsp;&nbsp;".$row[stele];
     if($row[zona]!="")
     {
      echo "<div class=zona>Zona:".$row[zona]."</div>";
      }
echo"</td></tr>";
echo "</table>";

//selectare descriere
$idhh=$row[id_hotel];
$sql1="SELECT * FROM descriere,poze WHERE descriere.id_hotel=$idhh AND poze.id_hotel=$idhh";
$rez1=mysql_query($sql1) or die($sql1."\n".mysql_error());
$row=mysql_fetch_assoc($rez1);
print_r($row); exit();

while($row=mysql_fetch_assoc($rez1))
{
echo "<table border=0 width=\"100%\">";
echo "<tr><td bgcolor=#FFE4C4><img src=\"/turcia/pics/$row[poza_ext]\" alt=\"$nume\" border=1><br><br>

                          <img src=\"/turcia/pics/$row[poza_int]\" alt=\"$nume\" border=1></td>";
echo "<td bgcolor=#FFE4C4>";
if($row[localizare]!="")
{
echo "<font size=2><span class=hdscr>Localizare:</span> ".$row[localizare]."<br><br>";
}
if($row[cazare]!="")
{
echo "<span class=hdscr>Cazare:</span><font size=2> ".$row[cazare]."<br><br>";
}
if($row[facilitati]!="")
{
echo "<span class=hdscr>Facilitati:</span><font size=2> ".$row[facilitati]."<br><br>";
}
if($row[servicii]!="")
{
echo "<span class=hdscr>Servicii:</span><font size=2> ".$row[servicii]."<br><br>";
}
if($row[plaja]!="")
{
echo "<span class=hdscr>Plaja:</span><font size=2> ".$row[plaja]."<br><br>";
}
if($row[masa]!="")
{
echo "<span class=hdscr>Masa:</span><font size=2> ".$row[masa]."<br><br>";
}
if($row[sporturi]!="")
{
echo "<span class=hdscr>Sporturi:</span><font size=2> ".$row[sporturi]."<br>";
}
echo "</td></tr>";
echo "</table>";
}
echo "<br>";
}
echo "<a href=\"prezentare_oras.php?id=$idh\" class=linkback>~ Inapoi ~</a>";


  ?>

Thanks. Any sugesstion regarding my code is wellcomed.

laserlight

$row[index] should be $row['index'], unless you have defined a constant named index.
e.g. use $row['id_hotel'] instead of $row[id_hotel]

When using array elements within strings, either concatenate or use the complex syntax.
e.g.

$array = array("place" => "world");
echo "hello " . $array['place'];
//or
echo "hello {$array['place']}";

Your idea of giving pseudocode is a good one.
Now sift through your actual code and come up with the smallest and simplest script that demonstrates your problem.

eoghain

Sorry, it's monday morning and I wasn't seeing things clearly. For some reason I was looking at your table names and thinking they were field names. Anyway I'll leave my below observations for you to read through, but it's probably not applicable.

Your problem is obvious.

$sql1="SELECT * FROM descriere,poze WHERE descriere.id_hotel=$idhh AND poze.id_hotel=$idhh";

That query doesn't select a field called 'nume' but you are trying to access it in the while clause.

My guess is that you want 'nume' from your first query still (seeing as your second query $sql1 is embedded in the while clause dealing with the results from your first query). Change your second queries while clause to place the rows into $row2 instead of $row and change all other references to the second queries results to use $row2 and your problem should be fixed.

pcwizzzz

yes, you got it. I thought so, cause it sees $row[field] and look into the second query not the first so that's why it doesn't see the name field. Regarding the $row['field'] that's not the problem, it works damn fine with $row[field] too.
Thanks alot, that was the problem. But still, i'm probably too curious....... does the second query overwrite the first one in $row array and that's why it doesn't see the field from the first query? This why i think.

eoghain

Any time you do an operation like:

$x = 123;

The variable $x is going to get the value that you are assigning to it, and any previous value is going to get destroyed.

So when you do:

$sql="SELECT * FROM hoteluri WHERE id_oras=$idh AND stele=$nrst"; 
$rez=mysql_query($sql);
while($row=mysql_fetch_assoc($rez))
{
// $row is now assigned to each successive
// value returned by the above query $sql

$sql1="SELECT * FROM descriere,poze WHERE descriere.id_hotel=$idhh AND poze.id_hotel=$idhh";
$rez1=mysql_query($sql1);

while($row=mysql_fetch_assoc($rez1))
{ 
// $row is now assigned each successive
// value returned by the new query $sql1
}
}

$row is a designation that you assign and a place where you put a row of data from your query. So everytime you put something into that variable it destroys the data that was in there previously.

pcwizzzz

Yes, i got you, you are perfectly right. Thanks alot for all the explinations.