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return problem
Posted: Mon Jul 01, 2002 1:47 am
by mrpaulfrank
while($row_Recordset1 = mysql_fetch_assoc($Recordset1))
{
$image = $row_Recordset1['image'];
if($genre = "comedy")
{
echo "<A href=backs/$image.jpg><IMG SRC=images/$image.jpg></A>"."<br>\n".
"Title: ".$row_Recordset1['title']."<br>\n".
"Format: ".$row_Recordset1['format']."<br>\n".
"Price: ".$row_Recordset1['make']."<br>\n\n";
}
}
how can i get the results (one result being a whole loop existing of an image, title, format, and make) to be placed horizontally instead of vertically?
Posted: Mon Jul 01, 2002 1:59 am
by garymailer
Mrpaulfrank
Code: Select all
I've simply taken out all the <br> and the newline characters (\n) tags and included one at then end
while($row_Recordset1 = mysql_fetch_assoc($Recordset1))
{
$image = $row_Recordset1ї'image'];
if($genre = "comedy")
{
echo "<A href=backs/$image.jpg><IMG SRC=images/$image.jpg></A>".
"Title: ".$row_Recordset1ї'title']."
"Format: ".$row_Recordset1ї'format']."
"Price: ".$row_Recordset1ї'make'].".
<br />";
}
}
Posted: Mon Jul 01, 2002 3:28 am
by mrpaulfrank
it doesnt work:( all that does is make the entire thing one long string of text. what i want is this:
image
title
format
make
then the next one displayed a little over horizontally, and so on...
is there some sort of a 'tab' tag i can use to do this?? im completely lost.
thanks
Posted: Mon Jul 01, 2002 3:32 am
by hob_goblin
piece of cake
Code: Select all
echo "<table cellpadding="0" cellspacing="5"><tr>";
while($row_Recordset1 = mysql_fetch_assoc($Recordset1))
{
$image = $row_Recordset1ї'image'];
if($genre = "comedy")
{
echo "<td><table cellpadding="0" cellspacing="2"><tr><td>"
."<A href=backs/$image.jpg><IMG SRC=images/$image.jpg></A></td></tr>".
"<tr><td>Title: ".$row_Recordset1ї'title']."
"</td></tr><tr><td>Format: ".$row_Recordset1ї'format']."
"</td></tr><tr><td>Price: ".$row_Recordset1ї'make']."</td></tr></table></td>";
}
}
echo "</tr></table>";
Posted: Mon Jul 01, 2002 4:23 am
by mrpaulfrank
wow, im extremely lost. from what i see youre trying to echo a table before you echo the info from mysql which will be inserted into the table?? i still cant get it to work. maybe if someone can explain to me what i should be trying to do...i dunno. thanks
Posted: Mon Jul 01, 2002 4:35 am
by twigletmac
What the code should do is create a large table where on each iteration of the while loop a new cell is created and another table inserted into it containing a few rows with the information from the database, here's the code again, laid out differently and with a couple of parse errors removed:
Code: Select all
<?php
echo <<<END
<table cellpadding="0" cellspacing="5">
<tr>
END;
while($row_Recordset1 = mysql_fetch_assoc($Recordset1)) {
$image = $row_Recordset1ї'image'];
if ($genre == 'comedy') {
echo <<<END
<td>
<table cellpadding="0" cellspacing="2">
<tr>
<td><a href="backs/$image.jpg"><img src="images/$image.jpg"></a></td>
</tr>
<tr>
<td>Title: {$row_Recordset1ї'title']}</td>
</tr>
<tr>
<td>Format: {$row_Recordset1ї'format']}</td>
</tr>
<tr>
<td>Price: {$row_Recordset1ї'make']}</td>
</tr>
</table>
</td>
END;
}
}
echo <<<END
</tr>
</table>
END;
?>
Mac
Posted: Mon Jul 01, 2002 6:09 pm
by mrpaulfrank
whats wrong with this code? im getting a parse error and its not creating the table within a table. im trying to make each loop a table (each table having 4 rows) in a larger table so that the results will be columns in one row of the larger table (make sense?):
Code: Select all
<table width="800" border="0" cellspacing="0">
<tr>
<?php
$genre = "comedy";
$query_Recordset1 = "SELECT * FROM test WHERE genre = '$genre'";
$Recordset1 = mysql_query($query_Recordset1, $db) or die(mysql_error());
while($row_Recordset1 = mysql_fetch_assoc($Recordset1))
{
$image = $row_Recordset1ї'image'];
if($genre = "comedy")
{
echo "<table width="200" border="0" cellspacing="0">"."<tr>".
"<td><A href=backs/$image.jpg><IMG SRC=images/$image.jpg></A>"."</td>".
"<td>Title: ".$row_Recordset1ї'title']."</td>".
"<td>Format: ".$row_Recordset1ї'format']."</td>".
"<td>Make: ".$row_Recordset1ї'make']."</td></tr>";
}
}
?>
</tr>
</table>
Posted: Tue Jul 02, 2002 1:55 am
by twigletmac
The code I posted was the code that you are now using with the obvious parse errors in the original code removed...
Anyway,
Code: Select all
<table width="800" border="0" cellspacing="0">
<tr>
<?php
$genre = 'comedy';
$query_Recordset1 = "SELECT * FROM test WHERE genre = '$genre'";
$Recordset1 = mysql_query($query_Recordset1, $db) or die(mysql_error());
while($row_Recordset1 = mysql_fetch_assoc($Recordset1))
{
$image = $row_Recordset1ї'image'];
if ($genre == 'comedy') {
echo <<<END
<td>
<table cellpadding="0" cellspacing="2">
<tr>
<td><a href="backs/$image.jpg"><img src="images/$image.jpg"></a></td>
</tr>
<tr>
<td>Title: {$row_Recordset1ї'title']}</td>
</tr>
<tr>
<td>Format: {$row_Recordset1ї'format']}</td>
</tr>
<tr>
<td>Make: {$row_Recordset1ї'make']}</td>
</tr>
</table>
</td>
END;
}
}
?>
</tr>
</table>
Please also note that although it won't cause any parse errors it will cause you a headache,
has to be
in order to work.
Mac
Posted: Tue Jul 02, 2002 2:31 am
by hob_goblin
sorry

it was rather early in the morning, when parse errors seem to roam my mind
Posted: Wed Jul 03, 2002 4:53 am
by mrpaulfrank
all thes problems were because i was not escaping the quotes! haha. thanks guys