Why won't this print my mysql table contents?

tracy · December 1, 2006

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN">
<html>
<head>
<title></title>
</head>
<body>
</td>
<td>
</td>
<td>
</td>
<td>
</td>
<td>
</td>
<td>
</td>
<td>
</td>
</tr><!--$i++;
}

echo "</table>";

?>
</body>
</html>

Psycho · December 2, 2006

It could be that you have no opening tag for the PHP code. You also have no error handling to check for errors. Try this:

[code]<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN">
<html>
<head>
<title></title>
</head>
<body>

<?php
include "connectionpage.php";
$query="SELECT * FROM 'Inventory'";
$result = mysql_query($sql) or die ("The query: ".$query." produced the following error: ".mysql_error());
if (!mysql_num_rows($result)) {
echo "there were no results.";
} else {
echo "<table border=1>";
$firstRecord = true;
while ($row = mysql_fetch_assoc($result)) {
if ($firstRecord) {
echo "<tr>";
foreach ($row as $key => $value) {echo "<td>".$key."</td>";}
echo "</tr>";
$firstRecord = false;
}
echo "<tr>";
foreach ($row as $value) { echo "<td>".$value."</td>"; }
echo "</tr>";
}
echo "</table> ";
}
?>
</body>
</html>[/code]

tracy · December 2, 2006

Thanks.

The query:
SELECT * FROM 'Inventory'
produced the following error:
Query was empty

I don't know what this means...there is data in the table...

ataria · December 2, 2006

[code]
<?php
include ("connectionpage.php");

$query= mysql_query("SELECT * FROM `Inventory`");

$num = mysql_num_rows($query);

if ($num == '0') {
echo "Nothing Exist.";
die();
}
else {

echo "<table border="0" cellspacing="2" cellpadding="2">
<tr>
<td>Stock#
</td>
<td>Year
</td>
<td>Make
</td>
<td>Model
</td>
<td>Price
</td>
<td>Miles
</td>
<td>Photo
</td>
</tr>";
while ($info = mysql_fetch_array($query)) {
$stock = $info['stock'];
$year = $info['year'];
$make = $info['make'];
$model = $info['model'];
$price = $info['price'];
$miles = $info['miles'];
$photo1 = $info['photo1'];
echo "
<tr>
<td> $stock
</td>
<td> $year
</td>
<td> $make
</td>
<td> $model
</td>
<td> $price
</td>
<td> $miles
</td>
<td> $photo1
</td>
</tr>"
}

echo "</table>";
}

?>[/code]

tracy · December 2, 2006

I tried that earlier...error line 17 which is where the table starts...

echo "<table border...

thanks, though...just tried it again and now error on line 17...same place.

ataria · December 2, 2006

echo "<table border="0" cellspacing="2" cellpadding="2">

change to..

echo "<table border=0 cellspacing=2 cellpadding=2>

tracy · December 2, 2006

now it says error line 56, right after the } symbol at the bottom, about three or four lines from the bottom...

jsladek · December 2, 2006

Put a [b];[/b] after the [b]</tr>"[/b]

</tr>";

Psycho · December 2, 2006

You can mess with the HTML output all you want, but if your query is failing you still won't get any output!

Either you are not connecting to the database or the table name is incorrect. I'm not sure that your include statement will work the way it is written. Try changing it to the correct format like this:

include ("connectionpage.php");

If that doesn't work, post the code from that included page.

kenrbnsn · December 2, 2006

The query should be:
[code]<?php
$query="SELECT * FROM Inventory";
?>[/code]

You don't need any quotes here.

Ken

tracy · December 4, 2006

After all your suggestions, here's the new code:

[code]
<?php
include ("link.php");

$query= mysql_query("SELECT * FROM 'Inventory'");

$num = mysql_num_rows($query);

if ($num == '0') {
echo "Nothing Exist.";
die();
}
else {

echo "<table border=0 cellspacing=2 cellpadding=2>
<tr>
<td>Stock#
</td>
<td>Year
</td>
<td>Make
</td>
<td>Model
</td>
<td>Price
</td>
<td>Miles
</td>
<td>Photo
</td>
</tr>";
while ($info = mysql_fetch_array($query)) {
$stock = $info['stock'];
$year = $info['year'];
$make = $info['make'];
$model = $info['model'];
$price = $info['price'];
$miles = $info['miles'];
$photo1 = $info['photo1'];
echo "
<tr>
<td> $stock
</td>
<td> $year
</td>
<td> $make
</td>
<td> $model
</td>
<td> $price
</td>
<td> $miles
</td>
<td> $photo1
</td>
</tr>";

?>
[/code]
Here's the new error:

Parse error: syntax error, unexpected $end in /home/inv/public_html/display3.php on line 58

I feel I am connecting to the database, and my server seems to require the ' before and after the database name. Thanks...

trq · December 4, 2006

You are missing the closing } to your while(). And no, you do NOT need quotes around the TABLE name.

tracy · December 4, 2006

I actually took it out. Now when I put it back in, I get the error which caused me to take it out...

Parse error: syntax error, unexpected $end in /home/inv/public_html/display3.php on line 58

Thanks for any suggestions...

tracy · December 4, 2006

Any thoughts? Thanks.

Psycho · December 5, 2006

You are missing two right curly braces ( } ). You need one to close the while loop and another to close the else condition.

tracy · December 5, 2006

Parse error: syntax error, unexpected '>' in /home/inv/public_html/display.php on line 16

This is what happens when the code looks like this with the right curly brackets...

<?php
include ("link.php");

$query= mysql_query("SELECT * FROM Inventory");

$num = mysql_num_rows($query);

if ($num == '0') {
echo "Nothing Exist.";
die();
}
else {

echo '<table border="0" cellspacing="2" cellpadding="2">'

<td>Stock#
</td>
<td>Year
</td>
<td>Make
</td>
<td>Model
</td>
<td>Price
</td>
<td>Miles
</td>
<td>Photo
</td>
</tr>";
while ($info = mysql_fetch_array($query)) {
$stock = $info['stock'];
$year = $info['year'];
$make = $info['make'];
$model = $info['model'];
$price = $info['price'];
$miles = $info['miles'];
$photo1 = $info['photo1'];
echo "
<tr>
<td> $stock
</td>
<td> $year
</td>
<td> $make
</td>
<td> $model
</td>
<td> $price
</td>
<td> $miles
</td>
<td> $photo1
</td>
</tr>"
}

echo "</table>";
}

?>

Psycho · December 6, 2006

Might be a problem in your include file.

tracy · December 6, 2006

the include file is working fine because I used it in some other php pages and they work fine. thanks though.

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Why won't this print my mysql table contents?

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