error in SQL syntax

triforce

With this code:"

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN">

<html>
<head>
<title>Beelden</title>
</head>
<body>
<br><br>
<?php

$conn = mysql_connect("localhost","root","crowcrow");
echo $conn;
$db = mysql_select_db('beelden',$conn);
echo "$db db<br>";
$sql = mysql_query("SELECT * FROM kunst") or die ("Query failed");
echo "$sql sql<BR>";

$result = mysql_query($sql) or die(mysql_error());
$resultaat = mysql_query($sql);

echo $resultaat;
echo $sql;

echo "<table border=1 width=100%>\n";
echo "<tr><td> naam </td><td> informatie</td></tr>\n";

/*while ($rij = mysql_fetch_row($resultaat) or die(mysql_error())

{
printf ("<tr><td>%s %s</td><td>%s</td></tr>\n",
$rij[1], $rij[2], $rij[3]);
}*/
echo "</table>\n";
mysql_close($conn)

//mysql_free_result($uitslag);

?>
</body>
</html>
"

I get this error:"
Resource id #11 db
Resource id #2 sql
You have an error in your SQL syntax. Check the manual that corresponds to your MySQL server version for the right syntax to use near 'Resource id #2' at line 1

"
De query works if I input it directly. I don't get what I did wrong.

laserlight

Might be better to test with:

$conn = mysql_connect("localhost","root","crowcrow")
	OR die("Could not connect");
$db = mysql_select_db('beelden',$conn)
	OR die("Could not select database");
$sql = mysql_query("SELECT * FROM kunst")
	OR die ("Query failed");

I must ask though, what are you trying to do with:
$result = mysql_query($sql) or die(mysql_error());
$resultaat = mysql_query($sql);

At this point $sql holds a resource result.
Using mysql_query on a resource does not make sense.

triforce

I used your code, and he " died" in non of the 3 cases, so connecting, selecting ans query-ing goes well.

I must ask though, what are you trying to do with:
$result = mysql_query($sql) or die(mysql_error());
$resultaat = mysql_query($sql);

At this point $sql holds a resource result.
Using mysql_query on a resource does not make sense.

I problobly forgot to delete "$resultaat = mysql_query($sql);" from some some old code. I deleted it.

laserlight

In that case do you still have any errors?

triforce

Still this error:"
You have an error in your SQL syntax. Check the manual that corresponds to your MySQL server version for the right syntax to use near 'Resource id #2' at line 1"

laserlight

Post your whole code.
Based on your error message, you are still trying to use mysql_query() on a resource.

triforce

Based on your error message, you are still trying to use mysql_query() on a resource.

You where right. Forgot te delete another line.:rolleyes:
Now I hope I can get information for the database on the screen. It will probleably take some post here to get that right too😃.

triforce

I've got another problem:
Code on one page:"

echo "<table border=1>\n";
while ($rij = mysql_fetch_object($resultaat))
{
echo "<tr> <td width=20%> <a href='http://localhost/beelden/kunstenaars.php?naam=$rij->naam'> $rij->naam</a> </td>\n";
}
echo "</table>\n";
"

This gives a nice table with 2 links. But when I click one of the links I get an error.

Code on second page:"
echo $naam;
$sql = "SELECT * FROM kunstenaar WHERE naam=".$naam;
"

Notice: Undefined variable: naam in c:\inetpub\wwwroot\beelden\kunstenaars.php on line 15

So the variable "naam" is empty, but I thought that when you add a ? and what variable needs to be fild with what, that variable is directly accessible in the linked page.

laserlight

I would use:

echo "<table border=1>\n";
while ($rij = mysql_fetch_object($resultaat))
{
echo '<tr> <td width=20%> <a href="http://localhost/beelden/kunstenaars.php?naam=' . $rij->naam . '"> ' . $rij->naam . '</a> </td>' . "\n";
}
echo "</table>\n";

and

if (!empty($_GET['naam'])) {
	$sql = "SELECT * FROM kunstenaar WHERE naam='" . $_GET['naam'] . '";
}

zzz

You may want to download a copy of ezSQL Database Class (1.24) at http://php.justinvincent.com.

Brilliantly written and cuts development time drastically. Well documented as well.

philipolson

The code above has a syntax error but the point is valid. The point is relying on the PHP directive register_globals has been out-of-style for years and instead you should use Superglobals. For GET data (the stuff after the ? in the URL, otherwise known as the QUERY_STRING), you may use the $GET superglobal. So, instead of attempting to use $naam you would use $GET['naam']. The manual has a lot of information on this, here's a good starting point:

http://www.php.net/variables.external

laserlight

The syntax error is easily corrected:

if (!empty($_GET['naam'])) {
	$sql = "SELECT * FROM kunstenaar WHERE naam='" . $_GET['naam'] . "'";
}

if (!empty($_GET['naam'])) {
    $sql = "SELECT * FROM kunstenaar WHERE naam='{$_GET['naam']}";
}

philipolson

This may seem like an obvious/simple typo but JIC the second example is missing a ' so we have:

if (!empty($_GET['naam'])) { 
    $sql = "SELECT * FROM kunstenaar WHERE naam='{$_GET['naam']}'"; 
}