What's Wrong???

WebmasterJoe

I'm trying to recieve information from a database, but keep getting this error:

Parse error: parse error, unexpected T_ENCAPSED_AND_WHITESPACE, expecting T_STRING or T_VARIABLE or T_NUM_STRING

The code that I'm having trouble with is here:

$result = mysql_query("SELECT * FROM 'table' WHERE cat='$cat'", $db);
while($row = mysql_fetch_row($result)){
echo("$row['name']");
}
?>

Thanks for the help,
WebmasterJoe

ultraslacker

 echo $row['name'];

WebmasterJoe

Thanks for that.
Now I'm getting this error:

Warning: mysql_fetch_row(): supplied argument is not a valid MySQL result resource

nozom

try this

$result = mysql_query("SELECT * FROM 'table' WHERE cat='$cat'", $db);

while ( $row = mysql_fetch_array($result))
{
echo($row["name"]);
}

WebmasterJoe

That just gives me this error:
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource

hmmmmmm

stolzyboy

generally when you get that error, you aren't getting anything with the query, try echoing out $result and see what that shows

WebmasterJoe

How would I fix this? When I tryed echoing $result I got nothing.

nozom

ok.

You have a MySQL table called NAMES with the fields ID, Name, Email

soo you need this in your header:
<?php
// Request the database
$result = mysql_query(
"SELECT * FROM NAMES");
if (!$result) {
echo("Error performing query: " .
mysql_error() . "");
exit();
}
?>

in your HTML body

<?php
while ( $row = mysql_fetch_array($result) )
{
echo($row["ID"]);
echo(" ");
echo($row["Name"]);
echo(" ");
echo($row["Email"]);
echo(" ");

}

make sure you have your table set up correctly or you could be getting errors that way.

cheers,
Hugh
?>

stolzyboy

if you aren't getting anything from your $result, then your select statement isn't returning any results, so check your result statement

WebmasterJoe

Ok, I'm lost. Here's the whole script I'm using. Maybe you can figure out what's wrong with it.

<?php
$db = mysql_connect ("localhost", "name", "pass");

mysql_select_db("dbname",$db);
$result = mysql_query("SELECT * FROM table WHERE cat='$cat'", $db);

while($row = mysql_fetch_array($result))
{ 
echo $row['name'];
}

?>

stolzyboy

yes we've got your script before, what i am saying is that when you run the line $result = mysql_query(), the select isn't selecting anything, try doing a simple select * from tablename and see if that gives you anything, maybe there is something wrong in your where clause

siwis

Basic of basic debugging:
Generate your sql outside of the mysql_query() and echo it to screen.
Use mysql_num_rows() to discover how many rows have been returned.
ONLY if you get 1 or more rows would a call to mysql_fetch_array() be valid and you ideally ought therefore to use the number generated by mysql_num_rows() to decide whether to actually use the array.

<?php
$sql = "SELECT * FROM table WHERE cat='$cat'";
echo $sql . " rows returned<br>";
$result = mysql_query("$sql", $db);
$num_results = mysql_num_rows($result);
echo $num_results . " rows returned<br>";
if ($num_results != "0") {
	while($row = mysql_fetch_array($result)) { 
		echo $row['name'];
	};
};	// End of if ($num_results != "0")
?>

Initially echoing the sql should confirm that your sql actually has a string or value which which to query the database. If that string/value never makes it to the sql, you'll soon see because it won't be in the echo'd sql.
Next, by checking how many results are returned before trying to loop through non existent results, your script don't break. It is perfectly feasable that the WHERE within your sql does not match any records.

Table and field names are cASe SensiTIve

WebmasterJoe

Thanks! The script is working now. I'm not exactly sure what the problem was, but I've made a few changes based on your comments and it's working well now. 🙂

Thanks again,
Joe