PHP Code Errors

[phpnewman]

Good Day.

I've been trying to sucessfully run a simple PHP script, but have
failed. It is what others call a simple code. How would I query
a mySql database and display the results back to the user? This is the code I'm using.

<?
	trim($searchterm);

if(!$searchtype || !searchterm)
{
	echo "You have not entered the required information!";
	exit;
}

$searchtype = addslashes($searchtype);
$searchterm = addslashes($searchterm);

@ $db = mysql_pconnect("localhost", "userID", "pwd");

if(!$db)
{
	echo "Error:  Could not connect to database.";
	exit;
}

mysql_select_db("books");
$query = "select * from books";
$result = mysql_query($query);
$num_results = mysql_num_rows($result)

echo "<p>Number of books found: ".$num_results."</p>";

for($i=0; $i <$num_results; $i++)
{
	$row = mysql_fetch_array($result);
	echo "<P><strong>".($i+1).". Title: ";
	echo htmlspecialchars( stripslashes($row["title"]));
	echo "</strong><br>Author: ";
	echo htmlspecialchars( stripslashes($row["author"]));
	echo "<br>ISBN: ";
	echo htmlspecialchars( stripslashes($row["isbn"]));
	echo "<br>Price: ";
	echo htmlspecialchars( stripslashes($row["price"]));
	echo "</P>";
}

?>

I usually get the error msg's:
Parse error: parse error, unexpected T_ECHO
or
T_FOR

Can someone guide me down the right path with the correct
way of doing this?

Thanks...

[edwardsbc]

Well, here's one error:

if(!$searchtype || !searchterm)

Shouldn't it be !$serchterm? (no $ in your code)

It is also a good idea to always use <?php to start a php block.

Finally, although not an error, you would be wise to use single quotes in echo statements unless you need "evaluation". Double quotes force php to evaluate the string (look for php variable to substitute). You can see a definite performance improvement for scripts with lots of echo'd text (think large while or for loops).

[Jedi_Legend]

$num_results = mysql_num_rows($result)

That should be:

$num_results = mysql_num_rows($result);

[phpnewman]

Thanks for the help. I've watered down my code to the following. I'm tring to start small and get something to work.
I get this error":

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

<?

@ $db = mysql_connect("localhost", "adamcz.513", "BlackRainStorm");

if(!$db)
{
	echo 'Error:  Could not connect to database.';
	exit;
}

$result = mysql_query('select * from books');
$num_results = mysql_num_rows($result);

for($i = 0; $i < num_results; $i++)
{
	$row = mysql_fetch_array($result);
	echo '<P><strong>".($i+1).". Title: ';
	echo htmlspecialchars( stripslashes($row["title"]));
	echo '</strong><br>Author: ';
	echo htmlspecialchars( stripslashes($row["author"]));
	echo '<br>ISBN: ';
	echo htmlspecialchars( stripslashes($row["isbn"]));
	echo '<br>Price: ';
	echo htmlspecialchars( stripslashes($row["price"]));
	echo '</P>';
}

?>

Thanks for the help. I appreciate it.
-SM

[gelwa]

<?

@ $db = mysql_connect("localhost", "adamcz.513", "BlackRainStorm");

if(!$db)
{
    echo 'Error:  Could not connect to database.';
    exit;
}

.....
try this
$db = @mysql_connect("localhost", "adamcz.513", "BlackRainStorm");
The @ is used to suppress the Error generated when parsing PHP
and is supposed to appear infront of the called function

[phpnewman]

Thanks for the suggestion. I tried it but recieved the same
error?

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

-Back to the drawing board...
Thanks.

[phpnewman]

I swear I tried this earlier! But it works fine now the error:

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

I was also getting other errors.

But the fix was stupid simple. When selecting the database with the function: mysql_select_db("db_name"), I was putting
the table.... I didn't realize this for a whole week!!! Dammmm...

A week to find that... To bad I couldn't bury the evidence!!!

Thanks,

Adam

[iceman42]

check this for($i = 0; $i < num_results; $i++) there should be a $ in num_results

$sql = "select * from books";
$result = mysql_query($sql);

try that ...

[Runnion]

You'll want to change your database password now that you've posted it....

[phpnewman]

Thanks... Taken care of...

One of those long days...
Not a good thing though...

[markdark]

You might consider if you want to use the function mysql_num_rows(); Because when your database contains about 18.000 records en you select everything in this database your query will return a value of a great number of bytes, but when you use the following it would be a few bits:

<?php

$qry = mysql_query("SELECT COUNT(1) AS total FROM books");
$result = mysql_fetch_object($qry);

echo 'Total of rows is:'.$result->total;

?>

[Traduim]

<?php

$qry = mysql_query("SELECT COUNT(1) AS total FROM books");
$result = mysql_fetch_object($qry);

echo 'Total of rows is:'.$result->total;

?>

Perhaps it would be better to use COUNT() instead of COUNT(1)?
I think COUNT(1) will count all the ocurrences of first field in table, and I'm not sure if COUNT(1) will be always the same as COUNT(). I may be just a bit confused.

[phpnewman]

Thanks for the reply. The error was simply from the database name, not being correct, I was usuing the table by mistake.

mysql_select_db("db_name")

I've tested this error and have replicated the same error in other instances.

Thanks,

-SM