Here's the example from the PHP manual:
The tutorial here are very helpfull:
http://www.melonfire.com/community/columns/trog/
-- David
<?php
// Connecting, selecting database
$link = mysql_connect("mysql_host", "mysql_login", "mysql_password")
or die("Could not connect");
print "Connected successfully";
mysql_select_db("my_database")
or die("Could not select database");
// Performing SQL query
$query = "SELECT * FROM my_table";
$result = mysql_query($query)
or die("Query failed");
// Printing results in HTML
print "<table>\n";
while ($line = mysql_fetch_array($result, MYSQL_ASSOC)) {
print "\t<tr>\n";
foreach ($line as $col_value) {
print "\t\t<td>$col_value</td>\n";
}
print "\t</tr>\n";
}
print "</table>\n";
// Closing connection
mysql_close($link);
?>
> Hello,
>
> I am extremely new to MySQL and have never managed to get working
> smoothly with PHP before. I am trying really hard to understand
> how to work it, and am almost there.
>
> I have a problem which I do not know how to resolve and was
> wondering if anybody could help me. I have no idea what is wrong
> with the code and why I am getting the error message;
>
> Warning: Supplied argument is not a valid MySQL result resource in
> C:\apache\htdocs\sams\chapter10\results.php on line 47
>
> I am currently using a book to aid me with MySQL, and this is an
> example from the book. It does not seem to work and I have no idea what
> I may have done wrong to obtain this warning.
>
> I have changed my login and password to question marks.
>
> <?
>
> if (!$searchtype || !$searchterm)
>
> {
> echo "You have not entered search details. Please go back and try
> again.";
>
> exit;
>
> }
>
>
> $searchtype = addslashes($searchtype);
>
> $searchterm = addslashes($searchterm);
>
> @ $db = mysql_pconnect("mesh", "bookorama", "bookorama123");
>
> if (!$db)
>
> {
> echo "Error: Could not connect to database. Please try again
> later.";
>
> exit;
>
> }
>
> mysql_select_db("booktest");
>
> $query = "select * from booktest where ".$searchtype." like
> '%".$searchterm."%'";
>
> $result = mysql_query($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>".($i+1).". Title: ";
>
> echo stripslashes($row["title"]);
>
> echo "</strong><br>Author: ";
>
> echo stripslashes($row["author"]);
>
> echo "<br>ISBN: ";
>
> echo stripslashes($row["isbn"]);
>
> echo "<br>Price: ";
>
> echo stripslashes($row["price"]);
>
> echo "</p>";
>
> }
>
> ?>
>
> The problem seems to be around the lines of code;
>
> $result = mysql_query($query);
>
> $num_results = mysql_num_rows($result);
>
> Any assistance is appreciated.
>
> Yours,
>
> GF.
>
> _________________________________________________________________
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>
>
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