PHP-mysql

[singboy]

<?php
trim($searchterm);
if (!$searchtype || !$searchterm)
{
echo "You have not entered search details. Please go back and try
again.";
exit;
}

$searchtype = addslashes($searchtype);
$searchterm = addslashes($searchterm);
$db = mysql_connect("localhost", "mmu", "mmu4midc");

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

mysql_select_db("books");
$query = "select * from books 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>".($i+1).". Title: \t</strong>".$rows[o]."\t"."<strong>Author: \t</strong>".$rows[1]."\t"."<strong>ISBN:\t</strong> ".$rows[2]."<strong>Price: \t</strong>".$rows[3]."</p>";
}

?>

• problem is:mysql_num_rows(): supplied argument is not a valid MySQL result resource
  Cananyone help me, pls...😕

[peppeto]

Try this hoop wil help

$query ="Select * from" ; //your query

$result=mysql_query($query);

if (!mysql_num_rows($result)==0){
while($row = mysql_fetch_array($result)){
$cl_id = $row["book_id"];
$name = $row["name"];
$text= $row["text"];

echo("<tr valign=middle>");

echo("<td>".$name."</td>");
echo("<td>". $text."</td></tr>");
}
}
else
{

echo "No results.";
}

mysql_free_result($result);

[singboy]

//this is search.html
...
<form action="results.php" method="post">
Choose Search Type:<br>
<select name="searchtype">
<option value="author">Author
<option value="title">Title
<option value="isbn">ISBN
</select><br>
Enter Search Term:<br>
<input name="searchterm" type=text>
<br>
<input type=submit value="Search">
...

is the searchterm & searchtype capable to transfer to the following result.php

//result.php
<?
trim($searchterm);
if (!$searchtype || !$searchterm)
{
echo "You have not entered search";
exit;
}

$searchtype = addslashes($searchtype);
$searchterm = addslashes($searchterm);
$db = mysql_connect("localhost", "mmu", "mmu4midc");

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

mysql_select_db("books");
$query = "select * from books where $searchtype Like
'%".$searchterm."%'";
$result = mysql_query($query);

if (!mysql_num_rows($result)==0){
while($row = mysql_fetch_array($result)){
....
else
{
echo "No result";
}
mysql_free_result($result);
?>

• some problem occur when i try to run it.
  mysql_num_rows(): supplied argument is not a valid MySQL result resource, no results

Where is the erroe, can somebody plz tell me, i have try for half day. but still same error occur.

[BeastRider]

Your query is in error

$query = "select * from books where $searchtype Like
'%".$searchterm."%'";

The $searchtype variable should be a field name from the table not a php variable. No query = no result.