probir Posted November 30, 2010 Share Posted November 30, 2010 i am try to make a name, address search system into my website from my database. but i got this msg [Warning: mysql_num_rows() expects parameter 1 to be resource, boolean given in C:\xampplite\htdocs\3\searchresult.php on line 54] my full php code i.e. searchresult.php is under...... what i have mistake..... searchresult.php <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" /> <title>Untitled Document</title> </head> <body> <?php // TAKE THE INFORMATION FROM FORM. $search = $_GET['search']; // IF THERE IS NOT A KEYWORD GIVE A MISTAKE. if (!$search) echo "You didn't enter a keyword"; else { echo "<td>You searched for: <strong>$search </strong></td>"; mysql_connect('localhost','root',''); mysql_select_db('search'); $id=@$_GET['id']; //QUERY IS THE CODE WHICH WILL MAKE THE SEARCH IN YOUR DATABASE. //I WROTE AN EXPLANATION ABOUT IT AFTER THE CODE. $sql = "CREATE TABLE searchform \n" ."(\n" ."ID int NOT NULL AUTO_INCREMENT ,\n" ."FirstName varchar( 255 ) NOT NULL ,\n" ."LastName varchar( 255 ) NOT NULL ,\n" ."Email varchar( 255 ) NOT NULL ,\n" ."PhoneNumber varchar( 255 ) NOT NULL ,\n" ."PRIMARY KEY ( ID ) )";; $result1 = MySQL_query($query); if(!$result1) { echo MySQL_error()."<br>$query<br>"; } if(MySQL_num_rows($result1) > 5) { echo "<table width='750' align='center' border='0' cellspacing='0' cellpadding='0'>"; while($result2 = MySQL_fetch_array($result1)) { //A short description from category. $description = $result2['category']; $searchPosition = strpos($description, $search); $shortDescription = substr($description, $searchPosition, 150); // I added a link to results which will send the user to your display page. echo '<tr><td><p><strong><a href="displayresults.php?id='.$result2['id'].'">'.$result2['title'].'</strong></p></td></tr>'; echo "<tr><td>$shortDescription ...</td></tr>"; echo "<td>{$result2['name']} {$result2['surname']}</td><tr/>"; } echo "</table>"; }else { echo "No Results were found in this category.<br>"; }echo "<br>"; } ?> </body> </html> and searchform.php <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" /> <title>Untitled Document</title> </head> <body> <form action="searchresult.php" method="get"> <div align="center"> <p> <input name="search" type="text" size="60"/> <input name="submit" type="submit" value="Search" /> </p> </div> </form> </body> </html> Quote Link to comment Share on other sites More sharing options...
Pikachu2000 Posted November 30, 2010 Share Posted November 30, 2010 This database query does not search for anything. It creates a new database table, as "CREATE TABLE" would imply . . . $sql = "CREATE TABLE searchform \n" ."(\n" ."ID int NOT NULL AUTO_INCREMENT ,\n" ."FirstName varchar( 255 ) NOT NULL ,\n" ."LastName varchar( 255 ) NOT NULL ,\n" ."Email varchar( 255 ) NOT NULL ,\n" ."PhoneNumber varchar( 255 ) NOT NULL ,\n" ."PRIMARY KEY ( ID ) )";; Quote Link to comment Share on other sites More sharing options...
Psycho Posted November 30, 2010 Share Posted November 30, 2010 I'm surprised your query isn't failing! You are trying to run the query in the variable $query - but it hasn't been defined. Instead you defined the query in the variable $sql. But, I am confused as to why you would be running mysql_num_rows() against the result since the query is to CREATE a table, so I don't see how there would be any rows returned. Quote Link to comment Share on other sites More sharing options...
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.