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Warning: mysql_num_rows() expects parameter 1 to be resource, boolean given in

probir

By probir
in PHP Coding Help

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probir Newbie

probir

Posted
Posted

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>

Pikachu2000 Prolific Member

Pikachu2000

Posted
Posted

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 ) )";;

Psycho Prolific Member

Psycho

Posted
Posted

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.

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