PHP查询数据库数据时出现错误

clearskyo 2012-03-08 04:45:42

PHP 查询数据库books中的books表时出现以下错误！

Notice: Trying to get property of non-object in D:\wamp\Apache\htdocs\php\results.php on line 33

Fatal error: Call to a member function fetch_assoc() on a non-object in D:\wamp\Apache\htdocs\php\results.php on line 35

代码如下：



<html>

	<head>

		<title> Book-O-Rama Catalog Search</title>

	</head>

	<body>

		<h1>Book-O-Rama Catalog Search</h1>

<?php

	//create short variable names

	$searchtype=$_POST['searchtype'];

	$searchterm=trim($_POST['searchterm']);	

	

	if (!$searchtype || !$searchterm)	{

		exit('You have not entered search detais,please go back again');

	}

	 

	if (!get_magic_quotes_gpc()){	

		

		$searchtype=addslashes($searchtype);

		$searchterm=addslashes($searchterm);

	}

	

	

	@ $db=new mysqli('localhost','root','password','books');

	

	if (mysqli_connect_errno()){

		echo "could not connect to database,please try again later.".mysqli_connect_errno();

		exit;

	}else {  

		$sql ="select * from books where".$searchtype."like'%".$searchterm."%'";

	

		$result=$db->query($sql);

		$num_result=$result->num_rows;

		for ($i=0;$i<=$num_result;$i++){

			$row=$result->fetch_assoc();

			echo "<p>".($i+1)."title:";

			echo htmlspecialchars(stripslashes($row['author']));

			echo "Author:".stripslashes($row->newsauthor);

			echo "ISBN:".stripslashes($row->newsISBN);

			echo "price:".stripslashes($row->newsprice)."</p>";

		}

	}



	$result->free();

	$db->colse();

?>

	</body>

</html>

上述问题那微微大虾能给个解决方案！

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coder 2012-03-08

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$sql ="select * from books  where  ".$searchtype."  like  '%".$searchterm."%'";



        if ($result=$db->query($sql)) {

			$num_result=$result->num_rows;

			for ($i=0;$i<=$num_result;$i++){

				$row=$result->fetch_assoc();

				echo "<p>".($i+1)."title:";

				echo htmlspecialchars(stripslashes($row['author']));

				echo "Author:".stripslashes($row->newsauthor);

				echo "ISBN:".stripslashes($row->newsISBN);

				echo "price:".stripslashes($row->newsprice)."</p>";

			}

		}else{

			echo "fail.";

			exit;

		}

$sql =" select * from books where ".$searchtype." like '%".$searchterm."%'";

字符串连在一块了。所以，语句不对。
if ($result=$db->query($sql)) {最好这样。
因为$db->query($sql)会返回一个false，导致没有num_rows这个属性。