Question

以下是相关代码：

来自index.php：

require_once('includes/DbConnector.php');

// Create an object (instance) of the DbConnector
$connector = new DbConnector();

// Execute the query to retrieve articles
$query1 = "SELECT id, title FROM articles ORDER BY id DESC LIMIT 0,5";
$result = $connector->query($query1);

echo "vardump1:";
var_dump($result);
echo "\n";

/*(!line 17!)*/ echo "Number of rows in the result of the query:".mysql_num_rows($result)."\n";
// Get an array containing the results.
// Loop for each item in that array


while ($row = $connector->fetchArray($result)){

echo '<p> <a href="viewArticle.php?id='.$row['id'].'">';
echo $row['title'];
echo '</a> </p>';

来自dbconnector.php：

$settings = SystemComponent::getSettings();

// Get the main settings from the array we just loaded
$host = $settings['dbhost'];
$db = $settings['dbname'];
$user = $settings['dbusername'];
$pass = $settings['dbpassword'];

// Connect to the database
$this->link = mysql_connect($host, $user, $pass);
mysql_select_db($db);
register_shutdown_function(array(&$this, 'close'));

} //end constructor

//*** Function: query, Purpose: Execute a database query ***
function query($query) {

echo "Query Statement: ".$query."\n";

$this->theQuery = $query;

return mysql_query($query, $this->link) or die(mysql_error());

}

//*** Function: fetchArray, Purpose: Get array of query results ***
function fetchArray($result) {

echo "<|";
var_dump($result);
echo "|> \n";

/*(!line 50!)*/$res= mysql_fetch_array($result) or die(mysql_error());

echo $res['id']."-".$res['title']."-".$res['imagelink']."-".$res['text'];

return $res;
}

输出：

Query Statement: SELECT id, title FROM articles ORDER BY id DESC LIMIT 0,5 vardump1:bool(true) 
PHP Error Message

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /*path to*/index.php on line 17

Number of rows in the result of the query: <|bool(true) |>

PHP错误消息

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /*path to*/DbConnector.php on line 50

Answer 1

你的问题已经出现了问题：

return mysql_query($query, $this->link) or die(mysql_error())

它应该是这样写的：

$result = mysql_query($query, $this->link);
if(!$result) die(mysql_error());
return $result;

动态语言中常见的是or首先返回object，其值为true，但在PHP中，X or Y的结果始终为 true 或<强>假

Answer 2

正如您可以从手册中学到的，mysql_query返回值类型不是布尔值而是资源你的代码转换它的东西

Answer 3

同意上述内容，您的mysql_query存在问题。它永远不应该返回True。无论是虚假还是资源。

重构：mysql_query（$ query，$ this-＆gt; link）或die（mysql_error（））

在查询后回显mysqlerror（）并查看错误消息是什么。您可能没有有效的连接。

mysql_query（）返回true，但mysql_num_rows（）和mysql_fetch_array（）给出“没有有效的资源错误

3 个答案: