PHP 搜索 MySQL 数据库错误

Question

这个问题在这里已经有了答案:





mysql_fetch_array()/mysql_fetch_assoc()/mysql_fetch_row()/mysql_num_rows etc... expects parameter 1 to be resource

(31 个回答)


7年前关闭。




我正在创建一个 PHP 搜索功能 -

$result=mysql_query($query);
$numItems = mysql_num_rows($result); //Number of items that resulted from the query
if ($numItems == 0) {
    $query = "select * FROM $table WHERE (name LIKE '%$termSafe%' OR tags LIKE '%$termSafe%' OR search_tags LIKE '%$termSafe%' OR ingredients_1 LIKE '%$termSafe%') AND active=1 $queryOptions ORDER BY name ASC";
    $result=mysql_query($query);
    $numItems = mysql_num_rows($result);
    if ($numItems == 0) {
        return('no results');
    } else {
        return $result;
    }
} else {
    return $result;
}

当我运行它时，我收到以下错误:

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

如果我拉出那条线 $numItems - 它运行正常吗？我错过了什么？

Answer 1

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

mysql_query仅当 false 时返回 bool 值.

您应该升级到 mysqli或 PDO .

至少使用 mysql_real_escape_string() 转义值.

你应该检查错误

$result = mysql_query($query);
if (!$result) {
    $message  = 'Invalid query: ' . mysql_error() . "\n";
    $message .= 'Whole query: ' . $query;
    die($message);
}