我正在尝试从MySQL表中选择数据,但收到以下错误消息之一:
mysql_fetch_array()期望参数1为资源,给定布尔值
这是我的代码:
$username = $_POST['username']; $password = $_POST['password']; $result = mysql_query('SELECT * FROM Users WHERE UserName LIKE $username'); while($row = mysql_fetch_array($result)) { echo $row['FirstName']; }
查询可能由于各种原因而失败,在这种情况下,mysql_ *和mysqli扩展名都false将从它们各自的查询函数/方法中返回。您需要测试该错误情况并进行相应处理。
false
mysql_ *扩展名 :
注: 该mysql_功能已被弃用,并在PHP版本7已被删除。
请先检查一下,$result然后再传递给mysql_fetch_array。您会发现这是false因为查询失败。请参阅mysql_query文档以获取可能的返回值以及有关如何处理它们的建议。
$result
mysql_fetch_array
mysql_query
$username = mysql_real_escape_string($_POST['username']); $password = $_POST['password']; $result = mysql_query("SELECT * FROM Users WHERE UserName LIKE '$username'"); if($result === FALSE) { die(mysql_error()); // TODO: better error handling } while($row = mysql_fetch_array($result)) { echo $row['FirstName']; }
mysqli扩展 程序样式 :
$username = mysqli_real_escape_string($mysqli, $_POST['username']); $result = mysqli_query($mysqli, "SELECT * FROM Users WHERE UserName LIKE '$username'"); // mysqli_query returns false if something went wrong with the query if($result === FALSE) { yourErrorHandler(mysqli_error($mysqli)); } else { // as of php 5.4 mysqli_result implements Traversable, so you can use it with foreach foreach( $result as $row ) { ...
oo风格 :
$username = $mysqli->escape_string($_POST['username']); $result = $mysqli->query("SELECT * FROM Users WHERE UserName LIKE '$username'"); if($result === FALSE) { yourErrorHandler($mysqli->error); // or $mysqli->error_list } else { // as of php 5.4 mysqli_result implements Traversable, so you can use it with foreach foreach( $result as $row ) { ...
使用准备好的语句:
$stmt = $mysqli->prepare('SELECT * FROM Users WHERE UserName LIKE ?'); if ( !$stmt ) { yourErrorHandler($mysqli->error); // or $mysqli->error_list } else if ( !$stmt->bind_param('s', $_POST['username']) ) { yourErrorHandler($stmt->error); // or $stmt->error_list } else if ( !$stmt->execute() ) { yourErrorHandler($stmt->error); // or $stmt->error_list } else { $result = $stmt->get_result(); // as of php 5.4 mysqli_result implements Traversable, so you can use it with foreach foreach( $result as $row ) { ...
这些示例仅说明 了 应该执行的操作(错误处理),而没有说明如何执行。or die输出HTML时不应使用生产代码,否则它将(至少)生成无效的HTML。另外,数据库错误消息不应显示给非管理员用户,因为它会显示过多信息。
or die