php - mysql_fetch_array()/mysql_fetch_assoc()/mysql_fetch_row()/mysql_num_rows等…期望参数1为资源或结果

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


要么


  mysqli_fetch_array（）期望参数1为mysqli_result，给定布尔值


要么


  在布尔值/非对象上调用成员函数fetch_array（）


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'];
}

同样适用于类似

$result = mysqli_query($mysqli, 'SELECT ...');
// mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given
while( $row=mysqli_fetch_array($result) ) {
    ...

和

$result = $mysqli->query($mysqli, 'SELECT ...');
// Call to a member function fetch_assoc() on a non-object
while( $row=$result->fetch_assoc($result) ) {
    ...

和

$result = $pdo->query('SELECT ...', PDO::FETCH_ASSOC);
// Invalid argument supplied for foreach()
foreach( $result as $row ) {
    ...

和

$stmt = $mysqli->prepare('SELECT ...');
// Call to a member function bind_param() on a non-object
$stmt->bind_param(...);

和

$stmt = $pdo->prepare('SELECT ...');
// Call to a member function bindParam() on a non-object
$stmt->bindParam(...);

最佳答案

查询可能由于各种原因而失败，在这种情况下，mysql_ *和mysqli扩展名都将从其各自的查询函数/方法中返回false。您需要测试该错误情况并进行相应处理。

mysql_* extension：


  注意mysql_ functions are deprecated和已在php版本7中删除。


将$result传递给mysql_fetch_array之前，请先进行检查。您会发现它是false，因为查询失败。有关可能的返回值和如何处理它们的建议，请参见mysql_query文档。

$username = mysql_real_escape_string($_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 extension
程序风格：

$username = mysqli_real_escape_string($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。生产代码在输出HTML时不应使用or die，否则它将（至少）生成无效的HTML。另外，数据库错误消息也不应显示给非管理员用户，如discloses too much information所示。

关于php - mysql_fetch_array()/mysql_fetch_assoc()/mysql_fetch_row()/mysql_num_rows等…期望参数1为资源或结果，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/40076656/