Mysqli_query返回意外输出 - mysqli_query returns unexpected output

mysqli_query returns unexpected output

本文关键字：意外输出返回 query Mysqli | 更新日期: 2023-09-27

我是一名初级程序员，我通过mysqli_query从数据库获取信息有点麻烦。我首先连接到数据库，然后尝试从数据库内部的表cbo中获取信息。然后打印出查询的结果，而不是表中的信息。而这是我得到的。

mysqli_result Object
(
    [current_field] => 0
    [field_count] => 8
    [lengths] => 
    [num_rows] => 12
    [type] => 0
)

这是我使用的代码。Dump只回显变量。

<?php
    $con = mysqli_connect("localhost", "root", "harvard", "cbo projections");
    if ( mysqli_connect_errno() ) {
        echo "Failed to connect to MySQL: " . mysqli_connect_error();
    }
    $result = mysqli_query($con,"SELECT * FROM cbo");
    dump( $result );
?>

$result只是一个包含结果集的对象。你必须从中获取数据。读取mysqli_fetch_assoc或mysqli_fetch_array

例子:

if ($result = mysqli_query($link, $query)) {
    while ($row = mysqli_fetch_assoc($result)) {
        //Display fields here according to your table structure
    }
    mysqli_free_result($result);
}

你可以这样写

    while ($row = mysqli_fetch_assoc($result)) {
        $records[]=$row;
    }

这将创建一个名为records的数组，它将包含您获取的所有行，然后您可以稍后访问该数组并相应地处理

这是mysqli对象，你想用它做什么?你应该阅读https://www.php.net/manual/mysqli-result.fetch-assoc.php, https://www.php.net/manual/mysqli-result.fetch-object.php或https://www.php.net/manual/mysqli-result.fetch-array.php。

由例子:

<?php
$con=mysqli_connect("localhost", "root", "harvard", "cbo projections");
if (mysqli_connect_errno())
{
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$result = mysqli_query($con,"SELECT * FROM cbo");
while ($row = mysql_fetch_assoc($result)) {
    var_dump($row);
}
?>

$result = mysqli_query($con,"SELECT * FROM cbo");
$rows = array();
while ($row = mysqli_fetch_assoc($result)) {
  $rows[] = $row;
}
print_r($rows);