Question

如果数据库不为空，我需要检查300多个数据库，并显示没有表格的数据库名称。

这是我的代码[编辑]：

<?php

    $db = ['db_1', 'db_2', 'db_3', 'db_4'];

    foreach ($db as $db_name) {
        $conn = mysqli_connect("host", "user", "pass", $db_name) or die("Error " . mysqli_error($conn));
    }

    if ($conn->connect_error) {
        die("Conect error: " . $conn->connect_error);
    }

    $sql = "SHOW TABLES FROM " . $db_name . "";

    $result = mysql_query($sql);

    if ($result == NULL) {
        foreach ($db as $db_name) {
            echo "Empty database:" . $db_name . "</br>";
        }
    } else {
        exit;
    }

    ?>

但代码返回所有数据库，

Answer 1

这应该适合你：

<?php

$db = ['db_1', 'db_2', 'db_3', 'db_4'];

$conn = mysqli_connect("host", "user", "pass") or die("Error connecting: " . mysqli_error($conn));

foreach ($db as $db_name) {

    $sql = "SHOW TABLES FROM " . $db_name;

    $result = mysqli_query($conn, $sql);

    if (!$result) {
        echo "Unable to get tables from:" . $db_name . "<br />";
    } else {
        if ( mysqli_num_rows($result) === 0 )
            echo "Empty database:" . $db_name . "</br>";
        }
    }

mysqli_close($conn);
?>

旁注：

如果您的PHP版本不支持数组的方括号，请尝试更改：

$db = ['db_1', 'db_2', 'db_3', 'db_4'];

到

$db = array('db_1', 'db_2', 'db_3', 'db_4');

使用PHP检查没有记录的数据库

1 个答案: