Question

因此，当在我的程序中使用if！isset语句时，它不会识别变量已被设置。我已经设置好了，如果变量$ q没有设置，那么从我的mysqli数据库中获取查询。此查询将从我的数据库中获取最新的数据条目。此数据等于名为$ row的变量。稍后调用$ row变量以检索随后显示在网页上的记录。问题是if语句再次运行（刷新/重新加载页面时），即使从上次运行.php文件时设置了$ q。这是代码：

$db_user = "root";
$db_pass = "";
$db_name = "music_reviews_database";
$db_host = "localhost";

$conn = mysqli_connect($db_host, $db_user, $db_pass, $db_name);
if (!isset($q)) {
    $q = mysqli_query($conn, 
            "SELECT album.Album, 
            artist.ArtistsName, 
            datereviewed.DateReviewed, 
            features.Features, 
            genre.Genre, 
            rating.Rating, 
            songname.SongName,
            link.link,
            songname.ID,
            comments.Comments
            FROM songname 
            INNER JOIN album ON album.ID = songname.ID 
            INNER JOIN artist ON artist.ID = songname.ID
            INNER JOIN datereviewed ON datereviewed.ID = songname.ID
            INNER JOIN features ON features.ID = songname.ID
            INNER JOIN genre ON genre.ID = songname.ID
            INNER JOIN rating ON rating.ID = songname.ID
            INNER JOIN link ON link.ID = songname.ID
            INNER JOIN comments ON comments.ID = songname.ID
            ORDER BY DateReviewed DESC LIMIT 1");
    $row = mysqli_fetch_array($q);
}

所以代码应该是这样的吗？

$q = mysqli_query($conn, 
        "SELECT album.Album, 
        artist.ArtistsName, 
        datereviewed.DateReviewed, 
        features.Features, 
        genre.Genre, 
        rating.Rating, 
        songname.SongName,
        link.link,
        songname.ID,
        comments.Comments
        FROM songname 
        INNER JOIN album ON album.ID = songname.ID 
        INNER JOIN artist ON artist.ID = songname.ID
        INNER JOIN datereviewed ON datereviewed.ID = songname.ID
        INNER JOIN features ON features.ID = songname.ID
        INNER JOIN genre ON genre.ID = songname.ID
        INNER JOIN rating ON rating.ID = songname.ID
        INNER JOIN link ON link.ID = songname.ID
        INNER JOIN comments ON comments.ID = songname.ID
        ORDER BY DateReviewed DESC LIMIT 1");
$row = mysqli_fetch_array($q);
file_put_contents(data.php, return var_export($row,true));
if (file_exists(data.php)) {
    $row = include(data.php);
}

Answer 1

您可以删除

if (!isset($q)) {

和结束括号。它们是不必要的。

如果您希望每次都避免运行查询以获得结果，请尝试缓存结果。

E.g。

从查询中获得结果后，将内容发送到文件系统中的某个文件（$ path）

file_put_contents($path, '<?php return '.var_export($mysql_row,true).';?>');

然后你可以把它读出来：

if (file_exists($path)) $mysql_row = include($path);

类似于：

if (file_exists(data.php)) {
    $row = include(data.php);
}else{
      $q = mysqli_query($conn, 
        "SELECT album.Album, 
        artist.ArtistsName, 
        datereviewed.DateReviewed, 
        features.Features, 
        genre.Genre, 
        rating.Rating, 
        songname.SongName,
        link.link,
        songname.ID,
        comments.Comments
        FROM songname 
        INNER JOIN album ON album.ID = songname.ID 
        INNER JOIN artist ON artist.ID = songname.ID
        INNER JOIN datereviewed ON datereviewed.ID = songname.ID
        INNER JOIN features ON features.ID = songname.ID
        INNER JOIN genre ON genre.ID = songname.ID
        INNER JOIN rating ON rating.ID = songname.ID
        INNER JOIN link ON link.ID = songname.ID
        INNER JOIN comments ON comments.ID = songname.ID
        ORDER BY DateReviewed DESC LIMIT 1");
        $row = mysqli_fetch_array($q);    
        file_put_contents(data.php, '<?php return '.var_export($row,true).';?>');
    }

Answer 2

问题是if语句再次运行（刷新/重新加载页面时），即使从上次运行.php文件时设置了$q。

每个PHP脚本处理单个HTTP请求。脚本完成后，其所有上下文（HTTP请求，HTTP响应，变量等）都将被销毁。新请求由脚本的新实例处理，该实例对其先前的调用一无所知。

如果您需要将PHP脚本执行的信息传递给下一个，您可以使用sessions。然而，解释会议如何运作是一个过于广泛的主题，无法回答。

Isset无法使用mysqli_query

2 个答案: