Question

在我网站的其中一个页面上，我使用了4个表（登录，统计，民意调查和宣布）的内部联接，如下所示：

$login_user = mysqli_real_escape_string($cxn, $_SESSION['user']);

$sqlAll = mysqli_fetch_assoc(mysqli_query($cxn, 
    "SELECT login.picture, login.statement,
    stats.wealth,
    MAX(poll.polltime) AS pollunixtime,
    MAX(announce.announcetime) AS announceunixtime 
    FROM login 
    INNER JOIN stats ON login.user = stats.user 
    INNER JOIN poll ON login.user = poll.user 
    INNER JOIN announce ON login.user = announce.source 
    WHERE (login.user = '$login_user' AND announce.announcetype = 'feedback')"));

只要表格宣布中的至少一行，并且为该用户填充表格民意调查中的一行，此内部联接就可以正常工作，例如，此表单输入的值定义为：

<form id="myForm"><input type="hidden" name="wealth" value="<?= $sqlAll['wealth'] ?>"></form>//"View Page Source" shows this value is defined and shows up as value="number"

但是，我注意到当用户的任何一个表都没有行时，查询失败，因为查询中的某些值未定义。例如，如果没有任何表的行，则此表单输入的值未定义：

<form id="myForm"><input type="hidden" name="wealth" value="<?= $sqlAll['wealth'] ?>"></form>//"View Page Source" shows this value is undefined and shows up as value=""

我尝试以各种方式将CASE引入我的查询，例如：

$sqlAll = mysqli_fetch_assoc(mysqli_query($cxn, 
    "SELECT login.picture, login.statement,
     stats.wealth,
     MAX(poll.polltime) AS pollunixtime,
     MAX(announce.announcetime) AS announceunixtime
     FROM login
     INNER JOIN stats ON login.user = stats.user
     INNER JOIN poll ON CASE WHEN poll.polltime IS NOT NULL THEN login.user = poll.user END 
     INNER JOIN announce ON CASE WHEN announce.announcetime IS NOT NULL THEN login.user = announce.source END 
     WHERE (login.user = '$login_user' AND announce.announcetype = 'feedback')"));

由于polltime和announcetime属于INT类型，我也尝试使用＆＃34;＆gt;＆＃34;如下：

$sqlAll = mysqli_fetch_assoc(mysqli_query($cxn, 
    "SELECT login.picture, login.statement,
    stats.wealth,
    MAX(poll.polltime) AS pollunixtime,
    MAX(announce.announcetime) AS announceunixtime 
    FROM login
    INNER JOIN stats ON login.user = stats.user 
    INNER JOIN poll ON CASE WHEN poll.polltime > 0 THEN login.user = poll.user END    
    INNER JOIN announce ON CASE WHEN announce.announcetime > 0 THEN login.user = announce.source END 
    WHERE (login.user = '$login_user' AND announce.announcetype = 'feedback')"));

我还尝试过使用民意调查和宣布放置CASE声明的变体，如下所示：

$sqlAll = mysqli_fetch_assoc(mysqli_query($cxn, 
    "SELECT login.picture, login.statement, 
     stats.wealth, 
     CASE WHEN MAX(poll.polltime) IS NOT NULL THEN 'pollunixtime' END, 
     CASE WHEN MAX(announce.announcetime) IS NOT NULL THEN 'announceunixtime' END 
     FROM login 
     INNER JOIN stats ON login.user = stats.user 
     INNER JOIN poll ON login.user = poll.user 
     INNER JOIN announce ON login.user = announce.source 
     WHERE (login.user = '$login_user' AND announce.announcetype = 'feedback')"));

我尝试的变化没有奏效。基本上，我需要一个CONDITIONAL INNER JOIN，以便在给定用户存在polltime和announcetime时执行第二和第三内部联接，否则不执行第二和第三内部联接。这可能吗？

请注意我已经有一个黑客解决方案，在这个系统中涉及将上面的一个查询分解为3个单独的查询，并使用ISSET测试给定用户的polltime和announcetime是否存在。我想比这个解决方案做得更好！

Answer 1

使用LEFT JOIN代替INNER JOIN，您可以采取与您所要求的相似的方式。

4个表的三个INNER JOIN，查询中的某些值可能为空，也可能不为空

1 个答案: