Question

请原谅我的无知，但我很难搞清楚这一点。

我正在尝试从一个mysql命令获取结果，并在另一个命令中使用它。

这是我的代码，它不起作用。

//select the event end date of event ID
$sql = "SELECT enddate FROM mm_eventlist_dates WHERE id = $id";
$result = mysql_query($sql);

//plug in the event end date, find event that starts the next day
$sql = "SELECT id FROM mm_eventlist_dates WHERE startdate = date_add($result, INTERVAL 1 DAY)";
$result = mysql_query($sql);
$row = mysql_fetch_array($result);
echo "Next Event ID" . $row['id'];

我迷路了。

请帮忙！

谢谢，尼克

Answer 1

如果我理解您要完成的任务，看起来您想要查找在给定事件后一天开始的所有事件。正确？在这种情况下，您要做的是自联接，即将表连接到自身。您需要为表至少出现一个别名，以便SQL可以区分它们。

所以也许是这样的：

SELECT e2.id
FROM mm_eventlist_dates e1
join mm_eventlist_dates e2 on e2.startdate = date_add(e1.enddate, INTERVAL 1 DAY)
where e1.id=$id

Answer 2

是否有理由不能将它们合并到一个查询中？

SELECT m1.id FROM mm_eventlist_dates m1  
JOIN mm_eventlist_dates m2 ON m1.startdate = date_add(m2.enddate, INTERVAL 1 DAY)  
WHERE m2.id = $id

Answer 3

mysql_query（）返回结果集，而不是实际的数据库项。要做你想做的事情，做一些类似的事情（不包括错误检查等）：

//select the event end date of event ID
$sql = "SELECT enddate FROM mm_eventlist_dates WHERE id = $id";
$result = mysql_query($sql);

$enddateRow = mysql_fetch_array($result);

//plug in the event end date, find event that starts the next day
$sql = "SELECT id FROM mm_eventlist_dates WHERE startdate = date_add('" . $enddateRow["enddate"] . "', INTERVAL 1 DAY)";
$result = mysql_query($sql);
$row = mysql_fetch_array($result);
echo "Next Event ID" . $row['id'];

Answer 4

您不能直接在date_add中使用$ result。调用mysql_fetch_array（稍后你会做几行），并使用$ row ['enddate']。

Answer 5

//select the event end date of event ID
$sql = "SELECT enddate FROM mm_eventlist_dates WHERE id = $id";
$result = mysql_query($sql);
$row = mysql_fetch_array($result);
$enddate = $row['enddate'];

//plug in the event end date, find event that starts the next day
$sql = "SELECT id FROM mm_eventlist_dates WHERE startdate = date_add($enddate, INTERVAL 1 DAY)";
$result = mysql_query($sql);
$row = mysql_fetch_array($result);
echo "Next Event ID" . $row['id'];

我认为

Answer 6

您无法在另一个查询中直接使用mysql_query的结果，您需要先获取该值。

而不是

$result = mysql_query($sql);

//plug in the event end date, find event that starts the next day
$sql = "SELECT id FROM mm_eventlist_dates WHERE startdate = date_add($result, INTERVAL 1 DAY)";

尝试

   $result = mysql_query($sql);
   $enddate = mysql_fetch_assoc($result);

    //plug in the event end date, find event that starts the next day
    $sql = "SELECT id FROM mm_eventlist_dates WHERE startdate = date_add($enddate, INTERVAL 1 DAY)";

Answer 7

试试这个

//select the event end date of event ID
$sql = "SELECT enddate FROM mm_eventlist_dates WHERE id = $id";
$result = mysql_query($sql);

$row = mysql_fetch_assoc($result)
//plug in the event end date, find event that starts the next day
$sql = "SELECT id FROM mm_eventlist_dates WHERE startdate = date_add(".$row['enddate'].", INTERVAL 1 DAY)";
$result = mysql_query($sql);
$row = mysql_fetch_array($result);
echo "Next Event ID" . $row['id'];

在Another Mysql Command中使用一个Mysql命令的结果？

7 个答案: