Question

我有以下MySQL的输出。

Array
(
[0] => stdClass Object
    (
        [id] => 19
        [date] => 2010-10-04 11:00:00
        [course] => Yoga
        [course_id] => 19
        [count(*)] => 2
        [capacity] => 20
    )

[1] => stdClass Object
    (
        [id] => 20
        [date] => 2010-10-04 13:00:00
        [course] => Spin
        [course_id] => 20
        [count(*)] => 1
        [capacity] => 24
    )
...
...

我可以获取日期、课程等，但我不知道如何使用PHP获取count（*）。

foreach ($bookingnum as $key => $list){
    echo "<tr valign= top >
";
    echo "<td align= center >".$list->date."</td>
";
            echo "<td align= center >".$list->course."</td>
";
          //  echo "<td align= center >".$list->count(*)."</td>
";
         // this is wrong. how to get count(*)?? 
 ....
 ...

Answer 1

将基本查询修改为

SELECT COUNT(*) AS `total` ...

那么您可以将其用作

$list->total

Answer 2

哈，奇怪的情况你在那里。。。好吧，由于count（*）显然不是一个有效的属性名称，您可以做3件事：

use $list->{"count(*)"}
change the mysql query to count(*) as count and then use $list->count
change request mode so you get an array as result instead of an stdClass Object

Answer 3

$total = count($bookingnum);
foreach ($bookingnum as $key => $list){
    echo "<tr valign= top >
";
    echo "<td align= center >".$list->date."</td>
";
    echo "<td align= center >".$list->course."</td>
";
    echo "<td align= center >".$total."</td>
";
}

哎呀！我的错，错误的回答，当我看到法布里克对他回答+1时，我抓住了要点

Answer 4

显然，您可以获得具有奇怪名称的属性，如$list->；｛count（*）｝。但理想情况下，您会希望更改查询。给计数一个别名，如＜code＞SELECT。。。计数（*）cnt，因为使用它会更容易。然后您可以获得$list->；cnt，而不是采用奇怪的语法。

Answer 5

Edit: If you are doing a pagination and want to figure out the total number of records available, you might want to adjust your query with SELECT SQL_CALC_FOUND_ROWS * FROM .. and after your main query is sent , send another query

SELECT FOUND_ROWS() as total_records

但是，请记住SQL_CALC_FOUND_ROWS具有性能影响。

Edit: Just realized i just got the question wrong, really silly answer.. taking the first part back :)

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