Question

我正在尝试将变量存储到我的数据库中。我已经在我的数据库表中创建了一个应该存储变量的字段。我创建的变量是$ total，它存储计算，如下面的代码所示：

<?php                           
$user_query=mysql_query("select * from book where status != 'Archive'")or die(mysql_error());
while($row=mysql_fetch_array($user_query)){
$id=$row['book_id'];  
$cat_id=$row['category_id'];
$book_copies = $row['book_copies'];
$borrow_details = mysql_query("select * from borrowdetails where book_id = '$id' and borrow_status = 'pending'");
$row11 = mysql_fetch_array($borrow_details);
$count = mysql_num_rows($borrow_details);

$total =  $book_copies  -  $count; 
/* $total =  $book_copies  - $borrow_details;

echo $total; */
?>

我试过使用下面的代码存储它

<?php mysql_query("INSERT INTO book (borrow_total) VALUES('$total')");?>

Answer 1

看起来您正在寻找书表的更新。在这种情况下，这样的事情是正常的： -

<?php                           
$user_query =   mysql_query("select * from book where status !  =   'Archive'")or die(mysql_error());
while($row  =   mysql_fetch_array($user_query))
{
    $id     =   $row['book_id'];  
    $cat_id     =   $row['category_id'];
    $book_copies    =   $row['book_copies'];
    $borrow_details =   mysql_query("select * from borrowdetails where book_id  =   '$id' and borrow_status =   'pending'");
    $row11  =   mysql_fetch_array($borrow_details);
    $count  =   mysql_num_rows($borrow_details);

    $total  =    $book_copies  -  $count; 

    mysql_query("UPDATE book SET borrow_total = $total WHERE book_id = $id");

}
?>

如果有唯一键，您可以使用INSERT进行更新。例如，假设book_id是书籍表的唯一键： -

<?php                           
$user_query =   mysql_query("select * from book where status !  =   'Archive'")or die(mysql_error());
while($row  =   mysql_fetch_array($user_query))
{
    $id     =   $row['book_id'];  
    $cat_id     =   $row['category_id'];
    $book_copies    =   $row['book_copies'];
    $borrow_details =   mysql_query("select * from borrowdetails where book_id  =   '$id' and borrow_status =   'pending'");
    $row11  =   mysql_fetch_array($borrow_details);
    $count  =   mysql_num_rows($borrow_details);

    $total  =    $book_copies  -  $count; 

    mysql_query("INSERT INTO book (book_id, borrow_total) VALUES('$id', '$total') ON DUPLICATE KEY UPDATE borrow_total = VALUES(borrow_total)");

}
?>

这是尝试执行插入，查找已存在的记录，因此使用您要插入的值更新borrow_total列。当您想要在1个语句中更新数百条记录时，这非常有用。

但你可以使用一个SQL语句完成整个循环/更新，如下所示： -

UPDATE book b
INNER JOIN 
(
    SELECT book_id, COUNT(*) AS borrowed_count
    FROM borrowdetails 
    WHERE borrow_status =   'pending'
    GROUP BY book_id
) bd
ON b.book_id    =   db.book_id
SET b.borrow_total  =   b.borrow_total - bd.borrowed_count
where b.status !=   'Archive'

注意，如上所述，MYSQL_ *函数已被弃用（即对它们的支持已基本消失，很快就会从php中删除）所以最好切换到MYSQLI_ *函数或pdo。我在这里使用了MYSQL_ *函数只是为了与你的问题保持一致。

Answer 2

虽然你的问题本身并不完整，但我想更新所选书籍的'borrow_total'字段。因此，您只需要运行UPDATE查询而不是INSERTING，而不是INSERTING：

<?php mysql_query("UPDATE book SET borrow_total=$total WHERE book_id=$id"); ?>

希望这会有所帮助：）

Answer 3

您应该使用mysqli_query();函数而不是mysql_query(); ...尝试mysqli_ query();函数来执行查询..

Answer 4

如果要在mysql表中插入新记录，那么

<?php mysql_query("INSERT INTO book (borrow_total) VALUES('$total')");?>

它应该可以工作，如果你想更新mysql表中的一行，那么

<?php mysql_query("UPDATE book SET borrow_total=$total WHERE book_id=$id"); ?>

如何将可变数据存储到我的数据库中

4 个答案: