0 
<?php include "koneksi.php";
$kdj=$_POST['kdj'];
$id=$_POST['id'];
$qty = $_POST['qty'];
$kodelagi = mysql_query("Select kd_jual from nota order by id_nota DESC LIMIT 1");

if((!empty($id)) && (!empty($kdj)) && (!empty($qty)))
{
$query = mysql_query("INSERT INTO nota (id_item,qty,kd_jual)
values ('$id','$qty','$kodelagi');");
?><script language="Javascript">;
document.location = 'transaksi.php' </script><?php
}else 
{
print "<script>alert('Maaf, tidak boleh ada field yang kosong !');
javascript:history.go(-1);</script>";
}?>

知道为什么这不起作用吗？我只想从我的数据库上的表中取出最后一条记录以插入另一条记录.. 它的工作但是当我看到数据库时，kd_jual它被“资源 id #5”填充，而不是我想要的......请帮助

我的表“nota”：

______________________________________________
|id_nota | id_item | qty | kd_jual            |
|1       | 381     | 3   | 09-10-201303:45:46 |
|2       | 11      | 5   | Resource id #5     |
4

1 回答 1

0

添加行以在评论中获取结果

<?php include "koneksi.php";
    $kdj=$_POST['kdj'];
    $id=$_POST['id'];
    $qty = $_POST['qty'];
    $kodelagi = mysql_query("Select kd_jual from nota order by id_nota DESC LIMIT 1");
    ////////////////// add these two lines to fetch the results
    $row = mysql_fetch_array($kodelagi);
    $kodelagi1 = $row['kd_jual'];
    //////////////////
    if((!empty($id)) && (!empty($kdj)) && (!empty($qty)))
    {
    $query = mysql_query("INSERT INTO nota (id_item,qty,kd_jual)
    values ('$id','$qty','$kodelagi1');");    ////// $kodelagi1 fetched value
    ?><script language="Javascript">;
    document.location = 'transaksi.php' </script><?php
    }else 
    {
    print "<script>alert('Maaf, tidak boleh ada field yang kosong !');
    javascript:history.go(-1);</script>";
    }?>
于 2013-10-09T03:28:20.333 回答