<table>
<tbody align="center" style="font-family:inherit">
<?php
$resultbooked ="SELECT *,b1.paid_amount as sub_total FROM booking_details b1,payment_history b2 WHERE b1.id='".$bookingid."' AND b2.booking_id='".$bookingid
我正在尝试提交给自己的页面,但出于某种原因,以下代码不工作。另外,在成功插入数据后,如何获取table1主键ID?我有一个子表需要这个ID。谢谢你的建议。
<?php
include('db_login.php');
$connection = mysql_connect( $db_host, $db_username, $db_password );
if (!$connection){
die ("Could not connect to the database: <br />". mysql_error());
}
// Sele
我们在mysql上有这样的信息:
**和要求返回,但有2个限制:**
1- `"SELECT * FROM table ORDER BY point DESC LIMIT 0,2"` //return 'google' & 'jsfiddle'
2- `"SELECT * FROM table ORDER BY point DESC LIMIT 2,2"` //return 'stackoverflow' & 'msn'
3- `"INSERT INT
如何解决这个问题,因为我想找到丢失的ID,并将我带回使用ID丢失,我做了这段代码,但没有工作,因为它什么也不返回。
$query = 'SELECT (folio+1) FROM detalles_devoluciones WHERE (folio+1) NOT IN (SELECT folio FROM detalles_devoluciones) limit 1;';
$resultado = mysql_query($query) or die (mysql_error());
$total = mysql_num_rows($resultado);
我正在试着为下面的逻辑写mysql程序,
select id, fullname from users where fullname like concat(lastname, ' ', firstname, ' (' , middlename, '%');
如果上述查询返回0条记录,则
select id, fullname from users where fullname like concat(lastname, ' ', firstname, '%');
.... few more querie
如果ID不存在于表中,为什么不将用户发送到user.php?
<?php
include 'db_connect.php';
$id_exists = false;
$url_id = mysql_real_escape_string($_GET['id']);
$sql = "SELECT id FROM members WHERE id='$url_id'";
$result = mysql_query($sql);
if ($result == $url_id)
{
$id_exists = tru
我的问题在于将值显示到DGV控件的单元格中,例如:
MySQL表:
ID - Name - Price - Type
1 Mouse Pad 2.85$ - a
2 Keyboard 10.50$ - a
3 Hard Disk 80.00$ - c
4 Web Cam 15.02$ - b
5 Printer 45.62$ - c
6 DVD Writer 20.00$ - b
我的DataGridView控件:
ID - Name - PriceA -
这是什么错误?
Unhandled rejection SequelizeDatabaseError: ER_EMPTY_QUERY: Query was empty
at Query.formatError (node_modules/sequelize/lib/dialects/mysql/query.js:159:14)
at Query._callback (node_modules/sequelize/lib/dialects/mysql/query.js:35:21)
at Query.Sequence.end (node_modules/mysql/lib