我试图在MySQL中执行PL/SQL语句。但是,当我试图创建一个表时,它会显示一个语法错误。我得到以下错误
CREATE TABLE supplier(supid NUMBER(5) PRIMARY KEY, suppname VARCHAR2(15));
ERROR 1064 (42000): You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'NUMBER(5) PR
我的菜单有以下PHP代码,它可以选择MySQL中的所有链接和名称:
<?php
$url = '';
//select all the top row items
$sql="SELECT * from website_menu where parent_top = '' and parent = '' order by menu_order ASC ";
$rs=mysql_query($sql,$conn) or die(mysql_error());
while($result=mysql_fetch_array($r
我正在创建一个循环遍历记录的函数,并希望返回条目数组。
if(!function_exists("TicketAttachments")) {
function TicketAttachments($update_sequence) {
global $conn, $pdo_conn;
$results = array();
$sql="SELECT * from ticket_updates where ticketnumber = '".$update_sequence."'
我有一个PHP代码,它列出了表中的行:
$sql="SELECT * from pages order by menu_order ASC ";
$rs=mysql_query($sql,$conn) or die(mysql_error());
while($result=mysql_fetch_array($rs))
{
echo '<a href="edit_page.php?pagename='.$result["pagename"].'">'.$resu
我在HQL中有以下查询:
update ProjectFile pf1
set pf1.validUntil.id =123
where pf1 = (
select pf from ProjectVersion pv, ProjectFile as pf
where pf.validFrom.sequence <= pv.sequence
and pf.validUntil.sequence >= pv.sequence
and pf.state <> 12
and pf.projectVersion.project.id = 1
and pv.project.
我在堆栈溢出和另一个编码示例站点中看到了本主题讨论的内容。但我找不到任何成熟的解决方案,有人修复这个错误吗?
错误信息:
ERROR:root:Can't connect to to MySQL server with error: (pymysql.err.InternalError) Packet sequence number wrong - got 1 expected 0
ERROR:root:Can't connect to to MySQL server with error: (pymysql.err.InternalError) Packet sequence
我试图处理一个损坏的数据库,但没有成功,当我试图修改这个表时,我给出了错误
Column count of mysql.proc is wrong
在线查找此错误给我的许多结果都是相同的变化,您必须运行
mysql_upgrade
然而,当我跑的时候,我受到了欢迎。
The mysql_upgrade client is now deprecated. The actions executed by the upgrade client are now done by the server.
To upgrade, please start the new MySQL binary wit
我在PHP中运行这个SQL查询:
$sql="SELECT * FROM customer
JOIN contacts ON (contacts.company_sequence = customer.sequence)
WHERE customer.resellerid = '".$_POST["rid"]."'
AND contacts.email = '".$email."'
AND contacts.password = '
我的MySQL 4.1Windows服务不时会崩溃,日志中有以下几行:
091218 9:31:25 InnoDB: Database was not shut down normally!
InnoDB: Starting crash recovery.
InnoDB: Reading tablespace information from the .ibd files...
InnoDB: Restoring possible half-written data pages from the doublewrite
InnoDB: buffer...
091218 9:31:33
我有一个网页,它从我的MySQL数据库中分页结果。下面的代码用于创建下拉菜单以导航结果。目前,它被设置为选择页n;这假定当我请求页30 (例如)时,相关记录存储在结果集中的第30行,但从某些测试来看,情况显然并非如此。我应该如何修改它,以使它返回第n页呢?我知道这句话需要调整,但我不太明白。
//…previous code…
$rowsPerPage = 1;
// by default we show first page
$pageNum = 1;
// if $_GET['page'] defined, use it as
我已经在Mysql中从Firebird数据库创建了一个数据库。在数据库中有一个公共顺序表(Common_ID),它为所有表(数据库中的45个表)生成序列。有些表格是
Table1 = Sequence (ID Auto-increment)-The Common Sequence Table
Table2 = Process (Sequence, Process_Number(Unique), Category_Name)
Table2 = Tasks (Sequence, Barcode(Unique), Process_Number(sequence from Process Table)
我通过输入以下命令提示符命令创建了一个反应性项目
mkdir mysql-test & cd mysql-test
npm init -y
安装mysql -save
然后,我创建了一个名为app.js的文件,并输入了以下代码:
// app.js
const mysql = require('mysql');
// First you need to create a connection to the db
const con = mysql.createConnection({
host: 'rds-mysql-wmcw.c