我是一个比较新的php.The问题,当我在我的数据库的'leave‘表中插入值时,我面临着这个问题。错误如下所示。
您的SQL语法中有一个错误;请查看与您的MySQL服务器版本对应的手册,了解在第1行'where lid = 4‘附近使用的正确语法。
下面是我的代码
<?php
include_once 'config.php';
$accept = "accepted";
mysql_query("insert into `leave` (`action`) values ('$accept') where li
嗨,我在写下面的代码,
**mysql-model.js**
var mysqlModel = require('mysql-model');
var model = function(){
var MyAppModel = mysqlModel.createConnection({
host : 'localhost',
user : "root",
password : ""
我想在mysql中的concat()中格式化数据。但是mysql在我尝试的时候禁止了我。以下是功能:
SELECT user_id as id, DATE(leave_from) as date_from, DATE(leave_to) as date_to, NULL as delayed_checkin_remarks, NULL as early_checkout_remarks,
CONCAT(
(' LeaveType : ',
SELECT title FROM leave_type
WHERE i
这是我的代码
$Qemaster="select * from emaster where `branch`='$bid' and `department`='$did' and `status`!='L'";
$Remaster=mysql_query($Qemaster);
while($Rowemaster=mysql_fetch_array($Remaster)){
$empcode=$Rowemaster[id];
$name=$Rowemaster[name];
$Tleave=0;
echo
如何查找和替换空行时使用此.
$Qemaster="select * from emaster where `branch`='$bid' and `department`='$did' and `status`!='L'";
$Remaster=mysql_query($Qemaster);
while($Rowemaster=mysql_fetch_array($Remaster)){
$empcode=$Rowemaster[id];
$name=$Rowemaster[name];
$Tleave=
我在shell中的一个变量中存储了一个值。如下所示
app=$(mysql -uroot -p123456 -e 'SELECT applicant FROM `leave` where status="Applied" and applying_date= curdate()' comviva|tail -1);
但是使用这个变量,如果我想要更新任何列,我就会遇到错误。下面是错误。更新命令:
mysql -uroot -p123456 -e "update leave set status=\"pending\" where appli
我使用的是laravel 8。我有一个mysql命令,我想把它转换成风格:
select allocation.*, leav_leave_types.leave_type_code
from (
select * from leav_employee_annual_leave_allocations
where leave_year_id = $year_id and employee_id = $user_id
) as allocation
left join leav_leave_types on (leav_leave_types.id = allocation.
我尝试按月份名称对mysql查询进行排序,如下所示:
January -- 5
February -- 2
March -- 5
and so on
这是我的查询,但它没有排序:
SELECT leave_balance.balance, MonthName(leave_balance.date_added) AS month
FROM leave_balance WHERE leave_balance.staff_id_staff = $iid
GROUP BY month, leave_balance.leave_type_id_leave_type
HAVING leav
我使用函数mysql_fetch_assoc返回数据库中的数据,但不知道如何在表中打印其结果。下面是执行此过程的代码。
$number_of_date= "SELECT SUM(number_of_date) AS number_of_date FROM emp_leaves WHERE emp_id='$userID' AND leave_category='Annual' AND apply_year='$year'";
$res_qry = mysql_query($number_of_date) or die ('
我正在为MySQL数据库编写PLSQL触发器。我试图声明变量。所以我在触发器中写了一个声明块。以下是我的代码
CREATE TRIGGER leave_approve_trigger
AFTER UPDATE ON leave_status
FOR EACH ROW
DECLARE //Syntax error
current_id integer;
BEGIN
if NEW.status == 'APPROVED'
THEN
select id into current_id from leave_request_table;
insert into update_t
当我试图将数据从sql数据库检索到j组合框时出错:
“SQL语法出现错误;请检查与MySQL服务器版本对应的手册,以查找正确的语法错误”
public class NewJFrame extends javax.swing.JFrame {
Connection con=null;
PreparedStatement pst=null;
ResultSet rs=null;
/** Creates new form NewJFrame */
public NewJFrame() {
initComponents();
fillcombo();
}
private vo
我正在尝试为数据库创建一个触发器,使用nodejs的mysql包。但我说错了
您的SQL语法有错误;请检查与您的MySQL服务器版本对应的手册,以便在第1行的“Leave\”插入后使用接近\‘分隔符$$创建触发器'NotificationTrigger’的正确语法。
var createNotificationTrigger = {
query:
'DELIMITER $$\
CREATE TRIGGER `NotificationTrigger` \
After INSERT ON `Leave` \
FOR EACH ROW
我用的是php。我正在用我的表获取两个数组。我不明白为什么它会复制表,但是它能够显示来自另一个查询的值。
以下是整个代码:
$select_department2 = mysql_query("SELECT DISTINCT work.`Department` FROM `work` INNER JOIN employee_leaves ON employee_leaves.ID_No = work.ID_No WHERE MONTH(employee_leaves.Date_Approved) LIKE '$month' AND Leave_Type LIKE
以下代码是可运行的,工作正常,但是如果我将$dbh->do("set names utf8");更改为$dbh->do("set names gbk");,我将收到一个语法错误:
use strict;
use warnings;
use DBD::mysql;
my $dbh = DBI->connect("DBI:mysql:database=test;host=localhost","root","password");
$dbh->do("set names utf8
在MySQL中,我有一个带有的存储过程
DELIMITER $$
CREATE PROCEDURE ABC()
BEGIN
DECLARE a INT Default 0 ;
simple_loop: LOOP
SET a=a+1;
select a;
IF a=5 THEN
LEAVE simple_loop;
END IF;
END LOOP simple_loop;
END $$
它总是打印1。MySQL循环的正确语法是什么?
我已经在MySQL中创建了一个数据库注册,其中已经有2个表users和employees。我想创建第三个表Leave,但它显示错误1064。我什么都试过了。请告诉我哪里出了问题。
create table leave(
employeecode int(11) primary key auto_increment,
description varchar(100) not null,
fromdate date not null,
todate date not null,
status varchar(100) not null
);
SELECT
E.`employee_id`,
E.`full_name`,
LE.`no_of_leaves` AS AllocatedLeaves,
MLLT.`leave_type` AS LeaveTypeName,
COUNT(SELECT * FROM leave_approval WHERE employee_id = 1) AS TotalLeavesTaken
FROM employee E
INNER JOIN leave_entitlement LE
ON E.`employee_id` = LE.`employee_id`
INNER JOIN `ml_leave_t
在创建触发器时,我会得到以下错误:
您的SQL语法出现了错误;请检查与您的MySQL服务器版本对应的手册,以获得在第3行附近使用的正确语法
这是我的疑问:
CREATE TRIGGER reserved BEFORE INSERT ON Reserv
FOR EACH ROW BEGIN
DECLARE trip_id integer;
DECLARE free integer;
DECLARE count integer;
set @free := (select place_free from Trips where id = trip_id);