if ($Funct == "BUILD" AND $Mode == "GROUND")
{
$sql = "SELECT * FROM $landCon WHERE North = '" . $North . "' AND West = '" . $West ."'" ;
$result = mysql_query($sql) or die("Error occurred in [$sql]: " . mysql_error());
$count = mysql_num_rows($result);
If ($Count == 0 )
{
$sql = "INSERT INTO $landCon (West,North,Type1) VALUES ($West,$North,'$Char')";
mysql_query($sql) or die("Error occurred in RootA as:[$sql]: " . mysql_error());
}
else
{
$result=mysql_query("UPDATE $landCon SET Type1= '$Char' WHERE North = …我需要做自己的arraylist课程.我正在尝试编写插入方法,如果在此索引中存在已存在的对象,我将面临将对象添加到索引0的问题.它适用于其他指数,但不适用于0.任何帮助将不胜感激？
public void insert(int index, Object object) throws IndexOutOfBoundsException {
if(index < 0) {
System.out.println("Niepoprawny index!");
throw new IndexOutOfBoundsException();
}
if(index > array.length - 1 || array[index] != null) {
// zwi?kszenie rozmiaru tablicy
// je?li jest zbyta ma?a lub na danym miejscu istnieje jaki? obiekt
Object[] temp = new Object[array.length + index];
System.arraycopy(array, 0, temp, 0, array.length);
array = temp;
}
// przesuni?cie o 1 w prawo wszystkich elementów
System.arraycopy(array, index - 1, array, index, array.length - index);
array[index] …注意到奇怪的事情.在我的数据库中,所有表只有奇数主键.
id | value
1 | 10
3 | 15
5 | NULL
7 | 1
在其他Yii网站一切都好.
通过不同方法添加到不同表中的行:
$model = new Tbl1;
$model->value = 10;
$model->save();
和
$command = Yii::app()->db->createCommand();
$sql = 'INSERT INTO `tbl2` SET `value` = 10';
$command->setText($sql);
$command->execute();
这很奇怪,是吗？
我正在尝试将信息插入到sql express 2008数据库的表中,当我尝试插入字段时,我收到错误"'附近的语法不正确',"有人可以向我解释我的代码中是什么导致这个问题？
protected void insertworkshop_Click(object sender, EventArgs e)
{
using (SqlConnection conn2 = new SqlConnection(@"Data Source=CIS489_3\WILDLIFE;Initial Catalog=WildLife_Education;Integrated Security=True;"))
{
SqlCommand CmdSql2 = new SqlCommand("INSERT INTO [tblWorkshop] ([WorkshopName], [WorkshopBeginingDate], [WorkshopEndingDate], [WorkshopLocation], [InstructorID])VALUES (@WorkshopName, @WorkshopBeginingDate, @WorkshopEndingDate, @WorkshopLocation, @InstructorID)", conn2);
conn2.Open();
CmdSql2.Parameters.AddWithValue("@WorkshopName", workshopinsertname.Text.ToString());
CmdSql2.Parameters.AddWithValue("@WorkshopBeginingDate", workshopinsertstart.Text.ToString());
CmdSql2.Parameters.AddWithValue("@[WorkshopEndingDate", workshopinsertend.Text.ToString());
CmdSql2.Parameters.AddWithValue("@WorkshopLocation", workshopinsertlocation.Text.ToString());
CmdSql2.Parameters.AddWithValue("@InstructorID", insertinstructorid.SelectedValue.ToString());
CmdSql2.Connection = conn2;
CmdSql2.ExecuteNonQuery();
conn2.Close();
UpdateInsertWorkshop.Update();
this.addnewworkshop_ModalPopupExtender.Hide();
}
我做了一些研究,发现获取新插入行的id的最佳解决方案是输出子句.我第一次使用insert语句输出子句而无法解决它.
这是我的sql代码.这个sql出了什么问题？
ALTER PROCEDURE [dbo].[MedBul_Insert_Message]
(@ParentID int, @MessageFrom varchar(500),@MessageTo varchar(500), @MessageSubject varchar(500),@MessageBody varchar(4000),@MessageIsRead int)
AS
BEGIN
SET LANGUAGE Turkish;
Declare @message_id table (MessageID int);
insert into Messages output INSERTED.ID INTO @message_id (ParentID ,MessageFrom ,MessageSubject ,MessageBody ,MessageIsRead ) values(@ParentID ,@MessageFrom ,@MessageSubject ,@MessageBody ,@MessageIsRead );
END
我是PL/SQL和存储过程的新手.我正在尝试编写一个将通过CallableStatement从Java程序执行的存储过程.该过程采用两个参数,获取最后一条记录的id,递增它并使用新增加的id插入新记录.我在网上发现了一些主要做同样事情的例子,但我无法解决错误.
CREATE OR REPLACE PROCEDURE insertEmployeeProcedure
(lastname IN VARCHAR, firstname IN VARCHAR) AS
BEGIN
lastEmpId NUMBER := SELECT COUNT(*)
INTO lastEmpId
FROM Employees;
INSERT INTO Employees(id, lname, fname) VALUES(lastEmpId + 1, lastname, firstname);
END insertEmployeeProcedure;
/
错误是:
Executed successfully in 0.018 s, 0 rows affected.
Line 1, column 1
Error code 984, SQL state 42000: ORA-00984: column not allowed here
Line 8, column 5
Error code 900, SQL state 42000: ORA-00900: invalid SQL statement
Line 9, column 1
Error code …我有一个脚本正在创建一个表,然后插入一行.这是我执行创建表的SQL代码:
CREATE TABLE polls (
id INT NOT NULL UNIQUE AUTO_INCREMENT,
name VARCHAR(255) NOT NULL UNIQUE,
author VARCHAR(255) NOT NULL,
created DATETIME NOT NULL,
expires DATETIME,
PRIMARY KEY(id)
)
这是我添加新行的地方:
INSERT INTO polls
VALUES ('TestPoll'),('Billy Bob'),('2013-05-01 04:17:31'),('2013-05-01 04:17:31')
要么
INSERT INTO polls
VALUES ('TestPoll','Billy Bob','2013-05-01 04:17:31','2013-05-01 04:17:31')
(无论如何我得到同样的错误)
我总是得到这个错误:
<class '_mysql_exceptions.OperationalError'>, OperationalError(1136, "Column count doesn't match value count at row 1"), <traceback object at 0x7f7bed982560>
我有一个表COMPONENTS3列(的Sno,Component,Quantity),其中我写Sno,并Component开始列,我要填写栏"量"使用一些表情(例如:((d1+d2)*d3)),涉及从另一个表变量SHEET(d1 int,d2 int ,d3 int,d4 int ,d5 int,d6 int).
在这里,我需要根据COMPONENTS表(components.sno)中的Sno列中的值将值写入数量列.
我曾经在'x'中保留表达式值并尝试插入COMPONENTS,如下所示:
insert into components(Quantity) values(x)
where components.sno='y'; [Y is inetger starting from 0 to 70]
但上面的查询显示错误 where
请建议我最好的SQL查询来实现这一目标!提前致谢..!
我在Delphi中使用TSQLQuery执行插入.数据插入完美,但程序显示标题消息错误.有任何想法吗 ？.这是我的代码:
With DMConnect.qryCrearProyecto do begin
ParamByName('cliente').AsString := Self.Edit2.Text;
ParamByName('obra').AsString := Self.Edit3.Text;
ParamByName('ubicacion').AsString := Self.Edit4.Text;
ParamByName('nroEstudio').AsInteger := StrToInt(Self.Edit5.Text);
ParamByName('sondeo').AsInteger := StrToInt(Self.Edit6.Text);
ParamByName('nivelFreatico').AsFloat := StrToFloat(Self.Edit7.Text);
Open;
Close;
end;
我目前正在通过教程学习PHP,我正在尝试运行以下代码但是得到了一个无法解释的语法错误,其中没有其他人似乎与它们一起出现.
错误消息 解析错误:语法错误,意外';' 在第12行的..../users.php中
第12行=
return (mysql_result(mysql_query("SELECT `user_id` FROM `users` WHERE `username` = '$username'"), 0, 'user_id');
整页代码
<?php
function user_exists($username) {
$username = sanitize($username);
return (mysql_result(mysql_query("SELECT COUNT(`user_id`) FROM `users` WHERE `username` = '$username'"), 0) == 1) ? true : false;
}
function user_active($username) {
$username = sanitize($username);
return (mysql_result(mysql_query("SELECT COUNT(`user_id`) FROM `users` WHERE `username` = '$username' AND `active` =1"), 0) ==1) ? true : false;
}
function user_id_from_username ($username) {
$username = sanitize($username);
return (mysql_result(mysql_query("SELECT `user_id` FROM `users` WHERE `username` …