标签: create-table

我在PRIMARY KEY附近遇到了一个错误,但无法确定究竟是什么,有人可以看看吗？

CREATE TABLE `reports` (

    `key`                   INT UNSIGNED AUTO_INCREMENT, 
    `role`                  VARCHAR(70), 
    `region`                VARCHAR(70),
    `inspection_type`       VARCHAR(70),
    `inspection_number`     VARCHAR(70),
    `customer_number`       VARCHAR(70),

    `report_date`           DATE DEFAULT NULL,  



    `order_date`            DATE DEFAULT NULL,  


    `customer`              VARCHAR(70),
    `customer_division`     VARCHAR(70),
    `location`              VARCHAR(70),
    `memo`                  VARCHAR(255),       
    `billingkey`            VARCHAR(70),

    PRIMARY KEY(key)

) ENGINE=InnoDB DEFAULT CHARSET=UTF8;

这是我到目前为止的MySQL代码:

CREATE DATABASE bankbase;

USE bankbase;

CREATE TABLE clienttable(
ClientID SMALLINT(15) NOT NULL DEFAULT 0,
ClientFirstName VARCHAR(30) NOT NULL DEFAULT "first name",
ClientLastName VARCHAR(30) NOT NULL DEFAULT "last name",
ClientPhone CHAR(10) NOT NULL, ClientEmail 

TINYTEXT(30) NULL,
ClientAddress TINYTEXT(128) NOT NULL,
PRIMARYKEY(ClientID)
);

CREATE TABLE branchtable(
BranchID SMALLINT(15) NOT NULL DEFAULT 0,
BranchCity TINYTEXT(30) NOT NULL DEFAULT city, 

BranchManagerFName VARCHAR(30) NULL DEFAULT "Branch Manager's First Name",
BranchManagerLName VARCHAR(30) NULL 

DEFAULT "Branch Manager's LAst Name",
BranchPhone CHAR(10) NOT NULL,
BranchEmail TINYTEXT(30) NULL …

我搜索了论坛,但没有找到正确答案对不起,如果我发帖,我想创建这个表

create table HUB_SHIFT_REPORT
(
  Da_te timestamp(8) not null
, Week_Day varchar2(10) not null constraint cnstrnt_day check (Week_Day in('Su','Mo','Tu','We','Th','Fr','Sa'))
, Shift varchar2(10) not null constraint cnstrnt_shift check (Shift in('Morning','Evening','Night'))
, Handled_by varchar2(60) not null constraint cnstrnt_handled check (Handled_by in('Artyom','Natan','Asaph','Yaniv','Fadi','AdiG','AdiF','Paul','Benjamin','Matan','Yosef'))
, Incident_Number number(10) not null PRIMARY KEY
, Centreon_Duration varchar2(10) not null
, Time_received varchar2(10) not null
, Time_restored varchar2(10) not null
, Time_closed varchar2(10) not null
, Product varchar2(60) not null constraint cnstrnt_product check (Product in ('Primo','MetaLib','SFX','bX','Primo Central','USTAT','Alma','Aleph','voayger','Marcit','ARC','Verde','Rosetta','Digitool','MetaLib+','NUUL','Not product related')) …

我有以下内容doGet():

protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
        MysqlDataSource ds = new MysqlConnectionPoolDataSource();
        ds.setServerName("localhost");
        ds.setPort(3306);
        ds.setUser("root");
        ds.setPassword("");
        try {
            Connection connection = null ; 
            connection =  ds.getConnection();
            Statement statement = connection.createStatement();
            // create the DB .. 
            statement.executeUpdate("CREATE DATABASE IF NOT EXISTS "
                            +"testSQLtable" )  ;

            // use DB ..

            statement.executeUpdate("USE testSQLtable") ; 

            // create table ...
            statement.executeUpdate("DROP TABLE IF EXISTS  account "
                    +"CREATE TABLE  account ( "
                     +"accountNum INT( 11 ),"
                     +"dateCreated DATE NOT NULL," …

我正在尝试使用PHP和MySQL创建一个表.第一个表不会在数据库上创建,但第二个表将创建.我认为它是我的参数/约束.这是我的代码:

if ($conn==true){
     $tablefriends = "CREATE TABLE IF NOT EXISTS friends (
        friend_id INT NOT NULL AUTO_INCREMENT PRIMARY KEY ,
        friend_email VARCHAR(64) NOT NULL PRIMARY KEY ,
        password VARCHAR(16) NOT NULL ,
        profile_name VARCHAR(32) NOT NULL ,
        date_started DATE NOT NULL ,
        num_of_friends INT UNSIGNED ZEROFILL NULL default '0'
        );";

    $tablemyfriends = "CREATE TABLE IF NOT EXISTS myfirends ( 
        friend_id1 INT NOT NULL , 
        friend_id2 INT NOT NULL 
        );";

    $resulttf = @mysqli_query($conn, $tablefriends);
    if($resulttf==false){
        echo "<p>Failed to create friends table</p>"; …

我在PHP中传递一个查询变量.我想把它作为表名,但我知道可能存在SQL语法错误.当我打印语句时,传递变量意味着它可以工作,但是数据库根本就没有创建.

这是我创建数据库的代码:

$DBName = "database_name";  
$sql = "CREATE TABLE '$DBName'.'$login' (  
    ClientID int NOT NULL AUTO_INCREMENT,   
    AgentClients varchar(15),  
    ClientTotal int  
    )";  
mysql_query($sql,$link);
where `$login = $_POST['login'];

此外,我现在并不担心安全漏洞,所以不要担心.

任何见解将不胜感激.

我收到这个错误:

您的SQL语法有错误; 检查与你的MySQL服务器版本相对应的手册,以便在''JOIN`-- shirt`link```-- shirt``````````````````````````````````````````````````````````````````````````

我现在已经梳理了近半个小时,无法弄明白.有人可以指出我吗？这是代码.

SELECT
`shirts`.`shirt_name`,
`shirts`.`men` AS `main_photo`,
GROUP_CONCAT ( `shirt_sizes`.`size_name` ) AS `sizes`
FROM
`shirts`
JOIN
`shirts_link` ON `shirts_link`.`shirt_id`=`shirts`.`id`
JOIN
`shirt_sizes` ON `shirt_sizes`.`id`=`shirts_link`.`size_id`
JOIN
`shirt_prices` ON `shirt_prices`.`id`=`shirts_link`.`price_id`
WHERE `men`!=''
GROUP BY
`shirt_prices`.`price_cat`

如果问题比这个脚本更深入,这里是我试图联系在一起的其他表.

衬衫表

CREATE TABLE shirts (
    id INT UNSIGNED AUTO_INCREMENT PRIMARY KEY NOT NULL,
    shirt_name VARCHAR(20) NOT NULL,
    men VARCHAR(10) NULL,
    women VARCHAR(10) NULL,
    boys VARCHAR(10) NULL,
    girls VARCHAR(10) NULL,
    babies VARCHAR(10) NULL,
)ENGINE=INNODB;

INSERT INTO shirts(shirt_name,men,women,boys,girls,babies) VALUES
    ('Crewneck Tee','me_crn','wo_crn','bo_crn','gi_crn','ba_crn'),
    ('V-Neck Tee','me_vnc','wo_vnc','','',''),
    ('Scoop Neck …

我是PHP和mysql的新手,我正在尝试用PHP创建一个事件表,但由于某种原因它没有被创建.我不确定我写的是否有不正确之处.

<?php include_once("php_includes/db_conx.php");
$tbl_events = "CREATE TABLE IF NOT EXISTS events (
              id INT(11) NOT NULL AUTO_INCREMENT,
              title VARCHAR(255) NOT NULL,
              privacylevel ENUM('a','b','c') NOT NULL DEFAULT 'a',
              imageurl VARCHAR(255) NULL,
              content TEXT NOT NULL,
              event_start_date DATETIME NOT NULL,
              event_end_date DATETIME NOT NULL,
              event_startloc_name VARCHAR(255) NULL,
              event_startloc_lat FLOAT( 10, 6 ) NULL,
              event_startloc_lon FLOAT( 10, 6 ) NULL,
              event_endloc_name VARCHAR(255) NULL,
              event_endloc_lat FLOAT( 10, 6 ) NULL,
              event_endloc_lon FLOAT( 10, 6 ) NULL,
              your_location_lat FLOAT( 10, 6 ) NULL,
              your_location_lon FLOAT( …

有谁能用简单的词语描述这张照片中的两个陈述？你的回答将是真正的赞赏.

谢谢

如何创建新表并插入另一个表的内容？