我试图用PHP脚本在同一个MySQL数据库中创建2个表:表'user'主键'user_id'和表'order'主键'order_id'和外键'user_id'来自'user'表(1对多关系).
表用户成功创建没有问题:
$sql="CREATE TABLE user(
user_id INT UNSIGNED NOT NULL AUTO_INCREMENT,
type ENUM('member','admin') NOT NULL,
username VARCHAR(30) NOT NULL,
email VARCHAR(80) NOT NULL,
pass VARBINARY(32) NOT NULL,
first_name VARCHAR(40) NOT NULL,
last_name VARCHAR(40) NOT NULL,
date_expires DATE NOT NULL,
date_created TIMESTAMP NOT NULL DEFAULT CURRENT_TIMESTAMP,
date_modified TIMESTAMP NOT NULL DEFAULT '0000-00-00 00:00:00',
PRIMARY KEY (user_id),
UNIQUE (username),
UNIQUE (email)
)ENGINE=InnoDB DEFAULT CHARSET=utf8";
但是,我无法创建表顺序:
$sql="CREATE TABLE order(
order_id INT UNSIGNED NOT NULL AUTO_INCREMENT,
user_id INT UNSIGNED NOT NULL,
transaction_id VARCHAR(19) NOT …我正在尝试用PHP做一个非常简单的UPDATE,如下所示:
$nlk = $lk + "1";
mysql_query("UPDATE posts SET like = '".$nlk."' WHERE id = '".$cid."'") or die(mysql_error());
$lk是从字段中获取的值like,这是默认值0.
$cid是一个id字段的值,它是打开的auto_increment.
我收到此错误:
您的SQL语法有错误; 查看与您的MySQL服务器版本对应的手册,以便在'like ='附近使用正确的语法1'WHERE id ='45''在第1行
这是什么问题？
$query = "INSERT INTO textbooks (uid, title, edition, author, isbn, condition, description, photo) VALUES (9, 'q', 'a', 'b', 'c', 'd', 'e', '$extension')";
$result = mysql_query($query) or die('Could not connect: ' . mysql_error());
Could not connect: 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 'condition, description, photo) VALUES (9, 'q', 'a', 'b', 'c', 'd', 'e', '.jpg')' at line 1
我一直在收到一个MySQL语法错误,但我不能为我的生活弄清楚什么是错的.我放置随机变量,字符串和int只是为了测试,但它不起作用.
- 标题,版本,作者,isbn,条件,描述,照片都是varchars
- uid是一个int
- 描述是文字
我尝试将简单数据更新为表名"订单"但我仍然得到错误.
我尝试了很多版本查询,但仍然相同;
first try :
$result = mysql_query("UPDATE order SET order_status_id=200 WHERE order_id=75") or die(mysql_error());
second try :
$result = mysql_query("UPDATE order SET order_status_id='200' WHERE order_id='75'") or die(mysql_error());
错误;
首先尝试:您的SQL语法有错误; 查看与您的MySQL服务器版本对应的手册,以便在第1行的'order SET order_id = 200 WHERE order_id = 75'附近使用正确的语法
第二次尝试:您的SQL语法有错误; 查看与您的MySQL服务器版本对应的手册,以便在第1行的'order SET order_status_id ='200'WHERE order_id ='75''附近使用正确的语法
表结构
order_id int(11)
order_status_id int(11)
我尝试更新其他表只是为了确保我的查询正确,所有表都可以更新.
*我使用Opencart,我的网站使用https.
谢谢.
所以我一直坚持这个SQL错误.
这是我正在使用的SQL行:
INSERT INTO images (image_name, orientation, restored, commercial, automotive, bespoke, before, after, date_added)
VALUES ('image4-after.jpg', 'portrait', '1', '1', '1', '1', '24', '5', '2012-07-08')
使用这种结构:
image_id - int(11) AUTO_INCREMENT
image_name - varchar(40)
orientation - varchar(4)
restored - tinyint(1)
commercial - tinyint(1)
automotive - tinyint(1)
bespoke - tinyint(1)
before - int(11)
after - int(11)
date_added - date
收到错误消息:
1064 - 您的SQL语法出错; 检查与MySQL服务器版本对应的手册,以便在'before,after,date_added'附近使用正确的语法VALUES('image4-after.jpg','portrait','1','1','1''在第1行
任何人都可以告诉我我做错了什么？
谢谢
我遇到了这个MySql查询的问题:
INSERT INTO groups (NAME, DESC, TIME, OWNER) VALUES ('$GNAME', '$DESC', '$TIME', '$UID')
本质上,该脚本将注册组的人员的组名,描述,时间和用户名添加到mysql数据库中.
完整脚本:
<?php
include_once('include/session.php');
$GNAME = $_POST['groupname'];
$DESC = $_POST['desc'];
$SPAM = $_POST['spam'];
$UID = $_POST['UID'];
$TIME = date('Y-m-d H:i:s');
if($SPAM == "queuee"){
$query ="INSERT INTO groups (NAME, DESC, TIME, OWNER) VALUES ('$GNAME','$DESC','$TIME', '$UID')";
$result = mysql_query($query) or die("There as been an Error! <hr>Error:<hr>".mysql_error() ."<br><hr><a href=\"../index.php\">Go Back</a>");
header("Location: ../group.php?id=$GNAME");
}else{
?>
The Security Question was wrong. Try Again.
<?
}
?>
我使用Phpmyadmin将Query直接运行到MySql本身.但它仍然犯了一个错误.错误是:
#1064 - You have an …我有2张桌子:PLAYER和GROUP.
我打电话的时候:
SELECT * FROM PLAYER
一切都好,但是当我打电话时:
SELECT * FROM GROUP
有错误:
#1064- 您的SQL语法有错误; 检查与MySQL服务器版本对应的手册,以便在第1行的"GROUP"附近使用正确的语法
我不明白为什么会有这样的错误.
mysql> use j****
Database changed
mysql> show tables;
Database setup was all right
+---------------------+
| Tables_in_j******** |
+---------------------+
| fos_user |
| goal |
| goal_type |
| industry |
| interval |
| target |
| user |
+---------------------+
7 rows in set (0.11 sec)
mysql> explain interval;
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 'inter
val' at …嗨我正在尝试从Mysql服务器加载数据库并继续收到此错误mysqli_num_rows()期望参数1为mysqli_result,布尔值在第15行的C:\ Users\Mirne\Desktop\Server\root\funkcie.php中给出.我检查了很少的帖子,但我没有发现我的错误.Ty求助
<?php
global $X,$Y,$height,$weight,$key,$type,$text,$position;
function load() {
$servername = "localhost";
$username = "root";
$password = "xxxx";
$dbname = "web builder";
$link = mysqli_connect($servername, $username, $password, $dbname);
if (!$link) {
die("Connection failed: " . mysqli_connect_error());
}
$sql = "SELECT x, y, height, weight, key , type , text , position FROM suradnice";
$result = mysqli_query($link, $sql);
if (mysqli_num_rows($result) > 0) { ////////line15///////
while($row = mysqli_fetch_assoc($result)) {
$X=$row["x"] ;
$Y=$row["y"] ;
$height=$row["height"] ;
$weight=$row["weight"];
$key=$row["key"];
$type=$row["type"] ;
$text=$row["text"] ;
$position=$row["position"] ; …它不断收到此SQL错误,但我非常确定查询中没有错误.我一直盯着它看太久了.还有其他可能导致这种情况的事吗？
INSERT INTO game_data (clue, image, answer, wrong, right) VALUES ('asdf', 'asdf', 'asdf', 'assdf', 'asdf')
Invalid Query: 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 'right) VALUES ('asdf', 'asdf', 'asdf', 'assdf', 'asdf')' at line 1
我用这个确切的代码运行其他查询,只是交换变量,它的工作原理.
并且为了更好的衡量,有一个PHP生成这个:
$sql = "INSERT INTO game_data (clue, image, answer, wrong, right) VALUES
('" . $clue . "', '" . $image . "', '" . $answer . "', …