我有以下代码:
$link=mysql_connect("localhost", "user", "password") or die(mysql_error());
mysql_select_db(raddacen_staff,$link) or die(mysql_error());
$query=mysql_query("INSERT IGNORE INTO database...............") or die(mysql_error());
if(mysql_affected_rows()>0){
e
这张桌子看起来像这样
mysql> DESCRIBE tenants;
+------------+--------------+------+-----+---------+----------------+
| Field | Type | Null | Key | Default | Extra |
+------------+--------------+------+-----+---------+----------------+
| id | int unsigned | NO | PRI | NULL
在存储过程中,声明变量有问题。我用的是MySQL。我的示例代码:
CREATE PROCEDURE `name`()
BEGIN
DECLARE varname INT;
SELECT * FROM `table` INTO var;
END
MySQL返回错误:
错误1064 -您的SQL语法有错误;请检查与您的MariaDB服务器版本相对应的手册,以便在第3行使用正确的语法。
我编写了下面的代码片段,其中使用了一个SQL查询,它使我获得了DB中特定存储的storeID:
private int inDB(String storename, String phone, int zipcode)
{
try
{
String select = "SELECT storeID FROM Store WHERE storename = ? OR phone = ? AND City_zipcode = ?;";
PreparedStatement pst = dbc.p
我有这部分代码:
$result = mysql_query("SELECT * FROM posts WHERE id = '$id'") or die (mysql_error());
$rec = mysql_fetch_array($result) or die (mysql_error());
$like = $rec['like'];
$like += 1;
mysql_query("UPDATE posts SET like = '$like' WHERE id = '$id'") o
mysql> select count(id) total from applicants;
+-------+
| total |
+-------+
| 0 |
+-------+
1 row in set (0.00 sec)
mysql> select count(id) total from jobs;
+-------+
| total |
+-------+
| 0 |
+-------+
1 row in set (0.00 sec)
mysql> select count(id) total from applicants union s
下面的PHP代码报告一个错误代码:
$id = $_SESSION['sno'];
$q = mysql_query("select * from messages where seen=0 and to=$id");
if(!$q){die("critical failure: ".mysql_error());}
报告的错误为:
critical failure: You have an error in your SQL syntax; check the manual that corresponds to your MySQL serv
我有MySQL 5.1.58,我执行了以下命令,用sqlmap对数据库进行指纹分析
python sqlmap.py -d "mysql://root:password@localhost:3306/northwind" --fingerprint'
输出是
[12:26:35] [INFO] the back-end DBMS is MySQL
[12:26:35] [INFO] actively fingerprinting MySQL
[12:26:35] [INFO] executing MySQL comment injection fingerprint
b
#1064 -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,以获取在第2行的) AS county FROM q_mem_tim附近使用的正确语法。
SELECT Count( * ) AS m,
SUM(CASE
WHEN y = '2011' THEN 1
ELSE 0 ) AS county
FROM q_mem_tim
我的代码出了什么问题?
我有一个变量和一个user_name,我想要在user_name的字符串(Function_description)中搜索它
这有什么问题:
$function_keywords = mysql_real_escape_string($_POST['function_keywords']);
if($function_keywords=="" || empty($function_keywords)){
redirect("show.php?functions=PHP");
}
//trim whitespace from the sto
我在PHP中使用带有ADOdb的MySQL,并且尝试将关键字作为绑定变量多次传递,而不必在MySQL数组中重复它,这可能吗?我被分号抛出了一个错误。
$queryParams = [$keyword, $row_ids];
$query = "
SET @myKeyword = ?;
SELECT *
FROM my_table
WHERE row_id IN (?)
AND (
row_desc1 = @myKeyword
OR row_des2 = @myKeyword
OR row_desc3 =
我正在尝试创建一个从mysql.slow_log表中分析数据的脚本。出于这些目的,我需要将我感兴趣的一些行复制到另一个临时表中,因此我将如下所示:
CREATE TABLE IF NOT EXISTS test.slow_log_to_analyze LIKE mysql.slow_log;
TRUNCATE TABLE test.slow_log_to_analyze;
INSERT INTO test.slow_log_to_analyze
SELECT * FROM mysql.slow_log WHERE start_time < (NOW() - INTERVAL 1 M
守则是:
DELIMITER $$
CREATE FUNCTION CHECK_AVABILITY(nama CHAR(30))
RETURNS INT(4)
DECLARE vreturn INT(4);
BEGIN
IF nama = 'ika' THEN
SET vreturn = 0;
ELSE
SET vreturn = 1;
END IF
RETURN vreturn;
END $$
错误信息是:
错误1064 (42000):您的sql语法出现了错误;检查与MySQL服务器版本对应的手册,以获得使用接近“声明check INT4”的正确语法;开始
我们很感激你的帮
我试图使用hibernate压缩字符串列表并将其以JSON格式存储在mysql中:
List<String> options = new ArrayList<>();
String first = "What is your name?";
options.add(first);
String option = objectMapper.writeValueAsString(options);
// option value is: ["what is your name?"]
我的桌子结构:
CREATE TABLE question
我正在尝试转换从mysql转储的表。下面是我从mysql获得的用于创建表的代码:
CREATE TABLE "tbl_profession_attributes" (
"id" bigint(20) NOT NULL,
"tbl_profession_attribute_id" bigint(20) DEFAULT '0',
"code" varchar(10) NOT NULL,
"name" varchar(100) NOT NULL,
"keyword" text NOT NU
我试图使用一个计数函数来显示有多少员工拥有“销售”这个头衔。
if( !$connect)
{
die("ERROR: Cannot connect to database $db on server $server
using user name $user (".mysqli_connect_errno().
", ".mysqli_connect_error().")");
}
else
{
$userQuery = "COUNT(empID) FROM personnel WHERE jobTitle='Sales'
实际上,我正在使用Spring引导,在数据库中创建表时出现了错误。我使用MySQL作为我的数据库
错误是
错误显示在Spring 中
java.sql.SQLSyntaxErrorException: 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 'like bigint not null, time_created time, primary key (blog_i
我的数据库中有许多表,其中一个名为group是由代码创建的:
CREATE TABLE IF NOT EXISTS `sc`.`group` (
`id` INT NOT NULL AUTO_INCREMENT,
`name` VARCHAR(45) NOT NULL,
`description` TEXT NULL,
`group_type_id` INT NOT NULL,
PRIMARY KEY (`id`),
CONSTRAINT `fk_group_group_type1`
FOREIGN KEY (`group_type_id`)
REFER
我得到了下面的错误,不知道问题出在哪里。
错误:您的SQL语法有错误;请查看与您的MySQL服务器版本对应的手册,了解正确的语法,以便在第1行使用接近'order (orderid,customerid,productid,brand,model,price,amount,totalcost) V‘
//connect to database
$connection = mysql_connect("localhost","root","") or die ("Can't connect");
mysql_sele