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New xampp security concept: Access Forbidden Error 403 - Windows 7 - phpMyAdmin

由 ￣綄美尐妖づ 提交于 2020-01-08 15:52:01
问题 I have downloaded and installed XAMPP 1.8.1 for Windows on Windows 7 Ultimate. I have set up XAMPP to run together with IIS as per these instructions All good so far, my PHP sites run locally and everything except phpMyAdmin is available from the XAMPP menu. However when I try to access phpMyAdmin I get this error: Access forbidden! New XAMPP security concept: Access to the requested directory is only available from the local network. This setting can be configured in the file "httpd-xampp

New xampp security concept: Access Forbidden Error 403 - Windows 7 - phpMyAdmin

由 与世无争的帅哥 提交于 2020-01-08 15:51:30
问题 I have downloaded and installed XAMPP 1.8.1 for Windows on Windows 7 Ultimate. I have set up XAMPP to run together with IIS as per these instructions All good so far, my PHP sites run locally and everything except phpMyAdmin is available from the XAMPP menu. However when I try to access phpMyAdmin I get this error: Access forbidden! New XAMPP security concept: Access to the requested directory is only available from the local network. This setting can be configured in the file "httpd-xampp

php-lnmp环境搭建

由 僤鯓⒐⒋嵵緔 提交于 2020-01-07 21:51:00
参考网站： http://www.liyblog.top/p/9 1.nginx和php基本安装 1.更新apt apt update 2.安装nginx apt install nginx 3.查看nginx状态 systemctl status nginx 4.安装php apt install php php-fpm apt-get install php7.2 php7.2-fpm 5.查看php版本（ubuntu 默认安装php7.2） php -v 2.nginx配置php-fpm 1.查看nginx具体位置，查找nginx配置文件位置（当前是在 /etc/nginx ） whereis nginx 2.进入站点配置目录 cd /etc/nginx/sites-available 3.配置站点信息 server配置注意： 1.listen 80; 设置监听端口 2.root /var/www/html 设置项目目录 3.index index.php 设置站点目录入口文件 4.server_name www.liyblog.top 设置站点域名 5.php配置见下图代码 6.fastcgi_pass unix:/var/run/php/php7.0-fpm.sock; （配置socket，要配置相对应版本的php） 7.删除/etc/nginx/sites

Linux centos7 安装 phpMyAdmin

由 被刻印的时光 ゝ 提交于 2020-01-07 16:41:06
yum install httpd php mariadb-server –y 搭建lamp运行环境之后安装phpMyAdmin遇到的一些问题记录一下 1.官网下载phpMyAdmin压缩包 wget -c https://files.phpmyadmin.net/phpMyAdmin/4.9.3/phpMyAdmin-4.9.3-all-languages.tar.gz 2.解压缩到/var/www/html/ tar -zxvf phpMyAdmin-4.9.3-all-languages.tar.gz -C /var/www/html/ 3.重命名 cd /var/www/html/ mv phpMyAdmin-4.9.3-all-languages phpMyAdmin 到这一步一般都可以正常通过localhost/phpMyAdmin访问了，但是有时候总会出现些其他问题。 4.进入到phpMyAdmin安装目录下复制phpmyadmin的简单配置文件config.sample.inc.php，作为默认配置文件config.inc.php 复制文件 cp config.sample.inc.php config.inc.php 编辑配置文件 vim config.inc.php 配置文件现在需要一个短语密码，找到 $cfg['blowfish_secret'] = '';

How to set a foreign key which is dependent on the relation of other two tables?

由 我怕爱的太早我们不能终老 提交于 2020-01-07 14:02:12
问题 I have 3 tables; teachers table, subjects table and events table. 1.There is one to many relationship between subjects and teachers, that is each teacher can teach only one subject but, many teachers can teach same subjects. 2.There is a many to many relationship between teachers and events. Example of subjects table id(PK) | name ------------------- 1 | php ------------------- 2 | java ------------------- 3 | python ------------------- 4 | c++ -------------------- 5 | c# Example of teachers

How to insert dynamic array into the database in php?

由 為{幸葍}努か 提交于 2020-01-07 09:51:15
问题 i am working on dynamic array i need to insert these array in the database.when i insert dynamic array into the database instead of displaying in one row its cover three row. coding for insertion this is my array $asma[]=GA::select($ga->population,'total',3); this is insertion query i need to insert the detail in six column as my output comprises of six values . $Voltage = array(); $Duration = array(); $Number = array(); foreach($asma as $key => $value) { foreach ( $value as $ind => $hObject

Creating relations between tables PHPMYADMIN

由 假装没事ソ 提交于 2020-01-07 09:36:12
问题 Please help me! Im creating a database to store data about patients in an A&E Department. However, Im unsure how to relate the tables. Table structure: PATIENTS (PatientID(PK),Forename, surname, gender, DOB, Address, History, illness, priority) A&E (ID(PK), PatientID(FK), address, city, postcode, telenumber) NURSE (NurseID(PK), forename, surname) CONDITION (ID(PK), PatientID(FK) symptoms, diagnosis, treatment) Basically the relationships between these are: PATIENT attends A&E PATIENT seen_by

WHERE clause effecting SQL query

由 a 夏天 提交于 2020-01-07 08:32:12
问题 I am trying to make this program where I can delete a thread if I am logged in. Now I already have the button linked and everything, I have it doing multiple tasks when pressed, but it seems to not run the SQL query I want it to. Now I have a variable called $forumid which is set in the URL and retrieved using $_GET['forumid']; I know this is setting properly, because I have done echo $forumid; and its been correct. But there is one line of code that doesn't run for some reason, and that is:

WHERE clause effecting SQL query

由 馋奶兔 提交于 2020-01-07 08:32:12
问题 I am trying to make this program where I can delete a thread if I am logged in. Now I already have the button linked and everything, I have it doing multiple tasks when pressed, but it seems to not run the SQL query I want it to. Now I have a variable called $forumid which is set in the URL and retrieved using $_GET['forumid']; I know this is setting properly, because I have done echo $forumid; and its been correct. But there is one line of code that doesn't run for some reason, and that is:

Can't upload text value to mysql by php [duplicate]

由 て烟熏妆下的殇ゞ 提交于 2020-01-07 08:06:31
问题 This question already has an answer here : Can't upload text value to mysql (1 answer) Closed 2 years ago . Every time I submit the form of my website for comment, It only creates a new ID.But all other value is empty, what should I do? Do I need create some attribute for the name, email and content? <?php error_reporting(E_ALL);ini_set('display_errors',1); $servername = "localhost"; $username = "seamaszhou"; $password = "123456"; $dbname = "guest"; $conn = new PDO("mysql:host=$servername
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