Is this correct? If not what is the correct syntax I am new to php hence trying to learn. <?php // Check browser for JavaScript support $jsSupport='true'; ?> <noscript><?php $jsSupport='false'; ?></noscript> <?php if ($jsSupport == 'false') { include ('no-script-layout.php'); } else { include ('re
它是否正确？ 如果不是什么是正确的语法 我是新来的PHP，因此试图学习。 <?php // Check browser for JavaScript support $jsSupport='true'; ?> <noscript><?php $jsSupport='false'; ?></noscript> <?php if ($jsSupport == 'false') { include ('no-script-layout.php'); } else { include ('regular-layout.php');
I got a docker-compose setup with two containers: One is the php/apache service and the other container is the database (mysql). Here is my docker-compose.yml version: '2' services: app: depends_on: - db links: - db:mysql build: . image: app ports: - "80:80" restart: always links: - db:db volumes: - ../:/var/www/html
我有两个容器的docker-compose设置：一个是php / apache服务，另一个容器是数据库（mysql）。 这是我的docker-compose.yml version: '2' services: app: depends_on: - db links: - db:mysql build: . image: app ports: - "80:80" restart: always links: - db:db volumes: - ../:/var/www/html/ db: image: mysql:latest rest
Before everyone starts butting in with "security risks" "cant be done" stop there and read the ENTIRE post I have a web server set up from a home laptop which is serving as a games web server im trying to create a GUI so its easier for us to maintain the server and im trying to use batch files to do the actions on the computer So to put this into perspective I have my inde
在所有人开始接触“安全隐患”之前，“不能做”停在那里并阅读整篇文章 我有一个从家用笔记本电脑设置的网络服务器，它充当游戏网络服务器，我尝试创建一个GUI，以便我们更容易维护服务器，并尝试使用批处理文件在计算机上执行操作 因此，为了使这个透视我有我的索引文件index.php <form method="post"> <input type="submit" name="startServer" value="Start Server"> </form> <? if(isset($_POST['
As noted on Reddit's LOL PHP sub, PHP 7 may use either the extended class or the base class when referring to self , in contrast to PHP 5 which always refers to the extended class. <?php class Foo { const A = "FooA"; const B = self::A . self::C; const C = "FooC"; } class Bar extends Foo { const A = "BarA"; const C = "BarC"; } var_dump(Bar::B); Try it online PHP
正如Reddit的LOL PHP sub中所提到的，PHP 7在引用self时可以使用扩展类或基类，而PHP 5通常指扩展类。 <?php class Foo { const A = "FooA"; const B = self::A . self::C; const C = "FooC"; } class Bar extends Foo { const A = "BarA"; const C = "BarC"; } var_dump(Bar::B); 在线试用 PHP 5 string(8) "BarABarC" PHP 7 string(8) "FooABarC" PHP 7的行为尤其令人担忧，因为似乎没有任
这里的代码！不工作，请帮助。 class user{ var idgen; function uid(){ //return uniqid (rand(), true); return "123"; } function gen() { $this->idgen=$this->uid(); //$this->idgen=udi();//even this dint work } function pint() { echo "The id is :".$this->idgen; } } $use= new user(); $use->gen(); $use->pint(); Change the second line of your c
这里的代码！不工作，请帮助。 class user{ var idgen; function uid(){ //return uniqid (rand(), true); return "123"; } function gen() { $this->idgen=$this->uid(); //$this->idgen=udi();//even this dint work } function pint() { echo "The id is :".$this->idgen; } } $use= new user(); $use->gen(); $use->pint(); 将代码的第二行更改为 private $i
For my application, written in PHP 5+, I have a common.php which is included from all other pages. Within that I have an include sanitize.php which aims to sanitise any input vars used in the URL. So, targetting $_GET[] values. This is just to have one place where I can tidy any vars, if used, and use them in the code later. There seems to be no tidy way, I've seen, to sanitise based on
对于我用PHP 5+编写的应用程序，我有一个common.php ，它包含在所有其他页面中。 在那里我有一个include sanitize.php ，它旨在清理URL中使用的任何输入变量。 所以，定位$_GET[]值。 这只是为了让我有一个地方可以整理任何变量（如果使用的话），然后在代码中使用它们。 基于预期/期望的投入，似乎没有一种整洁的方式来净化。 我最初看到的方法是这个sanitize.php有一个foreach循环遍历所有变量，查找所需的清理类型，然
I have an XML file with many member entries, formatted like so: <staff> <member> <name></name> <image></image> <title></title> <email></email> <phone></phone> <location></location> <info></info> <webTitle></webTitle>
我有一个包含许多成员条目的XML文件，格式如下： <staff> <member> <name></name> <image></image> <title></title> <email></email> <phone></phone> <location></location> <info></info> <webTitle></webTitle> <webURL>&
I'm looking at a PHP logic problem for me, but I can't fix it, so I ask you guys here. How can I call/combine the underneath PHP img with the var string method? I would like to call an image with PHP: <img src="<?php echo $this->getSkinUrl('images/prev.gif'); ?> But I would like to call the php from the following string: var str_buffer = new String ( "<table width="1
我正在为我寻找一个PHP逻辑问题，但我无法解决它，所以我在这里问你们。 我如何调用/结合下面的PHP img与var string方法？ 我想用PHP调用一个图像： <img src="<?php echo $this->getSkinUrl('images/prev.gif'); ?> 但我想从以下字符串中调用php： var str_buffer = new String ( "<table width="100%">n"+ "<tr>n <td>n"+ "<img src="prev.gif" width="16" height="16" border
var $foo = array('foo' => $bar); I am getting an UNEXPECTED T_VARIABLE error. I can't use variables when creating arrays? This declaration is inside a class, and I am running PHP v5.3.2 When removing the var, I get another error Parse error: syntax error, unexpected T_VARIABLE, expecting T_FUNCTION Thanks var is not PHP syntax... A simple $foo = array('foo' => $bar); would su
var $foo = array('foo' => $bar); 我收到一个意外的T_VARIABLE错误。 创建数组时，我不能使用变量？ 这个声明是在一个类里面的，我正在运行PHP v5.3.2 当删除var时，我得到另一个错误Parse error: syntax error, unexpected T_VARIABLE, expecting T_FUNCTION 谢谢 var不是PHP语法...一个简单的 $foo = array('foo' => $bar); 就足够了。 关键字var仅用于在类中声明变量（即实例变量），但即使这是PHP4语法
It is my code: <?php if (empty($_COOKIE["count"]) && empty($_COOKIE["date"])) { setcookie("count",1); setcookie("date",date("d.m.y H:i")); } else { var $c=$_COOKIE["count"]; } ?> And I get error: expecting statement before var $c. How can I solve my problem? The var keyword is deprecated and was for use in classes, which this code snippet is not. No
这是我的代码： <?php if (empty($_COOKIE["count"]) && empty($_COOKIE["date"])) { setcookie("count",1); setcookie("date",date("d.m.y H:i")); } else { var $c=$_COOKIE["count"]; } ?> 我得到错误： 期望 var $ c之前的声明 。 我如何解决我的问题？ var关键字被弃用，并用于类中，而这段代码片段不是。 注意：由于兼容性原因（作为public关键字的同义词），仍然支持使