How to make PHP or JavaScript spawn an HTML form

What is a way that I can make a button to spawn infinite forms? Any method would work, but I just can't think of any. This is kind of the layout I am thinking of Country | Province | Topic | China | Beijing | Flood | (+) China | Shaanxi | Flood | (+) In this imaginary example, the plus button would spawn another form where the values would be ident

如何让PHP或JavaScript产生一个HTML表单

什么是我可以制作按钮以产生无限形式的方式? 任何方法都可行,但我无法想到任何方法。 这是我想到的布局 Country | Province | Topic | China | Beijing | Flood | (+) China | Shaanxi | Flood | (+) 在这个虚构的例子中,加号按钮会产生另一种形式,其值将是相同的,但如果一个事件影响多个地方,则允许用户更改一个字段。 您可以使用jQuery随时创建表单元素,然后使用append

Using href in jquery with get method

i´m learning jquery and i have a question. On html, when i want to pass variables by get, i write: $subject=$_GET['subject']; $id_user=$_GET['id_user']; $id_pupil=$_GET['id_pupil']; <a href='process.php?subject=".$subject."&& id_user=".$id_user." && id_pupil=".$id_pupil."'>Add</a> But i want to do it on Jquery. My idea is: I have this part of code: while ($r

在jquery中使用href和get方法

我正在学习jquery,我有一个问题。 在html上,当我想通过get来传递变量时,我写道: $subject=$_GET['subject']; $id_user=$_GET['id_user']; $id_pupil=$_GET['id_pupil']; <a href='process.php?subject=".$subject."&& id_user=".$id_user." && id_pupil=".$id_pupil."'>Add</a> 但我想在JQuery上做。 我的想法是: 我有这部分代码: while ($row = mysql_fetch_array($result)) {

jquery: how to open an url in new tab with passing post data to that url

I have a php file localhost/~user/sample.php file which gets data from post method. <?php $you = $_POST["ids"]; $start= $_POST["start"]; echo $you."--".$start; I want to write a jquery code which will open the url "localhost/~user/sample.php" in a separate window on button click inside my html page and also pass the arguments required for it. I can use get method in php, but the

jquery:如何在新选项卡中打开一个URL并将发布数据传递给该URL

我有一个PHP文件localhost /〜user / sample.php文件,它从post方法获取数据。 <?php $you = $_POST["ids"]; $start= $_POST["start"]; echo $you."--".$start; 我想写一个jQuery代码,它将打开URL“localhost /〜user / sample.php”在一个单独的窗口中按钮点击我的HTML页面内,并传递它所需的参数。 我可以在php中使用get方法,但变量的数量更多 我可能会使用表单,如下所示: <form action="sample.php" method="po

Get Anchor tag class/id from table jQuery

This question already has an answer here: How to get the id of an anchor tag in jQuery? 2 answers use For getting class name $(".kimztableclass a").attr("class"); For getting Id $(".kimztableclass a").attr("id"); For getting text $(".kimztableclass a").text(); 使用attr()方法$('.mytableanchortagclass tr').find('a').attr('id') 这将返回类kimzanchorclass所有锚标记: $('a.kimzanchorclass

从表jQuery获取锚标记类/标识

这个问题在这里已经有了答案: 如何在jQuery中获取锚标记的ID? 2个答案 使用 获取类名 $(".kimztableclass a").attr("class"); 获取Id $(".kimztableclass a").attr("id"); 获取文本 $(".kimztableclass a").text(); 使用attr()方法$('.mytableanchortagclass tr').find('a').attr('id') 这将返回类kimzanchorclass所有锚标记: $('a.kimzanchorclass').attr('id');

Using a different MySql DB for write operations in CakePHP

I need to scale my mysql RDS instance. I can launch read replicas in amazon RDS, but problem is that the write operations won't work on read replicas. I need to redirect all the write operations (INSERT, UPDATE, DELETE) on master server somehow, and all the read operations on a read replica. RDS will take care of synchronizing the read replica with latest changes. is there any easy way

在CakePHP中使用不同的MySql DB进行写入操作

我需要扩展我的mysql RDS实例。 我可以在amazon RDS中启动只读副本,但问题是写操作无法在只读副本上工作。 我需要以某种方式重定向主服务器上的所有写操作(INSERT,UPDATE,DELETE),以及读副本上的所有读操作。 RDS将负责将读取副本与最新更改同步。 有没有简单的方法来修改蛋糕来执行不同的数据库上的所有写命令? 无需更改控制器或型号? 我现有的服务器无法处理连接和cpu峰值。 已经尝试升级它,但没有奏效。

laravel 4.2 model polymorphic function

so i am learning laravel and so far i have built an application with signup and sign in and some data which can be stored in the database. now i have 2 models: the users model which has its own validation set of rules, and another model contact which has its own set of rules.Taking a closer look at how laravel generates its models they both inherit from a base class called Eloquent. what i wou

laravel 4.2模型多态函数

所以我正在学习laravel,到目前为止,我已经构建了一个注册和登录应用程序以及一些可以存储在数据库中的数据。 现在我有两个模型:拥有自己的验证规则集的用户模型,以及具有自己的一组规则的另一个模型联系人。仔细研究laravel如何生成其模型,它们都从称为Eloquent的基类继承。 我想要的是一个多态的“isValid”函数,它只是验证用户输入与在函数被调用的类中创建的一组规则。但是现在,当我创建联系人模型时,我发现自己重复

PHPUnit code coverage in PhpStorm does not bootstrap Laravel

I'm having a strange problem on Laravel 5.5. I'm using PhpStorm and am trying to use it to make a phpunit call on my Vagrant host to PHPUnit like this: vagrant:///Users/mymachine/Webdev/project/usr/bin/php -dxdebug.coverage_enable=1 /home/vagrant/project/vendor/phpunit/phpunit/phpunit --coverage-clover /home/vagrant/.phpstorm_helpers/project_AdminLoginTest_coverage --bootstrap /home/va

PhpStorm中的PHPUnit代码覆盖不会引导Laravel

我在Laravel 5.5上遇到了一个奇怪的问题。 我正在使用PhpStorm,并试图使用它来将我的Vagrant主机上的phpunit调用改为PHPUnit,如下所示: 流浪汉:///用户/ mymachine上/ Webdev的/项目的/ usr / bin中/ PHP -dxdebug.coverage_enable = 1 /家庭/流浪者/项目/供应商/ PHPUnit的/ PHPUnit的/ PHPUnit的--coverage三叶草/home/vagrant/.phpstorm_helpers / project_AdminLoginTest_coverage --bootstrap /home/vagrant/project

Adding subdirectory to tests in Laravel; connection can't be resolved

I'm trying to add some structure into my /tests/ directory, at the moment files are structured like this: /tests /Models UserTest.php ExampleTest.php TestCase.php Both UserTest and ExampleTest are extended from TestCase. Running Exampletest works just fine (adds user to database), whereas UserTest fails with the following error: Fatal error: Call to a member function connection

将子目录添加到Laravel的测试中; 连接无法解决

我试图在我的/ tests /目录中添加一些结构,目前文件的结构如下所示: /测试 /楷模 UserTest.php ExampleTest.php TestCase.php UserTest和ExampleTest都是从TestCase扩展而来的。 运行Exampletest工作得很好(将用户添加到数据库),而UserTest失败并出现以下错误: 致命错误:调用第3340行的 Illuminate Datab ase Eloquent Model.php中的null成员函数connection() 这意味着这个函数接收null或没有参数

Laravel model event binding not called

I'm adding validation to a BaseModel class which all eloquent models will eventually extend in my application. The base class resembles: class BaseModel extends Eloquent { private $errors; private $validator; public function __construct() { $this->validator = new BaseValidator(get_class($this)); } public static function boot() { parent::boot(); static::creating(function($

未调用Laravel模型事件绑定

我将验证添加到BaseModel类中,所有雄辩的模型最终将在我的应用程序中扩展。 基类类似于: class BaseModel extends Eloquent { private $errors; private $validator; public function __construct() { $this->validator = new BaseValidator(get_class($this)); } public static function boot() { parent::boot(); static::creating(function($model) { return false; }); static::sa

Laravel model event saving is not firing

I'm trying to simulate what Ardent package is doing. Which is validating a model right before saving. I've created this BaseModel (According to Laravel Testing decoded book). And added this code : class BaseModel extends Eloquent { protected static $rules = []; public $errors = []; public function validate(){ $v = Validator::make($this->attributes, static::$ru

Laravel模型事件保存不会被解雇

我试图模拟Ardent软件包在做什么。 保存前正在验证模型。 我已经创建了这个BaseModel (根据Laravel Testing decoded书)。 并添加了以下代码: class BaseModel extends Eloquent { protected static $rules = []; public $errors = []; public function validate(){ $v = Validator::make($this->attributes, static::$rules); if($v->passes()) { return true;