php 将视图加载到变量中
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Load view into a variable
提问by Sml004
Is there any way that I can get content of a PHP file in to variable?
有什么方法可以将 PHP 文件的内容放入变量中？
I want to do this
我想做这个
$msg = $this->load->view('some_view');
but when I do this, $msgis NULL.
但是当我这样做时，$msg是NULL。
Is it possible?
是否可以？
回答by Yan Berk
It is possible:
有可能的：
$msg = $this->load->view('some_view', '', true);
回答by Madan Sapkota
There is a third optional parameter lets you change the behavior of the function so that it returns data as a string rather than sending it to your browser. This can be useful if you want to process the data in some way. If you set the parameter to true (boolean) it will return data. The default behavior is false, which sends it to your browser. Remember to assign it to a variable if you want the data returned:
还有第三个可选参数可让您更改函数的行为，使其以字符串形式返回数据，而不是将其发送到浏览器。如果您想以某种方式处理数据，这会很有用。如果您将参数设置为 true（布尔值），它将返回数据。默认行为是 false，它将它发送到您的浏览器。如果要返回数据，请记住将其分配给变量：
$msg = $this->load->view('some_view', '', true);
Source : http://codeigniter.com/user_guide/general/views.html
回答by Pavan Gosavi
$string = $this->load->view('myfile', '', TRUE);
https://codeigniter.com/user_guide/general/views.html#returning-views-as-data
https://codeigniter.com/user_guide/general/views.html#returning-views-as-data
