我有一个静态View
类,它从另一个类传递了一个字符串。当字符串作为变量传递时,它会起作用。当我将其更改为常量时,错误是:
[17-Feb-2016 19:08:48 欧洲/柏林] PHP 警告: include(): 失败 开放 '/Applications/MAMP/htdocs/its_vegan/scripts/back_end/views/template' 供收录 (include_path='.:/Applications/MAMP/bin/php/php7.0.0/lib/php') in /Applications/MAMP/htdocs/its_vegan/scripts/back_end/views/view.php on 23号线
class View {
/**
* -------------------------------------
* Render a Template.
* -------------------------------------
*
* @param $filePath - include path to the template.
* @param null $viewData - any data to be used within the template.
* @return string -
*
*/
public static function render( $filePath, $viewData = null ) {
// Was any data sent through?
( $viewData ) ? extract( $viewData ) : null;
ob_start();
include ( $filePath );// error on this line
$template = ob_get_contents();
ob_end_clean();
return $template;
}
}
class CountrySelect {
const template = 'select_template.php'; //the const is template
public static function display() {
if ( class_exists( 'View' ) ) {
// Get the full path to the template file.
$templatePath = dirname( __FILE__ ) . '/' . template; //the const is template
$viewData = array(
"options" => '_countries',
"optionsText" => 'name',
"optionsValue" => 'geonameId',
"value" => 'selectedCountry',
"caption" => 'Country'
);
// Return the rendered HTML
return View::render( $templatePath, $viewData );
}
else {
return "You are trying to render a template, but we can't find the View Class";
}
}
}
在国家选择中有什么工作:
$templatePath = dirname( __FILE__ ) . '/' . static::$template;
为什么模板必须是静态的?我可以把它变成一个静态常数吗?
你也可以
使用self::template
由于类常量是在每个类级别而不是每个对象定义的,因此static::template
将引用相同的内容,除非您有一个子类。 (见 https://secure.php.net/manual/en/language.oop5.late-static-bindings.php)
template
是指全局常数(例如define('template', 'value');
)
在此行
$templatePath = dirname( __FILE__ ) . '/' . template;
template
不是常量,因为常量template
类中声明的。此代码的工作方式类似
$templatePath = dirname( __FILE__ ) . '/template';
因此,请使用static::template