php 使用 Twig 生成 JSON
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Using Twig to generate JSON
提问by russellmania
I want to have a URL that returns a simple JSON object. I am trying to use Twig to generate the JSON object:
我想要一个返回简单 JSON 对象的 URL。我正在尝试使用 Twig 生成 JSON 对象：
{
"urls": [
{% for child in page.root %}
"{{ child.url }}"{% if not loop.last %},{% endif %}
{% endfor %}
]
}
The carriage returns will not remain in place though, and I keep getting a result that looks like this:
但是回车不会保留在原位，我不断得到如下所示的结果：
{'urls':['../ants/','../brick-report/','../the-pollution-intervention/','../barclay/','../broken-advertising/','../aldat-n-densom/','../thisisart/','../there-she-goes-again/']}
which Jquery will not parse with it's ajax or getJSON methods. It's totally ignoring this JSON. How might I convince Twig to put the right whitespace in place? I've looked at the manual and it only seems concerned with NOT inserting whitespace.
Jquery 不会用它的 ajax 或 getJSON 方法解析。它完全忽略了这个 JSON。我如何说服 Twig 放置正确的空格？我看过手册，似乎只关心不插入空格。
回答by biera
That works for me (twig template):
这对我有用（树枝模板）：
var parsedJSON = JSON.parse('{{ ['one', 'two', 'three']|json_encode|e('js') }}');
And that:
然后：
console.log(parsedJSON);
will output:
将输出：
Array ["one", "two", "three"]
in FF console.
在 FF 控制台中。
回答by DasBaconfist
Twig has a filter for this.
Twig 有一个过滤器。
json_encode, it uses PHP json_encode function.
json_encode，它使用 PHP json_encode 函数。
for your case:
对于您的情况：
{{ {'urls': page.root}|json_encode }}
will output
会输出
{"urls":["..\/ants\/","..\/brick-report\/","..\/the-pollution-intervention\/","..\/barclay\/","..\/broken-advertising\/","..\/aldat-n-densom\/","..\/thisisart\/","..\/there-she-goes-again\/"]}
the code is tested and works. For more information take a look at the Twig Documentation for json_encode.
代码经过测试并有效。有关更多信息，请查看json_encode的 Twig 文档。
回答by Alain Tiemblo
Don't use Twig to generate your json response.
不要使用 Twig 来生成你的 json 响应。
In your controller, use:
在您的控制器中，使用：
return new Response(json_encode($var));
Sample:
样本：
public function sampleAction()
{
$urls = array('../test', '../something', '../sample');
return new Response(json_encode($var));
}
If URLs are generated from Symfony2 routes, you can use:
如果 URL 是从 Symfony2 路由生成的，您可以使用：
public function sampleAction()
{
$urls = array(
$this->generateUrl('my_test'),
$this->generateUrl('my_something'),
$this->generateUrl('my_sample'),
);
return new Response(json_encode($var));
}
回答by Ravi Chauhan
Thats easy if you extend twig.
如果你延长树枝，那很容易。
First, create a class that will contain the extension:
首先，创建一个包含扩展的类：
<?php
namespace Acme\DemoBundle\Twig\Extension;
use Symfony\Component\DependencyInjection\ContainerInterface;
use \Twig_Extension;
class VarsExtension extends Twig_Extension
{
protected $container;
public function __construct(ContainerInterface $container)
{
$this->container = $container;
}
public function getName()
{
return 'some.extension';
}
public function getFilters() {
return array(
'json_decode' => new \Twig_Filter_Method($this, 'jsonDecode'),
);
}
public function jsonDecode($str) {
return json_decode($str);
}
}
回答by vivek
Basically the $.getJson()method requires json but ther is a string so you can use $.get()to get the response and use the parser to parse the string to JSON
基本上$.getJson()方法需要 json 但有一个字符串，因此您可以使用$.get()获取响应并使用解析器将字符串解析为 JSON
$.get("ffff",function(data){
// user parser
JSON.parse();
});
