Symfony2形式为JSON结构 [英] Symfony2 form to JSON structure
问题描述
Symfony2表单如何转换为JSON数据结构?例如:
$ builder
- > add('name','text')
- > add('password','password')
;
会导致类似的情况:
{
字段:{
名称:{
类型:'text'
},
密码:{
类型:'密码'
}
}
}
在 $ form = $ this-> createForm(new FormType(),new Entity())
之后遍历每个元素并没有帮助,找不到某些属性可以在表单构建器中定义。
我假设你想在控制器中获得这些信息,在这种情况下,您可以很容易地从表单对象中获取底层实体,如下所示:
$ entity = $ form-> ;的getData();
此时,您可以手动将所需的字段拖放到数组中,并将 json_encode()
,或者...实现 JsonSerializable
接口,然后直接使用 json_encode()
对象本身。
例如:
<?php
namespace FooApp / BarBundle /实体;
使用JsonSerializable;
class Baz实现了JsonSerializable
{
private $ name;
私人$密码;
// ...
函数jsonSerialize()
{
return [
'fields'=> [
'name'=> ['type'=> $ this-> name],
'password'=> ['type'=> $ this->密码],
],
];
然后,在你的控制器中:
$ entity = $ form-> getData();
$ json = json_encode($ entity);
调用 json_encode()
会自动调用<
$ strong>更新2016年6月23日
我偶然发现了这个问题 - 并且...我意识到我没有回答您的实际题。
您不希望将表单的底层实体转换为JSON,而是希望将表单结构表示为数据。我很抱歉误会 - 希望我可以通过更新来纠正它。
这是一个概念验证,应该适用于非嵌套形式(尽管它应该直截了当地为这种情况创建递归版本或其他东西)。但是,假设已经实例化了一个表单,其中包含字段 name
和 password
,如下所示:
$ form = $ this-> createForm(FooType :: class,$ foo);
然后应该遍历实例并派生出结构的表示;例如:
$ fields = ['fields'=> []];
foreach($ form-> all()as $ field){
$ name = $ field-> getName();
$ type = $ field-> getConfig() - > getType() - > getBlockPrefix();
$ fields ['fields'] [$ name] = ['type'=> $类型]
}
echo json_encode($ fields,JSON_PRETTY_PRINT);
收益率:
<
字段:{
name:{
type:text
},
password:{
type:password
}
}
}
希望这有助于:)
How can the Symfony2 form be transformed to JSON data structure? Looking for proper bundle gave me no results;
Example:
$builder
->add('name', 'text')
->add('password', 'password')
;
Would result in something like that:
{
fields: {
name: {
type: 'text'
},
password: {
type: 'password'
}
}
}
Iterating over each element in form after $form = $this->createForm(new FormType(), new Entity())
was not helpful, could not find some properties that could be defined in form builder.
I assume that you want to get this information in a controller once you have posted the form, in which case you can easily get the underlying entity from the form object, like so:
$entity = $form->getData();
At this point you can either manually pull out the fields you want into an array and json_encode()
that, or... implement the JsonSerializable
interface in your entity and then directly json_encode()
the object itself.
For example:
<?php
namespace FooApp/BarBundle/Entity;
use JsonSerializable;
class Baz implements JsonSerializable
{
private $name;
private $password;
// ...
function jsonSerialize()
{
return [
'fields' => [
'name' => ['type' => $this->name],
'password' => ['type' => $this->password],
],
];
}
}
Then, in your controller:
$entity = $form->getData();
$json = json_encode($entity);
Calling json_encode()
will automatically invoke Baz::jsonSerialize()
and return the array structure you defined, which in turn is JSON-encoded.
Update 2016-06-23
I happened across this question again by chance - and... I realise that I didn't answer your actual question.
You didn't want to convert the form's underlying entity to JSON - instead you want to represent form structure as data. My apologies for misunderstanding - hopefully I can rectify that with an update.
This is a proof-of-concept that should work for a non-nested form (although it should be straightforward to create a recursive version or something for that case). But, assuming a scenario where you have instantiated a form, comprising of fields name
and password
, like so:
$form = $this->createForm(FooType::class, $foo);
It should then possible to iterate over the instance and derive a representation of the structure; e.g:
$fields = ['fields' => []];
foreach ($form->all() as $field) {
$name = $field->getName();
$type = $field->getConfig()->getType()->getBlockPrefix();
$fields['fields'][$name] = ['type' => $type];
}
echo json_encode($fields, JSON_PRETTY_PRINT);
Yields:
{
"fields": {
"name": {
"type": "text"
},
"password": {
"type": "password"
}
}
}
Hope this helps :)
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