我可以在JSONstring的第一个深度拆分，而不需要重新编码吗？
我可以在JSONstring的第一个深度拆分，而不需要重新编码吗？
提问于 2018-06-08 21:22:57
因为我不知道behat是否可以使用字符串以外的任何参数来注入Behat FeatureContext函数参数，所以我想知道是否可以以这样一种方式拆分字符串，使我只剩下一个json_objects数组。
我已经用json_decode和json_encode做到了这一点，但这感觉有点重复，因为我首先对对象字符串进行解码，然后将其编码回单个对象。
每个示例都具有以下Behat功能：
Feature: Provide a consistent standard JSON API endpoint
In order to build interchangeable front ends
As a JSON API developer
I need to allow Create, Read, Update, and Delete functionality
Background:
Given there are Albums with the following details:
"""
[{
"artist":"Pink Floyd",
"title":"The Dark Side of the Moon",
"songs":[
{"title":"Speak to Me", "length":254}
{"title":"Breathe", "length":192}
{"title":"On the Run", "length":188}
]
},
{
"artist":"AC/DC",
"title":"Back to Black",
"songs":[
{"title":"Hells Bells", "length":205}
{"title":"Shoot to Thrill", "length":302}
{"title":"What Do You Do for Money Honey", "length":244}
]
}]
"""
And the "Content-Type" request header is "application/json"以及FeatureContext.php中的以下函数：
...
public function thereAreAlbumsWithTheFollowingDetails(string $jsonString) {
$albums = json_decode($jsonString, true);
foreach ($albums as $album) {
$albumJson = json_encode($album);
$this->apiContext->setRequestBody($albumJson);
$this->apiContext->requestPath("/api/album", "POST");
}
}
...回答 2
Stack Overflow用户
回答已采纳
发布于 2018-06-25 21:59:07
我遵循了一个教程，该教程通过REST添加了第一个测试对象，但这并不是必须发生的。
我还了解到，Behat的嵌套表不是一个好主意。
https://joebuschmann.com/specflow-nested-tables-a-bad-idea/
behat.yml
default:
suites:
default:
contexts:
- App\Features\Bootstrap\FeatureContext:
container: '@service_container'
entityManager: "@doctrine.orm.default_entity_manager"
- Imbo\BehatApiExtension\Context\ApiContext
extensions:
Imbo\BehatApiExtension:
apiClient:
base_uri: http://127.0.0.1:8000
Behat\Symfony2Extension:
kernel:
bootstrap: config\behat\bootstrap.php
class: App\Kernel下面的代码并不完全正确，但已经足够接近了。
FeatureContext.php
...
/**
* @Given there are Albums with the following details:
*/
public function thereAreAlbumsWithTheFollowingDetails(TableNode $table) {
foreach ($table->getColumnsHash() as $row) {
$album = new Album();
$album->setTitle($row['title']);
$album->setReleaseDate(new \DateTime($row['release_date']));
array_push($this->entities, $album);
}
$this->em->flush();
}
/**
* @Given albums with the following Songs:
*/
public function albumsWithTheFollowingSongs(TableNode $table) {
$i = 0;
foreach ($table->getColumnsHash() as $row) {
$song = new Song($row['title']);
$this->entities[$i]->setSong($song);
$i = $i + 1;
}
}
/**
* @When they are saved into the database
*/
public function theyAreSavedIntoTheDatabase() {
foreach ($this->entities as $entity) {
$this->em->persist($entity);
}
$this->em->flush();
}
...Stack Overflow用户
发布于 2018-06-09 05:49:44
据我所知，您希望添加一些数据来设置场景，在这种情况下，我将不再使用json，因为这是一个实现细节：
Given there are Albums with the following details:
| artist | title | songs |
| Pink Floyd | The Dark Side of the Moon | Speak to Me, Breathe, On the Run |
| AC/DC | Back to Black | Hells Bells, Shoot to Thrill, What Do You Do for Money Honey |
...然后在FeatureContext上将数据转换为json (如果需要的话)，但就我个人而言，如果您这样做是正确的，那么我将只注入应该在/api/ Albums控制器中用来创建专辑的相同服务。
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原文链接：
https://stackoverflow.com/questions/50761612
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