phramework / validate-filler
填充由 validate (JSON 规范) 生成表单
1.3.0
2022-05-13 12:20 UTC
Requires
- php: ^7.2.0 | ^8.0.0
- ext-json: *
- fzaninotto/faker: ^1.5
- phramework/validate: ^v0.12.0
Requires (Dev)
- codacy/coverage: ^1.0
- php-coveralls/php-coveralls: ^2.1
- phpunit/phpunit: ^9.3.0
- squizlabs/php_codesniffer: ^3.5.2
This package is auto-updated.
Last update: 2024-09-25 12:19:56 UTC
README
填充由 validate (JSON 规范) 生成表单
用法
使用 composer 安装包
composer require phramework/validate-filler
示例:从 json 解析 schema
<?php $validator = \Phramework\Validate\ObjectValidator::createFromJSON('{ "type": "object", "properties": { "a": { "type": "string", "enum": [ "1", "2", "3" ] }, "b": { "type": "string", "enum": [ "i", "ii", "iii" ] } }, "required": ["a"] }'); $value = (new \Phramework\ValidateFiller\Filler()) ->fill($validator); var_dump($value);
示例输出
class stdClass#1381 (1) { public $a => string(1) "2" }
class stdClass#1381 (2) { public $a => string(1) "3" public $b => string(2) "ii" }
- 将始终包含属性
"a"
,因为它需要 - 有时会包含属性
"b"
(概率性)
示例:使用 ObjectValidator 构造函数
<?php $validator = new \Phramework\Validate\ObjectValidator( (object) [ 'a' => new \Phramework\Validate\EnumValidator([ '1', '2', '3' ]), 'b' => new \Phramework\Validate\EnumValidator([ 'i', 'ii', 'iii' ]) ], ['a'], false ); $value = (new \Phramework\ValidateFiller\Filler()) ->fill($validator);
开发
安装依赖项
composer update
测试和代码格式检查
composer test
composer lint
生成文档
composer doc
许可证
版权 2015-2017 Xenofon Spafaridis
根据 Apache 许可证 2.0 版(“许可证”)许可;除非适用法律要求或经书面同意,否则不得使用此文件。您可以在以下位置获得许可证的副本:
https://apache.ac.cn/licenses/LICENSE-2.0
除非适用法律要求或书面同意,否则在许可证下分发的软件按“原样”提供,不提供任何明示或暗示的保证或条件。有关许可证的具体语言管理权限和限制,请参阅许可证。