英文:
Search in json column in framework laravel
问题
你想获取这个记录吗？
英文:
I have a column which contains
[
{
"type": "content",
"data": {
"content": "<p>\u043a\u0432\u0435\u0440\u0438\u0431\u0438\u043b\u0434\u0435\u0440<\/p>"
}
}
]
How can I get this record?
My code is:
$q = $request->get('q');
$posts = Post::query()
->where('active', '=', true)
->where('published_at', '<=', Carbon::now())
->latest('published_at')
->where(function($query) use ($q) {
$query->whereJsonContains("content", [['type' => 'content', 'data' => ['content' => "%$q%"]]]);
})
->paginate(10);
答案1
得分: 1
Sure, here are the translated parts:
1 将您的 JSON 字段添加到 cast
protected $casts = [
'content' => 'array',
];
2 我使用俄语编码时，从数组编码为 JSON 时使用标志：
`$data['content'] = json_encode($data['block'], JSON_UNESCAPED_UNICODE);`
3 我的数据库查询如下：
`$posts = Post::query()->where("content", 'like', "%$someValue%")->paginate(10);`
英文:
My solution:
1 Add your json field to cast
protected $casts = [
'content' => 'array',
];
2 I work with russian language, when i encode to json from array i use flag:
`$data['content'] = json_encode($data['block'], JSON_UNESCAPED_UNICODE);`
3 My query in db looks like
`$posts = Post::query()->where("content", 'like', "%$someValue%")->paginate(10);`
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