Question

我的数据库中有下表。

Schema::create('projects', function (Blueprint $table) {
    $table->increments('id');
    $table->string('title');
    $table->string('url');
    $table->string('requirements');
    $table->string('coverImage');
    $table->string('domain');
    $table->text('feedbacks');
    $table->text('technologies_id');
    $table->timestamps();
});

我想在我的technologies_id中放入一个数组，例如：

technologies_ids: [1,2,3]

我尝试了以下方法。

$project = new Project;
$project->title = $request->title;
$project->url = $request->url;
$project->requirements = $request->requirements;
$project->coverImage = $request->coverImage;
$project->customer_id = $request->customer_id;
$project->domain = $request->domain;
$project->feedbacks = $request->feedbacks;
$project->technologies_id = json_encode($request->technologies_id);
$project->services_id = $request->services_id;

错误

数组到字符串的转换（SQL：插入projects

Answer 1

在项目模型中添加

protected $casts=['technologies_id'=>'array']

删除json_encode

$project->technologies_id = $request->technologies_id;

并在您的刀片文件中

technologies_id必须是复选框中的数组，或选择

 name="technologies_id[]"

存储数据时，从数据库获取数据时将自动进行编码和解码

在文档中阅读

https://laravel.com/docs/5.8/eloquent-mutators#array-and-json-casting

Answer 2

您可以序列化数组，使其可以是字符串。

$technologies_id = serialize(json_encode($request->technologies_id);

然后，如果要再次将其转换为数组，请使用unserialize($array)

希望有帮助！

Answer 3

将数据存储在列中时，请使用implode()

$project->technologies_id = implode(",",$request->technologies_id);

然后，当您从列中获取或获取数据时，请使用explode()

$project = Project::find(1);
$techonolgy_ids = explode(',',$project->technologies_id);

如何将数组保存在一个列中？

3 个答案: