How to insert array in database column that is type string?

Question

I am having multiple checkboxes in my form, and when I select some of them and submit form I get error 'Array to string conversion'. My database column user_preferences is type string so that is the reason of error. I just don't know how to insert multiple values in that column. Any help is appreciated. Here is my code.

UserController.php

public function update(StoreUserInfo $request, User $user)
{
    //return $request->userPreferences;
    if ( !($user->id == Auth::user()->id)) {
        abort(404);
    }

    $request->validated();

    $user->where('id', $user->id)->update(
        [
            'first_name' => $request->first_name,
            'last_name' => $request->last_name,
            'email' => $request->email,
            'city' => $request->city,
            'type' => $request->type,
            'user_preferences' => $request->userPreferences,
            'updated_at' =>  Carbon::now()
        ]
    );
        return redirect()->back()->with('message', 'User information updated');
}

edit.blade.php

<div class="row page-hero d-flex align-items-center justify-content-center">
    <label for="preferences" class="text-center">Select your preferences</label>         
</div>
<div class="row">                
    <div class="col-md-1" style="margin-right:15px; margin-left:60px;">
        <div class="custom-control custom-checkbox">
            <input id="kuca" name="userPreferences[]" value="kuca" type="checkbox" class="custom-control-input">
            <label class="custom-control-label" for="kuca">Kuca</label>
        </div>
    </div>
    <div class="col-md-1" style="margin-right:15px;">
        <div class="custom-control custom-checkbox">
            <input id="stan" name="userPreferences[]" value="stan" type="checkbox" class="custom-control-input">
            <label class="custom-control-label" for="stan">Stan</label>
        </div>
    </div>
    <div class="col-md-1" style="margin-right:50px;">
        <div class="custom-control custom-checkbox">
            <input id="apartman" name="userPreferences[]" value="apartman" type="checkbox" class="custom-control-input">
            <label class="custom-control-label" for="apartman">Apartman</label>
        </div>
    </div>
    <div class="col-md-1" style="margin-right:15px;">
        <div class="custom-control custom-checkbox">
            <input id="soba" name="userPreferences[]" value="soba" type="checkbox" class="custom-control-input">
            <label class="custom-control-label" for="soba">Soba</label>
        </div>
    </div>
    <div class="col-md-1" style="margin-right:15px;">
        <div class="custom-control custom-checkbox">
            <input id="lokal" name="userPreferences[]" value="lokal" type="checkbox" class="custom-control-input">
            <label class="custom-control-label" for="lokal">Lokal</label>
        </div>
    </div>
    <div class="col-md-1" style="margin-right:15px;">
        <div class="custom-control custom-checkbox">
            <input id="plac" name="userPreferences[]" value="plac" type="checkbox" class="custom-control-input">
            <label class="custom-control-label" for="plac">Plac</label>
        </div>
    </div>
    <div class="col-md-1" style="margin-right:15px;">
        <div class="custom-control custom-checkbox">
            <input id="garaza" name="userPreferences[]" value="garaza" type="checkbox" class="custom-control-input">
            <label class="custom-control-label" for="garaza">Garaza</label>
        </div>
    </div>           
</div>

You can serialize the values. Or you can make a relational mapping of 2 tables in mysql. Either works. — Andrei
– Andrei, Commented Jul 24, 2019 at 10:43
You can use php serialize function to to store as string and then unserialize the value while getting.....or you can also use json_encode to store and then json_decode the value while getting....but i would like to suggest you to store the multiple value in saparate table — Rajat Sharma
– Rajat Sharma, Commented Jul 24, 2019 at 10:50

Harun Yilmaz · Accepted Answer · 2019-07-24 11:03:07Z

3

You can use $casts property (https://laravel.com/docs/5.8/eloquent-mutators#array-and-json-casting):

class User extends Model
{

    protected $casts = [
        'user_preferences' => 'array',
    ];
}

By doing this, Laravel will automatically serialize array to JSON and vice versa.

edited Jul 24, 2019 at 11:03

answered Jul 24, 2019 at 10:49

Harun Yilmaz

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Comments

MOrez · Accepted Answer · 2019-07-24 10:47:11Z

2

Use json_encode

You can encode in your Update Query or you cast the field in the Model.

answered Jul 24, 2019 at 10:47

MOrez

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Comments

Latheesan · Accepted Answer · 2019-07-24 10:50:12Z

1

You could use php implode() function to join the string array into one string like this:

$user->where('id', $user->id)->update(
    [
        'first_name' => $request->first_name,
        'last_name' => $request->last_name,
        'email' => $request->email,
        'city' => $request->city,
        'type' => $request->type,
        'user_preferences' => implode(',', $request->userPreferences),
        'updated_at' =>  Carbon::now()
    ]
);

https://php.net/implode

Then when you are displaying the values in your show.blade.php, you can then use $userPreferences = explode(',', $row->userPreferences); to get the array back.

https://php.net/explode

answered Jul 24, 2019 at 10:50

Latheesan

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3 Comments

mrmar Over a year ago

It works, but know when I try and change first_name to update it I get this error 'implode(): Invalid arguments passed' . Can you help?

Latheesan Over a year ago

well, you should update your view to restore the selected values so that when you edit the first name, you are re-submitting the previously selected values back.

mrmar Over a year ago

I'm not sure I know how to do that. Can you help a bit?

Suraj15689 · Accepted Answer · 2019-07-24 10:50:20Z

1

You can implode the value with comma(,) separated and save the string.

For Example.

$preferences = implode(',',$request->userPreferences);
$user->where('id', $user->id)->update(
    [
        'first_name' => $request->first_name,
        'last_name' => $request->last_name,
        'email' => $request->email,
        'city' => $request->city,
        'type' => $request->type,
        'user_preferences' => $preferences,
        'updated_at' =>  Carbon::now()
    ]
);

answered Jul 24, 2019 at 10:50

Suraj15689

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Comments

Matthias S · Accepted Answer · 2019-07-24 10:50:51Z

1

I think the easiest solution is to store your value in JSON. There is a php function that allows you to encode and decode objects and arrays to JSON, namely json_encode and json_decode respectively.

In your case, you could simply change your update statement to:

$user->where('id', $user->id)->update(
        [
            'first_name' => $request->first_name,
            'last_name' => $request->last_name,
            'email' => $request->email,
            'city' => $request->city,
            'type' => $request->type,
            'user_preferences' => json_encode($request->userPreferences),
            'updated_at' =>  Carbon::now()
        ]
    );

When you get your data from the model, you need to decode the string back to an array. You can do that like this:

$user = User::find($user->id);
if ($user) {
    $preferences = json_decode($user->user_preferences);
}

As the other answers here stated, Laravel can automate this for you with Array & JSON Casting

answered Jul 24, 2019 at 10:50

Matthias S

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1 Comment

mrmar Over a year ago

If I do this, it works, but when I update for example later just first_name my values in user_preferences become null. How can I solve that?

Sagar Gautam · Accepted Answer · 2019-07-24 10:51:31Z

1

You can Use json_encode function while inserting the data to database.

After that you can type cast in Users Model so that you will get normal array while fetching records. Add following function to User Model.

public funtion getUserPreferencesAttribute()
{
     return json_decode($this->attributes['user_preferences']);
}

answered Jul 24, 2019 at 10:51

Sagar Gautam

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6 Comments

mrmar Over a year ago

If I do this, it works, but when I update for example later just first_name my values in user_preferences become null. How can I solve that?

Sagar Gautam Over a year ago

@Marko If you don't send the preferences then it will be null. you have to add these preferences while updating

mrmar Over a year ago

but I don't want every single time I update something in profile to have to click on those preferences. I want them to persist in database and to be checked.

Sagar Gautam Over a year ago

@Marko Then you need conditional update. If request has userPreferences then update the preferences column otherwise don't update this one.

Sagar Gautam Over a year ago

@Marko, It's not a column with single value, it contains array of preferences value, so you need to loop over it and check in blade.

|

ColinMD · Accepted Answer · 2019-07-24 10:52:15Z

1

If you are using Laravel and a more recent version of MySQL you may be better to create a field type JSON and cast it to an array in your eloquent model.

Then Laravel will automatically encode and decode the field for you.

It will also mean you can query the field by using model::where('preference->type', 'thing').

It may be possible to do this with a string field type but I'm not 100% on that but it will work with JSON.

answered Jul 24, 2019 at 10:52

ColinMD

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