How to make dynamic link from array value in laravel?

Question

The purpose that I want to make a link from an array value is the link will show detail's modal.

So I have this array:

array:1 [▼
  42 => "00285",
  43 =? "00123"
]

I found some method like this:

foreach ($certificate->pluck('number', 'id')->toArray() as $href => $text) {
    if ($href != $text) {
        echo '<a href="?page=' . $href . '">' . $text . '</a>, ';
    } else {
        echo $text;
    }
}

But in some case, that my controller will break to the next grid "echo" must be "return",

Example grid that I wanted:

But the above code will show like this:

Not showing the values inside the grid at first picture

EDIT:

My code from picture 1 - [laravel-admin]:

$grid->column('Sertifikat')->display(function () {
    $skips = ["[", "]", "\""];
    $certificate = LandCertificate::where('data_order_id', $this->id);
    $costumerName = Costumer::where('id', $this->costumer_id);

    $certificateNumberArray = $certificate->pluck('number') ?? null;
    $temp1 = str_replace($skips, '', $certificateNumberArray);
    $certificateNumber = str_replace(',', ', ', $temp1);

    if ($certificate->pluck('owner') != $costumerName->pluck('name')) {
        return "<span class='text-danger'>$certificateNumber</span>";
    } else
        return $certificateNumber;
});

Actually, that code only shows the result at picture number 1 without link like picture number 2

Blade code:

https://github.com/z-song/laravel-admin/blob/master/resources/views/grid/table.blade.php

Can you show your blade file? How are you calling the method? — aceraven777, Nov 27 '19 at 01:53
Can you show your blade file? How are you displaying the `$grid`? Please provide the complete details specifically the code. — aceraven777, Nov 27 '19 at 02:15
The grid class can be found here: https://github.com/z-song/laravel-admin/tree/master/resources/views/grid — vreedom18, Nov 27 '19 at 02:28

score 0 · Answer 1 · answered Nov 27 '19 at 02:08

0

You must use html_entity_decode

if ($certificate->pluck('owner') != $costumerName->pluck('name')) {
     return html_entity_decode("<span class='text-danger'>$certificateNumber</span>");
} else {
    return $certificateNumber;
}

answered Nov 27 '19 at 02:08

Poldo

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The code without using html_entity_decode is fine, the problem that I want to return array values to link – vreedom18 Nov 27 '19 at 02:32

score 0 · Accepted Answer · answered Nov 28 '19 at 03:13

I found the solution to make this problem fixed.

Try getting array value then return it back with modification like this:

My Controller:

$grid->column('Sertifikat')->display(function () {
    $skips = ["[", "]", "\""];
    $certificate = LandCertificate::where('data_order_id', $this->id);
    $costumerName = Costumer::where('id', $this->costumer_id);

    foreach ($certificate->pluck('number', 'id')->toArray() as $href => $text) {
        if ($href != $text) {
            if ($certificate->pluck('owner') != $costumerName->pluck('name')) {
                $certificateDetail[] = "<a href='#' data-toggle='modal' data-target='#messageModal' data-id='{$href}' data-from='{$href}' data-title='{$href}' data-message='{$href}' data-time='{$href}'><span class='text-danger'>$text";
            } else
                $certificateDetail[] = "<a href='#' data-toggle='modal' data-target='#messageModal' data-id='{$href}' data-from='{$href}' data-title='{$href}' data-message='{$href}' data-time='{$href}'>$text";
        } else {
            $certificateDetail[] = $text;
        }
    }

    $certificateNumberArray = $certificate->pluck('number') ?? null;
    $temp1 = str_replace($skips, '', collect($certificateDetail));
    $certificateNumber = str_replace(',', ', ', $temp1);

    return $certificateNumber;
});

Result:

How to make dynamic link from array value in laravel?

EDIT:

2 Answers2