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refactor(subscriptions): use BackUrl::withBack for links on index

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This commit is contained in:
萝卜
2026-03-14 20:14:33 +00:00
parent 55095d3d4b
commit c37e7a8fa5
1 changed files with 3 additions and 7 deletions
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10
resources/views/admin/site_subscriptions/index.blade.php
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@@ -158,7 +158,6 @@
@php
$q = [
'order_type' => 'renewal',
'back' => $selfWithoutBack,
];
if ((int) ($filters['merchant_id'] ?? 0) > 0) {
$q['merchant_id'] = (int) $filters['merchant_id'];
@@ -166,7 +165,7 @@
if ((int) ($filters['plan_id'] ?? 0) > 0) {
$q['plan_id'] = (int) $filters['plan_id'];
}
$createOrderFromSubIndexUrl = '/admin/platform-orders/create?' . \Illuminate\Support\Arr::query($q);
$createOrderFromSubIndexUrl = \App\Support\BackUrl::withBack('/admin/platform-orders/create?' . \Illuminate\Support\Arr::query($q), $selfWithoutBack);
@endphp
<a class="btn btn-sm" href="{!! $createOrderFromSubIndexUrl !!}">创建续费订单（带当前筛选）</a>
</div>
@@ -199,9 +198,7 @@
<td>{{ $subscription->id }}</td>
<td>
@php
$subShowUrl = '/admin/site-subscriptions/' . $subscription->id . '?' . \Illuminate\Support\Arr::query([
'back' => $back,
]);
$subShowUrl = \App\Support\BackUrl::withBack('/admin/site-subscriptions/' . $subscription->id, $back);
@endphp
<a href="{!! $subShowUrl !!}">{{ $subscription->subscription_no }}</a>
@php
@@ -211,7 +208,6 @@
'site_subscription_id' => $subscription->id,
'quantity' => 1,
'remark' => $remarkPrefix . $subscription->subscription_no,
'back' => $back,
];
if ((int) ($subscription->merchant_id ?? 0) > 0) {
$q['merchant_id'] = (int) $subscription->merchant_id;
@@ -219,7 +215,7 @@
if ((int) ($subscription->plan_id ?? 0) > 0) {
$q['plan_id'] = (int) $subscription->plan_id;
}
$renewOrderUrl = '/admin/platform-orders/create?' . \Illuminate\Support\Arr::query($q);
$renewOrderUrl = \App\Support\BackUrl::withBack('/admin/platform-orders/create?' . \Illuminate\Support\Arr::query($q), $back);
@endphp
<div class="mt-4">
<a class="btn btn-secondary btn-sm" href="{!! $renewOrderUrl !!}">续费下单</a>