symfony - 如何限制 knp_paginator 扩展？

Question

我想永远限制分页框。假设加载了 100 个事件 -> 每页显示 3 个事件；这样分页框[1][2][3][4]....[40]就不会继续了……

Answer 1

在config.yml添加这个：

knp_paginator:
   page_range: 5 # number of links showed in the pagination menu (e.g: you have 10 pages, a page_range of 3, on the 5th page you'll see links to page 4, 5, 6)

如果您需要更改滑动，请使用以下任何一种：

template:                                        
    pagination: '@KnpPaginator/Pagination/sliding.html.twig'

@KnpPaginator/Pagination/sliding.html.twig (by default)
@KnpPaginator/Pagination/twitter_bootstrap_v3_pagination.html.twig
@KnpPaginator/Pagination/twitter_bootstrap_pagination.html.twig
@KnpPaginator/Pagination/foundation_v5_pagination.html.twig

Answer 2

我猜 knp 分页器会为您执行此操作，但如果没有，您可以尝试修改上面的任何模板以执行以下操作：

设置控制变量

{% if pageCount > maxNumberOfBoxes %}
        {% set breakpointAdded=true %}
{% endif %}

然后找到看起来像这样的循环

{% for page in pagesInRange %}
            {% if page != current %}
                <span class="page">
                <a href="{{ path(route, query|merge({(pageParameterName): page})) }}">{{ page }}</a>
            </span>
            {% else %}
                <span class="current">{{ page }}</span>
            {% endif %}

        {% endfor %}

并用这样的东西替换它（当然用你自己的逻辑）

{% for i in range(0,pageCount) %}
                {% if i > xBreakpoint and i < yBreakpoint and breakpointAdded == false %}
                    <span class="dots">...</span>
                    {% set breakpointAdded = true %}
                {% else %}
                    {% if page != current %}
                        <span class="page">
                            <a href="{{ path(route, query|merge({(pageParameterName): page})) }}">{{ page }}</a>
                       </span>
                    {% else %}
                        <span class="current">{{ page }}</span>
                    {% endif %}
                {% endif %}
            {% endfor %}

如果其中任何一项有效，您可以尝试从 SlidingPagination 类修改 getPaginationData 函数。

什么对你有用。