Question

我正在使用多选择和GET方法。我希望在提交后根据url parametres.i中的$ _GET方法重新加载表单时选择了多个选项，相关的URL参数为cuisine%5B%5D=indian&cuisine%5B%5D=thai。实际上多选约为cuisine。

我的代码如下：

 <select name="cuisine[]" class="selectpicker show-tick form-control" data-selected-text-format="count > 3" data-done-button="true" data-done-button-text="OK" multiple>
    <?php
        $selected_cuisine = $_GET['cuisine'];
        // Get all cuisines list by get_terms() function.Its built in wordpress
        $restaurant_cuisines = get_terms('cuisine', array('hide_empty' => false));
        $cuisines = array();
        foreach ($restaurant_cuisines as $restaurant_cuisine) {
            // echo $restaurant_cuisine;
            array_push( $cuisines, $restaurant_cuisine->slug );
            // echo $cuisines_list;
        }
        print_r ($selected_cuisine);
        print_r($cuisines);
        if(array_intersect($selected_cuisine, $cuisines)){
            $selected = 'selected';
        }else{
            $selected = '';
        }
        foreach ($restaurant_cuisines as $cuisine) {
            echo '<option value="'. $cuisine->slug .'" '. $selected .' >'. $cuisine->name .'</option>';
        }
    ?>
</select>

但问题在于每个选项都会被选中。实际上总共有3 cuisines : indian, thai & chainese和2 of them are selected - ＆gt; indian and thai。但问题是选择了3个选项。：/

Answer 1

请记住，当您使用array_intersect时，设置为true的两个数组中至少有一个匹配。所有内容都被选中。您可以试试这个：

<select name="cuisine[]" class="selectpicker show-tick form-control" data-selected-text-format="count > 3" data-done-button="true" data-done-button-text="OK" multiple>
<?php
    $selected_cuisine = $_GET['cuisine'];
    // Get all cuisines list by get_terms() function.Its built in wordpress
    $restaurant_cuisines = get_terms('cuisine', array('hide_empty' => false));
    $cuisines = array();
    foreach ($restaurant_cuisines as $restaurant_cuisine) {
        array_push( $cuisines, $restaurant_cuisine->slug );
    }
    foreach ($restaurant_cuisines as $cuisine) {
        if(in_array($cuisine->slug, $selected_cuisine)){
          $selected = 'selected';
        }else{
          $selected = '';
        }
        echo '<option value="'. $cuisine->slug .'" '. $selected .' >'. $cuisine->name .'</option>';
    }
?>

如果值存在于主数组中，则多选添加所选属性 - php

我的代码如下：

1 个答案: