如何在循环中在php中添加一个类？

Question

我尝试在1循环后添加2个bootstrap类col-md-8 ，然后在php循环中添加2个col-md-4类。 这个过程在完整循环时应该是相同的。 所以结果看起来像这样：

我目前的代码是吼叫，但它没有显示我需要的结果，无法知道循环将如何！

完整代码：

<div class="row text-center">
    <h2>What we offer</h2>
    <hr class="separator">
    <?php $get_menu_class=m ysqli_query($conn, "SELECT pcat_name, pcat_image FROM product_category ORDER BY pcat_id DESC"); $x=0; while($menu_class_result=m ysqli_fetch_array($get_menu_class) ) { $menu_class_name=h tmlspecialchars($menu_class_result[ 'pcat_name']); $menu_class_image=h tmlspecialchars($menu_class_result[ 'pcat_image']); if($x & 1) { $col='8' ; }else { $col='4' ; } ?>
    <div class="col-sm-<?php echo $col; ?>">
        <div class="we-offer">
            <a href="area">
                        <img src="<?php echo IMG_DIR."/menu_class/$menu_class_image"; ?>" alt="" class="img-responsive center-block">
                        <h3><?php echo ucfirst($menu_class_name); ?></h3>
                    </a>
        </div>
    </div>
    <?php $x++; } ?>

</div>

最新代码：

<div class="row text-center">
    <h2>What we offer</h2>
    <hr class="separator">
    <?php $get_menu_class=m ysqli_query($conn, "SELECT pcat_name, pcat_image FROM product_category ORDER BY pcat_id DESC"); $x=0; while($menu_class_result=m ysqli_fetch_array($get_menu_class) ) { $menu_class_name=h tmlspecialchars($menu_class_result[ 'pcat_name']); $menu_class_image=h tmlspecialchars($menu_class_result[ 'pcat_image']); $col=( (($x+1)/2)%2)? "8": "4"; ?>
    <div class="col-sm-<?php echo $col; ?>">
        <div class="we-offer">
            <a href="area">
                <img src="<?php echo IMG_DIR."/menu_class/$menu_class_image"; ?>" alt="" class="img-responsive center-block">
                <h3><?php echo ucfirst($menu_class_name); ?></h3>
            </a>
        </div>
    </div>
    <?php $x++; } ?>

</div>

目前的结果：

[！[在此处输入图片说明] [2]] [2]

Answer 1

请尝试这个逻辑： $col = ((($i+1)/2)%2)?"8":"4";

https://3v4l.org/GHGVp

如您所见，它会输出所需的结果。

The col for loop 0 is col4
The col for loop 1 is col8
The col for loop 2 is col8
The col for loop 3 is col4
The col for loop 4 is col4
The col for loop 5 is col8
The col for loop 6 is col8
The col for loop 7 is col4
The col for loop 8 is col4
The col for loop 9 is col8
The col for loop 10 is col8
The col for loop 11 is col4
The col for loop 12 is col4
The col for loop 13 is col8
The col for loop 14 is col8
The col for loop 15 is col4
The col for loop 16 is col4
The col for loop 17 is col8
The col for loop 18 is col8
The col for loop 19 is col4

您只需要在代码中替换

if($x & 1) {
    $col = '8';
}else {
    $col = '4';
}

同

$col = ((($x+1)/2)%2)?"8":"4";

Answer 2

我只是跟踪显示的变量。 如果您首先将其定义为1，那么您将只显示第一次所需的第一列，然后它将切换到另一列。

您不需要跟踪$x 。 我不会称这是一个简单的数学问题，而这里的另一个OK答案不适合您的用例。

<div class="row text-center">
    <h2>What we offer</h2>
    <hr class="separator">

    <?php
        $get_menu_class = mysqli_query($conn, "SELECT pcat_name, pcat_image FROM product_category ORDER BY pcat_id DESC");
        $col = 4;
        $displayed = 1;

        while($menu_class_result = mysqli_fetch_array($get_menu_class) ) {
            $menu_class_name = htmlspecialchars($menu_class_result['pcat_name']);
            $menu_class_image = htmlspecialchars($menu_class_result['pcat_image']);
    ?>
            <div class="col-sm-<?php echo $col; ?>">
                <div class="we-offer">
                    <a href="area">
                        <img src="<?php echo IMG_DIR."/menu_class/$menu_class_image"; ?>" alt="" class="img-responsive center-block">
                        <h3><?php echo ucfirst($menu_class_name); ?></h3>
                    </a>
                </div>
            </div>
    <?php
            if($displayed){
                switch($col){
                    case 4:
                        $col = 8;
                        break;
                    case 8:
                        $col = 4;
                        break;
                    default:
                        $col = 4;
                }

                $displayed = 0;
            } else {
                $displayed = 1;
            }
        }
    ?>

</div>