Question

下面的PHP代码总共生成240个DIV。我想在7,10,21,135和201 DIV之后插入不同的DIV。我该怎么办？

这些是我想要展示的DIV：

<div class="col-xs-3">field</div>
<div class="col-xs-3">field</div>
<div class="col-xs-3">field</div>
<div class="col-xs-3">field</div>
<div class="col-xs-3">field</div>
<div class="col-xs-3">field</div>
<div class="col-xs-3">field</div>
<div class="div7">div 7 description goes here</div>
<div class="col-xs-3">field</div>
<div class="col-xs-3">field</div>
<div class="col-xs-3">field</div>
<div class="div10">div 10 description goes here</div>
<div class="col-xs-3">field</div>
...goes on

这是PHP代码：

<div class="row">
    <?php foreach ($this->fields as $field) { ?>
        <div class="col-xs-3 EditFormClass">
            <?php echo $field[0]; ?>
        </div>
        <div class="col-xs-3 EditFormClass">
            <?php echo $field[1]; ?>
        </div>
    <?php } ?>
</div>

Answer 1

您可以将异常情况放在数组中，然后运行循环：

    <?php
    $difDiv = [];
    $difDiv[7] = 'div 7 description goes here';
    $difDiv[10] = 'div 10 description goes here';
    ?>
    <div class="row">
    <?php $i = 0; ?>
    <?php foreach ($this->fields as $field) { ?>
        <div class="col-xs-3 EditFormClass">
            <?php echo $field[0]; ?>
        </div>
        <div class="col-xs-3 EditFormClass">
            <?php echo $field[1]; ?>
        </div>
        <?php 
            $i++;
            if (isset($difDiv[$i])) { ?>
                <div class="div<?php echo $i ?> "><?php echo $difDiv[$i]; ?></div>
            <?php } ?>

    <?php } ?>
</div>

使用PHP在x DIV之后插入不同的DIV

1 个答案: