PHP中如何通过关联数组在一个变量中传递两个值

PHP中如何通过关联数组在一个变量中传递两个值,php,arrays,csv,associative-array,Php,Arrays,Csv,Associative Array,我使用的是UberGallery.php 对于每个图像弹出窗口,从UberGallery.php获取的图像底部都会显示图像名称(例如4488826 6f061c99ec b d) // Loop through array and add additional info foreach ($dirArray as $key => $image) { // Get files relative path

我使用的是UberGallery.php

对于每个图像弹出窗口,从UberGallery.php获取的图像底部都会显示图像名称(例如4488826 6f061c99ec b d)

  // Loop through array and add additional info
            foreach ($dirArray as $key => $image) {
                // Get files relative path
                $relativePath = $this->_rImgDir . '/' . $key;

                $galleryArray['images'][htmlentities(pathinfo($image['real_path'], PATHINFO_BASENAME))] = array(
                    'file_title'   => str_replace('_', ' ', pathinfo($image['real_path'], PATHINFO_FILENAME)),
                    'file_path'    => htmlentities($relativePath),
                    'thumb_path'   => $this->_createThumbnail($image['real_path'])
                );
            }
上述代码中的文件标题正在影响要更改的标题。在defaultGallery.php中显示的图像如下所示

 <?php if (!empty($images) && $stats['total_images'] > 0): ?>

        <ul id="galleryList" class="clearfix">

            <?php foreach ($images as $image): ?>
                <li><a href="<?php echo html_entity_decode($image['file_path']); ?>" title="<?php echo $image['file_title']; ?>" rel="<?php echo $relText; ?>"><img src="<?php echo $image['thumb_path']; ?>" alt="<?php echo $image['file_title']; ?>"/></a></li>
            <?php endforeach; ?>

        </ul>

    <?php else: ?>

        <p>No images found.</p>

    <?php endif; ?>
我应该如何将姓名和电子邮件传递到上述文件\u title以获得所需结果? 还是有其他方法可以做到这一点

请帮我找出解决办法。

像这样试试这个

 $galleryArray['images'][htmlentities(pathinfo($image['real_path'], PATHINFO_BASENAME))] = array(
                'file_title'   => ' ',
                'file_path'    => htmlentities($relativePath),
                'thumb_path'   => $this->_createThumbnail($image['real_path'])
            );

如果我们保持文件标题为空,则不会显示任何内容。我想显示我在问题中提到的两个字段。@priya786:你知道我该怎么做吗?你想用名称和电子邮件来代替titleyaah,我想用图像名称和电子邮件更改图像弹出窗口底部的标题(请查看我提供的链接演示)。我正在尝试从csv文件中获取并在那里显示。但是你做不到。@priya786:你的回答还不够OP的要求。
 $galleryArray['images'][htmlentities(pathinfo($image['real_path'], PATHINFO_BASENAME))] = array(
                'file_title'   => ' ',
                'file_path'    => htmlentities($relativePath),
                'thumb_path'   => $this->_createThumbnail($image['real_path'])
            );