PHP：如果指定的图像不存在，如何显示默认图像？ [英] PHP: How to display a default Image if the specified one doesn't exists?

问题描述

我正在开展一个小项目，我需要在数据库中显示作者的每个作者的图像。窥视下面的代码：

I am working on a little project and I need to display the author's image for each author in my database. peep the code below:

- 查询 -

$getboth_sql = mysql_query(

"SELECT lyrics.lyrics_id, lyrics.lyrics_title, lyrics.lyrics_text,artists.artist_nane,artists.artist_id
FROM lyrics,artists
WHERE lyrics.artist_id = artists.artist_id
ORDER BY lyrics.lyrics_title");

while ($both = mysql_fetch_assoc($getboth_sql)) {

    $lyrics_id = $both[lyrics_id];
    $lyrics_title = $both[lyrics_title];
    $lyrics_text = $both[lyrics_text];
    $artist_name = $both[artist_nane];
    $artist_id = $both[artist_id];

    ?>
    <div class="artimg"><img src="images/artists/<?php echo $artist_name;?>.jpg"  height="50px" width="50px"/></div>        



    <?php
}//END While

上述代码可以正常工作，但如果我没有在'artists'目录中保存图像，则不会显示该艺术家的图像。

The above code work but if I have not saved an image in the 'artists' directory no image show up for that artist.

<div class="artimg"><img src="images/artists/<?php echo $artist_name;?>.jpg"  height="50px" width="50px"/></div>

问题：如果找不到指定的on，显示默认图像的最佳方法是什么。

QUESTION: What is the BEST way to display a default image if the specified on is not found.

我一直在玩空文件和存在的函数，但我没有把它弄好。请帮忙。

I have been playing around with the empty and file exists functions but i haven't gotten it right. Please Help.

PS如果这个问题看起来很愚蠢，我不是专业人士！

PS I'm not a pro if this question seems stupid!

推荐答案

有很多方法可以做到这一点。

There are many ways to do this.

<?php

if(file_exists("images/artists/$artist_name.jpg"))
    $fileName = "$artist_name.jpg";
else
    $fileName = "default.jpg";
?>

<div class="artimg"><img src="images/artists/<?php echo $fileName;?>"  height="50px" width="50px"/></div>

这篇关于PHP：如果指定的图像不存在，如何显示默认图像？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！