dongxian8048 2018-05-07 19:16
浏览 82
已采纳

指定url的图像是否为null

I am designing a site that displays the icon of any song, on the server. I have created a php file that returns the image of the song, if it exists. My code in HTML:

<img src="/image.php?song=SongName" width="100" height="100">

If the given song has a cover icon, then the img element shows the icon. If it doesn't, I get an empty 100x100 img box.

Now, I want to specify in javascript, if the url address: '/image.php?song=SongName' returns null (if there is not an image cover on the song), in order to hide the img. How to achieve it?

Thanks.

PS. The code I've tried so far (there's a bit of php for the img url formation):

var picElement = document.getElementById("picture");
picElement.src = "/image.php?song=$nameFile";
if(picElement.src == null){
    picElement.style.display = "none";
}else if(picElement.src !=null){
    picElement.src = "/image.php?song=$nameFile";
    picElement.style.height = "20vh";
    picElement.style.width = "20vh";
    picElement.style["boxShadow"] = "3px 3px 10px 1px #e2b674";
}

The php code under image.php:

<?PHP
$song= $_GET["song"];
require_once("getID3-1.9.15/getid3/getid3.php");
$Path= .$song.".mp3";

$getID3 = new getID3;
$OldThisFileInfo = $getID3->analyze($Path);
if(isset($OldThisFileInfo['comments']['picture'][0]))
{
    header('Content-Type: ' . $OldThisFileInfo['comments']['picture'][0]['image_mime']);
    echo $OldThisFileInfo['comments']['picture'][0]['data'];
    die;
}?>
  • 写回答

1条回答 默认 最新

  • dongtui6347 2018-05-07 19:27
    关注

    If you want to handle displaying the default image using only markup, you can add an attribute onerror="this.src='/image.php?song=Default'" or similar to each image:

    <img src="/image.php?song=SongName" onerror="this.src='/image.php?song=Default'" width="100" height="100"/>
    

    This assumes that /image.php will properly return a 404 error when an image is not found, rather than displaying a default image itself.

    Update

    Following your PHP code, you might want to just echo the song name to HTML instead of to JavaScript:

    <img src="/image.php?song=<?PHP echo $nameFile; ?>" onerror="this.style.display='none'" width="100" height="100" />
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论

报告相同问题?

悬赏问题

  • ¥15 关于#matlab#的问题:在模糊控制器中选出线路信息,在simulink中根据线路信息生成速度时间目标曲线(初速度为20m/s,15秒后减为0的速度时间图像)我想问线路信息是什么
  • ¥15 banner广告展示设置多少时间不怎么会消耗用户价值
  • ¥16 mybatis的代理对象无法通过@Autowired装填
  • ¥15 可见光定位matlab仿真
  • ¥15 arduino 四自由度机械臂
  • ¥15 wordpress 产品图片 GIF 没法显示
  • ¥15 求三国群英传pl国战时间的修改方法
  • ¥15 matlab代码代写,需写出详细代码,代价私
  • ¥15 ROS系统搭建请教(跨境电商用途)
  • ¥15 AIC3204的示例代码有吗,想用AIC3204测量血氧,找不到相关的代码。