Question

我有来自外部网址的下载图片代码。

 $content = file_get_contents("http://shoecompany.in/wp-content/uploads/2015/06/new.jpg");

 $fp = fopen("assets/img/image.jpg", "w");
 fwrite($fp, $content);
 fclose($fp);

如果外部网址中没有图片，那么我想显示错误信息。那我该怎么办？

Answer 1

<强> file_get_contents

该函数返回读取数据或失败时返回FALSE。

因此您可以将错误检查为

 $content = file_get_contents("http://shoecompany.in/wp-content/uploads/2015/06/new.jpg");
if ($content === FALSE) {// check here
// handle error here... 
} else {
    $fp = fopen("assets/img/image.jpg", "w");
    if ($fp) {// fopen check here
        fwrite($fp, $content);
        fclose($fp);
    } else {
        // handle error here... 
    }    
}

Answer 2

if(@$content = file_get_contents("http://shoecompany.in/wp-content/uploads/2015/06/new.jpg") !== false){
        $content;
        //Store in the filesystem.
        $fp = fopen("assets/img/image.jpg", "w");
        fwrite($fp, $content);
        fclose($fp);
    }else{
        echo "error";
    }

Answer 3

此函数在字符串的情况下返回文件，在失败时返回false。函数返回布尔值为FALSE但也可能返回非布尔值，这就是为什么你必须使用＆＃34; ===＆＃34;

     $result=file_get_contents("http://shoecompany.in/wp-content/uploads/2015/06/new.jpg"); 

    if ($result === false) 
    { 
        print "Not Found. You Can handle error here";
    } 
    else 
    { 
        print " handle good case ";
    }

Answer 4

尝试使用if条件

if(file_get_contents("http://shoecompany.in/wp-content/uploads/2015/06/new.jpg"))
{
$content = file_get_contents("http://shoecompany.in/wp-content/uploads/2015/06/new.jpg");

 $fp = fopen("assets/img/image.jpg", "w");
 fwrite($fp, $content);
 fclose($fp);
}
else
{
...Some content to display

}

在外部网址中找不到文件

4 个答案: