Question

我试图这样做：加载http://example.com/nickname/picture（昵称作为你访问的网址中的var加载，这是有效的）解码到json 这个网站是

{
"picture": "https://example.com/hf8329yrh8oq.jpg"
}

加载图片我尝试了什么：

function curlGet($url)
   {
   $crl = curl_init();
   $timeout = 5;
   curl_setopt ($crl, CURLOPT_URL,$url);
   curl_setopt ($crl, CURLOPT_RETURNTRANSFER, 1);
   curl_setopt ($crl, CURLOPT_CONNECTTIMEOUT, $timeout);
   $status = curl_exec($crl);
   curl_close($crl);
   return $status;
   }
$profPicCurl = curlGet('https://example.com/'.urlencode($_GET['nickname']).'/picture')
$profPic = json_decode($profPicCurl,true);
echo file_get_contents($profPic.["picture"]);

我知道我没有在这个脚本中处理错误和内容，但我希望它能够先用真实图像处理。

所以平均问题：如何从解码的json网站显示和成像？

Answer 1

你真的需要curl吗？您也可以在我的代码中将file_get_contents替换为curlGet

<?php
  $profPic  = json_decode( file_get_contents( 'https://example.com/'.urlencode( $_GET['nickname'] ).'/picture' ) );
  if ( $profPic ) { // is a valid json object
    if ( isset( $profPic->picture ) ) { // profile picture exists
      $profPic  = $profPic->picture;
      $extension  = strtolower( pathinfo( $profPic, PATHINFO_EXTENSION ) ); // get the image extension
      $content  = file_get_contents( $profPic ); // get content of image
      if ( $content ) {
        header( "Content-type: image/".$extension ); // set mime type
        echo $content; // output the content
        exit();
      }
    }
  }
  echo "File not found"; // there is some errors :\
?>

Answer 2

function getUserImage($nick, $imgUrl = false) {
        $jsonString = file_get_contents("http://example.com/".$nickname."/picture");
        $json = json_decode($jsonString);
        if ($imgUrl){
            return $json->picture;
        } else {
            return file_get_contents($json->picture);
        }
};

然后使用

<?php
header("Content_Type: image/jpeg");
echo getUserImage("quagmire");

或

<img src="<?= getUrlImage("quagmire", true) ?>"/>

PHP从另一个URL（json）加载URL的内容（图片）

2 个答案: