我正在从包含链接到照片的url的数据库中提取tweet。
我可以在我的网站上显示照片，但是它们太大了。
代码如下:
foreach ( $entities->media as $media ) {
$tweet_text =str_ireplace($media->url, '<a href="'.$media->expanded_url.'">'
.$media->display_url.'</a>', $tweet_text);
{
$media_html = '';
$url = $media->media_url_https;
$link = $media->url;
$width = $media->sizes->w;
$height = $media->sizes->h;
$media_html = "<a href='"" . $url . "'" target='_blank'>";
$media_html .= "<img src='"" . $url . "'" width='"" .$width.
"'" height='"" .$height. "'" />";
$media_html .= "</a><br />";
$media_html .= $tweet_text;
}
return $media_html;
我试过了:
$width = $media->sizes->w;
$height = $media->sizes->h;
$width = ($width)/2;
$height = ($height)/2;
之后就不显示了。我尝试了很多变化，我唯一能做的就是添加
$width = $media->sizes->w+100;
$height = $media->sizes->h+100;
但这只是将w和h更改为100，并且如您所知，大多数图片都不是完全平方!
你们觉得怎么样?
添加
$RESULT = list($width, $height) = getimagesize($url);
我能够返回值。它是一个数组，所以如果不先加上$width或$height，我就不能除它们。
{
$media_html = '';
$url = $media->media_url_https;
$link = $media->url;
$width = $media->sizes->w;
$height = $media->sizes->h;
$RESULT = list($width, $height) = getimagesize($url);
$width = ($width)/2;
$height = ($height)/2;
$media_html = "<a href='"" . $url . "'" target='_blank'>";
$media_html .= "<img src='"" . $url . "'" width='"" .$width.
"'" height='"" .$height. "'" />";
$media_html .= "</a><br />";
$media_html .= $tweet_text;
}
return $media_html;