Image Groovy从URL下载图像
我想知道从这个RUL下载图像的正确方法是: 我试图下载它的方式使文件的格式未知。我测试的当前代码段是:Image Groovy从URL下载图像,image,url,groovy,download,Image,Url,Groovy,Download,我想知道从这个RUL下载图像的正确方法是: 我试图下载它的方式使文件的格式未知。我测试的当前代码段是: public void download(def address) { def file = new FileOutputStream(address.tokenize("/")[-1]) def out = new BufferedOutputStream(file) out << new URL(address).openStream() ou
public void download(def address) {
def file = new FileOutputStream(address.tokenize("/")[-1])
def out = new BufferedOutputStream(file)
out << new URL(address).openStream()
out.close()
}
公共无效下载(def地址){
def file=new FileOutputStream(address.tokenize(“/”[-1]”)
def out=新的BufferedOutputStream(文件)
这有用吗?我相信它应该:
public void download(def address) {
new File("${address.tokenize('/')[-1]}.png").withOutputStream { out ->
out << new URL(address).openStream()
}
}
公共无效下载(def地址){
新文件(${address.tokenize('/')[-1]}.png”)。withOutputStream{out->
out谢谢蒂姆,我也发现你的答案非常有用,只是一个小提示:
看起来您还没有关闭URL流。我只是从Groovy开始,我听说它在退出关闭时会关闭流,所以我们可以这样更改代码:
public void download(def address) {
new File("${address.tokenize('/')[-1]}.png").withOutputStream { out ->
new URL(address).withInputStream { from -> out << from; }
}
}
公共无效下载(def地址){
新文件(${address.tokenize('/')[-1]}.png”)。withOutputStream{out->
新URL(地址)。使用InputStream{from->out可以从其内容类型或字节数组中获取图像类型:
content="http://www.google.ru/images/logo.png".toURL().getBytes()
ext=URLConnection.guessContentTypeFromStream(new ByteArrayInputStream(content)).replaceFirst("^image/","")
new File("logo."+ext).setBytes(content)