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    <p>For some reason, the http response body when requesting that image contains a mime part (mime parts are useful for putting multiple files into a single response). In this response, there is only one mime part, so it is mostly useless.</p> <p>There is code in the javax.mail package that you might be able to use to parse this properly if you want, but it's not a very good api, imho.</p> <p>Alternatively, there are a bunch of ways you could hackishly fix this in code yourself. Since there's only one mime part, you can just throw away data from the beginning of your input stream until you see two newline characters in a row (bytes equal to 10). That should work since mime headers are supposed to be 7-bit ascii, iirc, so there's no character encoding to worry about.</p> <p>Here is some sample code:</p> <pre><code>URLConnection conn = urlNorthView.openConnection(); InputStream in = conn.getInputStream(); String contentType = conn.getHeaderField("Content-Type"); if (!"image/jpeg".equals(contentType)) { // hack: assuming it's mime if not a raw image int one = in.read(); if (one == -1) { // stop?? } int two = in.read(); while (two != -1 &amp;&amp; !(two == 10 &amp;&amp; one == 10)) { one = two; two = in.read(); } } // if it was mime, we've stripped off the mime headers // and should now get the image inputStream2Disk(in); </code></pre> <p>Edit: crap, instead of two \n, you'll see two \r\n, or the bytes 0x0d, 0x0a, 0x0d, 0x0a. Throwing away data until you see this pattern is left as an exercise to the reader ;)</p>
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    1. POHow can I save a JPEG from a URL to a file?
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    1. USJeremy Huiskamp
    1. USJeremy Huiskamp
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    1. POHow can I save a JPEG from a URL to a file?
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    1. COWorks like a charm :) int two = in.read(); int three = in.read(); int four = in.read(); while (two != -1 && !(one == 13 && two == 10 && three == 13 && four == 10 )) { one = two; two = three; three = four; four= in.read(); }
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    2. CO@M.R. nice, but this is too low-level manipulation. You'd generally want to avoid such.
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