@Deprecated
    public static String encodeUri(String uri) {
        try {
            return URIUtil.encodePathQuery(uri);
        } catch (URIException ex) {
            throw new EsHadoopIllegalArgumentException("Cannot escape uri [" + uri + "]", ex);
        }
    }
@ Deprecated public static String encodeUri ( String uri ) { try { return URIUtil . encodePathQuery ( uri ) ; } catch ( URIException ex ) { throw new EsHadoopIllegalArgumentException ( "Cannot escape uri [" + uri + "]" , ex ) ; } }
Splits the given string on the first '?' then encodes the first half as a path (ignoring slashes and colons)
and the second half as a query segment (ignoring questionmarks, equals signs, etc...).

@deprecated Prefer to use {@link HttpEncodingTools#encode(String)} instead for encoding specific
pieces of the URI. This method does not escape certain reserved characters, like '/', ':', '=', and '?'.
As such, this is not safe to use on URIs that may contain these reserved characters in the wrong places.