Sed get xml attribute value

Question

I have next xml file:

<AutoTest>
 <Source>EBS FX</Source>
 <CreateCFF>No</CreateCFF>
 <FoXML descriptor="pb.fx.spotfwd.trade.feed" version="2.0">
   <FxSpotFwdTradeFeed>
     <FxSpotFwd feed_datetime="17-Dec-2014 10:20:09" 
       cpty_sds_id="EBS" match_id="L845586141217" original_trade_id_feed="L80107141217" 
       value_date="20141218" trade_id_external="001-002141880445/5862" match_sds_id="EBSFeedCpty" 
       counter_ccy="USD" trade_id_feed="107" trade_type="S" feed_source_id="80"    quoting_term="M" 
       deal_ccy="GBP" rate="1.5" trade_date="20141217" modified_by="automation"    cpty_side="B" counter_amt="1500000"
       smart_match="0" booking_status_id="10" trade_status_id="22" deal_amt="1000000"  trade_direction="B">
       <Notes />
     </FxSpotFwd>
 </FxSpotFwdTradeFeed>
 <TestCases />
 </FoXML>
</AutoTest>

How to get value of trade_id_external attribute by using sed?
I tried with this expression: sed -n '/trade_id_external/s/.*=//p' ./file.xml but no luck

Solution

You don't even need a pattern /trade_id_external/ before the s///

$ sed -n 's/.*trade_id_external="\([^"]*\).*/\1/p' file
001-002141880445/5862

In basic sed , \(...\) called capturing groups which was used to capture the characters you want to print at the final.

Through grep,

$ grep -oP 'trade_id_external="\K[^"]*' file
001-002141880445/5862

-P would turn on the Perl-regex mode in grep. So we could use any PCRE regexes in grep with -P param enabled. \K in the above regex would discard the previously matched characters, that is, it won't consider the characters which are matched by the pattern exists before the \K