i using following code show referer link page on website. how can modify code shows part of link. i.e if website url www.example.com/?s=printing want extract printing. , should happen if format www.example.com/?s=aaa , not if format else www.example.com/printing. code: <?php session_start(); if ( !isset( $_session["origurl"] ) ) $_session["origurl"] = $_server["http_referer"]; echo $_session["origurl"] ?> i figured out , following code works: <?php session_start(); if ( !isset( $_session["origurl"] ) ) $_session["origurl"] = $_server["http_referer"]; $mysearchterm = $_server["http_referer"]; $whatiwant = substr($mysearchterm, strpos($mysearchterm, "=") +1); echo $whatiwant; ?>
i using eclipse juno in windows 7. doing programming on android 4.2 platform. want receive udp packets in gstreamer application. giving uri udp://127.0.0.1:1234 , sending file vlc player same computer. emulator giving message could not get/set setting from/on resource while in case of rtp, giving message no uri implementer found rtp ideas how receive streaming media in android?
i have ec2 instance mongodb 3.2.0 installed. have user admin access. can connect user , perform operation remotely. 1 can connect database test remote if knows ip mongo 11.11.11.11. although not able make operations, want restrict access user credentials can connect server. first need create users , grant roles users, example: use admin db.createuser( { user: "myuseradmin", pwd: "abc123", roles: [ { role: "useradminanydatabase", db: "admin" } ] } ) then need start server --auth parameter mongod --auth more information here
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