bravo14 Posted January 13, 2021 Share Posted January 13, 2021 Hi I am trying to pass 2 variables to a URL using typeahead, but the URL that it gerjneates is incorrectly formatted. The code I am using on the function is $('#txtLocation').typeahead({ source: function (query, process) { var country = $('#txtCountry').val(); var url = query + '&c='+country; //var uri = encodeURI(url); //alert(uri); return $.get('ajax/search_locations.php', { query: encodeURI(url) }, function (data) { data = $.parseJSON(data); return process(data); }); }, //showHintOnFocus:'all' }); The URL in the debugger is /ajax/search_locations.php?query=b%26c%3DUnited%2520Kingdom it should be /ajax/search_locations.php?query=b&c=United%2520Kingdom Any help or pointers on where I am going wrong would be fantastic Quote Link to comment https://forums.phpfreaks.com/topic/312001-typeahead-passing-2-variables-to-url/ Share on other sites More sharing options...
requinix Posted January 13, 2021 Share Posted January 13, 2021 { query: encodeURI(url) } That is the place that lists out the values to pass. You are only passing one because you crammed the query+c+country into the value. Keep the "query" being the query and add a "c" for the country. Quote Link to comment https://forums.phpfreaks.com/topic/312001-typeahead-passing-2-variables-to-url/#findComment-1583770 Share on other sites More sharing options...
bravo14 Posted January 13, 2021 Author Share Posted January 13, 2021 I have added the &c='+country that is where the problem is occurring, in this line var url = query + '&c='+country; Quote Link to comment https://forums.phpfreaks.com/topic/312001-typeahead-passing-2-variables-to-url/#findComment-1583771 Share on other sites More sharing options...
requinix Posted January 13, 2021 Share Posted January 13, 2021 Yeah. And I'm saying don't do that. It doesn't work. It's incorrect. Quote Link to comment https://forums.phpfreaks.com/topic/312001-typeahead-passing-2-variables-to-url/#findComment-1583772 Share on other sites More sharing options...
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