Definitions of Haney - OneLook

Definitions from Wiktionary (Haney)
noun:  A surname.
noun:  A town in Crawford County, Wisconsin, United States.Pressley, Lockett, Connelly, Scanlon, Turley, Pickard, Handley, Colvin, Hannon, Fraley, more...


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"; sn = document.getElementById("snippets"); sn.innerHTML = html_str; if (move_snippet_to_top) { document.getElementById("snippets_top").innerHTML = "
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"; move_snippet_to_top = false; } current_snippet = i; } var current_snippet = 0; function randSnippet() { if (rz_snippets.length > 0) { current_snippet = (current_snippet + 1) % rz_snippets.length; updateCurrentSnippet(current_snippet); } } function getTargetFromReferrer() { var query = document.referrer; var vars = query.split('&'); for (var i = 0; i < vars.length; i++) { var pair = vars[i].split('='); var name = decodeURIComponent(pair[0]); if (name.indexOf("Word", name.length - 4) !== -1) { return decodeURIComponent(pair[1]); } } return ""; } function getTargetFromRequest(paramKey) { var vars = window.location.href.slice(window.location.href.indexOf('?') + 1).split('&'); for (var i = 0; i < vars.length; i++) { var pair = vars[i].split('='); var name = decodeURIComponent(pair[0]); if (name.indexOf(paramKey, name.length - 5) !== -1) { // huh!? return decodeURIComponent(pair[1]); } } return ""; } if (show_all_snippets) { organizeSnippets(); if (matchCounts.length > 1 && rz_snippets.length > 4) { outputHeader(); } var target = getTargetFromRequest("mWord"); if (target == "") { showAllSnippets("All", "All"); } else { target = norm(target); var matches = showAllSnippets(target, "All"); if (matches == 0) { showAllSnippets("All", "All"); } else { updateSel = true; $('#selrhy').val(target); $('#selrhy').selectmenu("refresh"); updateSel = false; } } } else { var target = getTargetFromReferrer(); current_snippet = -1; if (target !== "" && target !== getTargetFromRequest("Word")) { referrer_target = true; target = target.toLowerCase(); rz_snippets.sort(function(a, b) { if (a==b) return 0; var q1=((a["t"]===target)?3:((a["m"]===target)?2:1)); var q2=((b["t"]===target)?3:((b["m"]===target)?2:1)); return q2-q1; }); referrer_target_match_count = 0; for (var i=0; i 0) { move_snippet_to_top = true; } } if (!move_snippet_to_top) { // prioritize artist of the day for (var i = 0; i < rz_snippets.length; i++) { x = rz_snippets[i]; if (x["title"] && x["title"].includes("Sweeney")) { current_snippet = i-1; found_featured = true; break; } } } randSnippet(); }

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