如何匹配keyup事件中的數組中的字符串？

Question

請首先閱讀我的要求，我只是解釋你。

我有一個數組（大小= 8264或長度），它存儲所有的數據。

現在我有一個輸入框，其中寫任何字符好 我想實現匹配一個字符串，它是由用戶輸入。

case 1: 
     if user input key then after search. 
    case 2 : 
     if user change text then searching loop will be stop 
     and again search with new string in array. 
//my code on keyup event 
var stationName = ["LAKHO - LAK", "LUSHALA - LAL", "LOTANA - LAN", "LOHAPUR - LAP", "LALITPUR - LAR", "LATHI - LAT", "LADNUN - LAU", "LAUL - LAUL", "LABHA - LAV", "LAMBHUA - LBA", "LIMBODRA - LBD", "LIMBGAON - LBG", "LILABARI - LBI", "LABAN - LBN", "LABPUR - LBP", "LUCKNOW CITY - LC", "LALBAGH CRT RD - LCAE", "LINCH - LCH", "LACHMANPUR ROAD - LCME", "LALPUR CHANDRA - LCN", "DEORI - DOE", "DHODRA MOHAR - DOH", "DEVGARH MADRIYA - DOHM", "DHOLKA - DOK", "DUKHERI - DOKY", "DHOLI - DOL", "DHOLIKUA - DOLK", "DUNDARA - DOR", "DEHRI ON SONE - DOS", "JHANJHARPUR - JJP", "JEJURI - JJR"]; 
stationName.sort(); 

function searchStationName(obj, event) { 
    if ((!obj) || (!event)) 
     return; 
    if (obj.value.length == 0) 
     return; 
    var tempText = '', regVar = ''; 
    var tempLength = 0; 
    var statNameArray = []; 
    var arrayKeyCode = [144, 9, 20, 17, 91, 32, 13, 46, 18, 16]; 
    var keycodeVal = event.keyCode; 
    for (var i = 0; i < 10; i++) { 
     if (keycodeVal == arrayKeyCode[i]) 
      return; 
    } 
     tempText = obj.value; 
     tempLength = stationName.length; 
     regVar = tempText.toUpperCase(); 
     for (var i = 0; i < tempLength; i++) { 
       if (stationName[i].match(regVar)) { 
        statNameArray[count] = stationName[i]; 
        } 
     } 
} 

那麼，如何實現這個請大家幫我

注：純代碼在JavaScript

Answer 1

<div id='arraymatched'> 

</div> 

arr_0 = ["0","2","3"] 
arr_2 = ["2","8","8"]; 

document.getElementById("arraymatched").innerHTML = arraymatch(arr_0,arr_2); 

function arraymatch(arr_0,arr_2) 
{ 
var arrmatch = []; 
for(i = 0; i <= arr_0.length; i++){ 
for(j = 0; j <= arr_2.length; j++){ 
    if(arr_0[i] == arr_2[j]){ 
     arrmatch.push(arr_0[i]); 
    } 
} 
} 
return arrmatch; 
}