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array 1:: store_id: card1(0 for no, 1 for yes), card2,....cardN; array2: store_id. working variale0: int stores; working varialble2: int VISA; working variable3: MASTER.... working variableN: nnnncard.

flyingegg(O'`)

#initilize working variables.
working_variable0= working_variable1=.... = working_variable0=0;
for (a_store_id in array2){
      working_variable0++;
      if (array1[a_store_id].card1) working_variable1++;
      if (array1[a_store_id].card2) working_variable2++;
      if (array1[a_store_id].card3) working_variable3++;
      ...
      if (array1[store_id].cardN) working_variableN++;
}

if (workong_varialbe0 == working_varialbe1) show_up(card1);
if (workong_varialbe0 == working_varialbe2) show_up(card2);
...
if (workong_varialbe0 == working_varialbeN) show_up(cardN);

(#3753807@0)
Last Updated: 2007-6-20
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工作学习 / 专业技术讨论 / 请教，请教。问一个javascript的问题： -luoboyang(萝卜秧); 2007-6-20 {499} (#3753686@0)
问题比较简单，描述起来比较麻烦。简单地说，如果有多个商家，每个商家接受不同的信用卡，如下：

store A visa, mc, AE
store B visa, AE
store C visa, mc, discover

这样，当用户选择不同的商家的时候，需要找出他们的common payment type，比如，如果选了A and C, 则visa, mc应该被摘出来；如果选了A and B, then visa,AE should be selected。
这些都要求在javascript里面做，我弄了几个数组了，不过，现在把自己给绕进去了。：））

希望找个最简洁的方法。。。。。

谢谢大家。：）

能不能这样 -ready4u(就等你了); 2007-6-20 {259} (#3753767@0)
1、建立一个新数组，用来记录最后的信用卡名单。

2、扫描所有选中的商家，第一个商家的所有信用卡都放进新数组。以后的商家里面的每个信用卡都拿出来和新数组里面的比较，如果没有发现，就将此卡从新数组中剔除。

3、重复到最后，新数组里面应该就只有大家都有的信用卡了。

当然这么做过。有问题的。 -luoboyang(萝卜秧); 2007-6-20 (#3753782@0)

array 1:: store_id: card1(0 for no, 1 for yes), card2,....cardN; array2: store_id. working variale0: int stores; working varialble2: int VISA; working variable3: MASTER.... working variableN: nnnncard. -flyingegg(O'`); 2007-6-20 {603} (#3753807@0)

#initilize working variables.
working_variable0= working_variable1=.... = working_variable0=0;
for (a_store_id in array2){
      working_variable0++;
      if (array1[a_store_id].card1) working_variable1++;
      if (array1[a_store_id].card2) working_variable2++;
      if (array1[a_store_id].card3) working_variable3++;
      ...
      if (array1[store_id].cardN) working_variableN++;
}

if (workong_varialbe0 == working_varialbe1) show_up(card1);
if (workong_varialbe0 == working_varialbe2) show_up(card2);
...
if (workong_varialbe0 == working_varialbeN) show_up(cardN);

Create a JSON object. It's the easiest way. -wind_cloud(风逝云栖); 2007-6-20 (#3753858@0)
javascript 支持位运算吗？两个整数and运算就出来了 -jj999ca(jj999ca); 2007-6-20 (#3753963@0)
JavaScript Operators.. -jj999ca(jj999ca); 2007-6-20 (#3753983@0)

谢谢。水缸里也这么提醒我了，正在考虑。：） -luoboyang(萝卜秧); 2007-6-20 (#3754117@0)

... -iloveflowers(新园丁); 2007-6-20 {361} (#3755106@0)
function getCardList(store) {
var cardList;
if (store== 'A') {
cardList = {'visa', 'mc', 'AE'};
} else if ....
}
return cardList;
}
function populateCardBox(cardList) {...}

function onStoreSelect(obj){
var store = obj[selectedIndex].selectedValue;
populateCardBox( getCardList(store));
}

<select ... onchange="onStoreSelect(this);">
easy -digitworm(digitworm); 2007-6-21 {744} (#3756209@0)
<script type ="text/javascript" >
var VISA = 0
var MC = 1
var AE = 2
var DISCOVER = 3

var A = new Array{true, true, true, false};
var B = new Array{true, false, true, false};
var C = new Array{true, true, false, true};

if (A[VISA] && B[VISA] && C [VISA]) {
// VISA is the common type
}

if (A[MC] && B[MC] && C [MC]) {
// MC is the common type
}

if (A[DISCOVER] && B[DISCOVER] && C [DISCOVER]) {
// MC is the common type
}

if (A[DISCOVER] && B[DISCOVER] && C [DISCOVER]) {
// MC is the common type
}
</script>

better one -digitworm(digitworm); 2007-6-21 {1332} (#3756315@0)
本文发表在 rolia.net 枫下论坛<html xmlns="http://www.w3.org/1999/xhtml" >
<head>
<title>Untitled Page</title>
<script type ="text/javascript" >
function show(){
var VISA = 0;
var MC = 1;
var AE = 2;
var DISCOVER = 3;

var A = new Array(true, true, true, false);
var B = new Array(true, false, true, false);
var C = new Array(true, true, false, true);
var common_type = new Array(4);

for (var i = 0; i<4; i++){
common_type[i] = A[i] && B[i] && C[i];
}

var str = '';
if ( common_type[VISA] ) {
// VISA is the common type
str += 'VASIA ';
}

if ( common_type[MC] ) {
// MC is the common type
str += 'MC ';
}

if ( common_type[AE] ) {
// AE is the common type
str += 'AE ';
}

if ( common_type[DISCOVER] ) {
// DISCOVER is the common type
str += 'DISCOVER ';
}
document.getElementById ('result').innerText = str;
}
</script>
</head>
<body>
<input id="Button1" type="button" value="button" onclick ='show();' />
<span id='result'></span>

</body>
</html>更多精彩文章及讨论，请光临枫下论坛 rolia.net
呵呵，如果里面都是动态的，就没有那么easy了。anyway,我最后用了3个数组，3个循环给弄出来了。：）thanks. -luoboyang(萝卜秧); 2007-6-21 (#3756324@0)

@Florida

array 1:: store_id: card1(0 for no, 1 for yes), card2,....cardN; array2: store_id. working variale0: int stores; working varialble2: int VISA; working variable3: MASTER.... working variableN: nnnncard.

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