在這裏,我只是反覆重複相同的代碼,只是遞增變量名稱。與RSS2JSON實用程序的feed url調用相同。 我可以以某種方式寫這個,所以我不必重複?我實際上有8個不同的提要我想使用。而不是重複代碼塊,更有效的方式來寫這個JavaScript?
<script type="text/javascript">
var cpacontent = document.getElementById('cpa');
function showFeed(data){
if(data.status == 'ok'){
var feedlength = data.items.length;
var output = '';
for(var i=0;i<1;++i){
output += '<p><a href="' +
data.items[i].link + '" target="_blank" >' +
data.items[i].title + '</a>';
}
cpacontent.innerHTML = output;
}
}
var fscontent = document.getElementById('fs');
function showFeed2(data){
if(data.status == 'ok'){
var feedlength2 = data.items.length;
var output2 = '';
for(var i=0;i<1;++i){
output2 += '<p><a href="' +
data.items[i].link + '" target="_blank" >' +
data.items[i].title + '</a>';
}
fscontent.innerHTML = output2;
}
}
var wealthcontent = document.getElementById('wealth');
function showFeed3(data){
if(data.status == 'ok'){
var feedlength3 = data.items.length;
var output3 = '';
for(var i=0;i<1;++i){
output3 += '<p><a href="' +
data.items[i].link + '" target="_blank" >' +
data.items[i].title + '</a>';
}
wealthcontent.innerHTML = output3;
}
}
</script>
<script type="text/javascript" src="http://rss2json.com/api.json?callback=showFeed1&rss_url=http%3A%2F%2Ffeedurlplaceholder1"></script>
<script type="text/javascript" src="http://rss2json.com/api.json?callback=showFeed2&rss_url=http%3A%2F%2Ffeedurlplaceholder2"></script>
<script type="text/javascript" src="http://rss2json.com/api.json?callback=showFeed3&rss_url=http%3A%2F%2Ffeedurlplaceholder3"></script>
_Can我莫名其妙地寫這篇文章,所以我不必重複_當然可以!你只需要/應該有1個函數showFeed() –