2015-01-07 37 views
0

以下TaffyDB腳本成功輸出整個數據庫:如何在Taffydb中集成.filter()?

db().each(function (name){ 
var bodyType=(name["bodyType"]); 
var category=(name["category"]); 
var name2=(name["name2"]); 
var photo=(name["photo"]); 
var resource=(name["resource"]); 
var caption=(name["caption"]); 
output = document.getElementById("display"); 
output.innerHTML+="<div class='segment'><p class='title'>" + bodyType + ": " + name2 + "</p>" + 
"<img class='pic' src='" + photo + "' width='150px'>" + 
"<p class='caption'>" + caption + "</p>" + 
"<p class='more'><a href='" + resource + "'>More</a></p></div>"; 
}); 

我的問題是關於如何過濾內容。該網站有這樣的代碼:

db().filter({column:value}); 

如何更改上面的代碼進行整合呢?

db().filter({bodyType:buggy}); 

這不起作用:

db().each(function (name){ 
var bodyType=(name["bodyType"]); 
var category=(name["category"]); 
var name2=(name["name2"]); 
var photo=(name["photo"]); 
var resource=(name["resource"]); 
var caption=(name["caption"]); 
db().filter({bodyType:buggy}); 
output = document.getElementById("display"); 
output.innerHTML+="<div class='segment'><p class='title'>" + bodyType + ": " + name2 + "</p>" + 
"<img class='pic' src='" + photo + "' width='150px'>" + 
"<p class='caption'>" + caption + "</p>" + 
"<p class='more'><a href='" + resource + "'>More</a></p></div>"; 
}); 

回答

0

更改的第一行是這樣的:

db({bodyType:"Buggy"}).each(function (name){