我们可以使用javascript对多个字段进行数据排序吗

Question

In my travel website (just like airbnb ), user can sort record by Availability ,Price ,Location etc...

My website data ( record ) , where in json format only.I can't fetch record from the database.

My Data Record :

var json = [
        {"id":"1","tagName":"apple",'lat':"37.764121333563985","lng":"-122.43083545083414",'cat':'bus','price':50},
        {"id":"2","tagName":"orange",'lat':"37.760390258066806","lng":"-122.40842161915589",'cat':'bus','price':50},
        {"id":"3","tagName":"banana",'lat':"37.76821074033546","lng":"-122.39946165364097",'cat':'car','price':70},
        {"id":"4","tagName":"watermelon",'lat':"37.76903501995947","lng":"-122.43674595206681",'cat':'bus','price':50},
        {"id":"5","tagName":"pineapple",'lat':"37.775329253488984","lng":"-122.42896304538613",'cat':'car','price':50},
        {"id":"6","tagName":"pineapple",'lat':"37.76808825514635","lng":"-122.42957791610439",'cat':'bus','price':30},
        {"id":"7","tagName":"pineapple",'lat':"37.764609542480706","lng":"-122.43126917941493",'cat':'bus','price':50},
        {"id":"8","tagName":"pineapple",'lat':"37.77486256468261","lng":"-122.40143099838457",'cat':'car','price':20},
        {"id":"9","tagName":"pineapple",'lat':"37.770304718740945","lng":"-122.42489358892846",'cat':'bus','price':50},
        {"id":"10","tagName":"pineapple",'lat':"37.7628338732316","lng":"-122.42631843030665",'cat':'car','price':10}
    ];

My Javascript Code :

$(json).each(function(i,marker){
    if (cat == marker.cat || price <= marker.price) || ( price <= marker.price && cat == marker.cat){
     var content = 'MY DATA CONTENT USING MARKER';
     $('.result').append(content);  
   }
});

My if condition is not working.

Can any one help?

Answer 1

如果您的条件括号全部弄乱了，请尝试以下操作：

if ((cat == marker.cat || price <= marker.price) || ( price <= marker.price && cat == marker.cat))

Answer 2

$(json).each(function(i,marker){
    if (cat == marker.cat || price <= marker.price) || ( price <= marker.price && cat == marker.cat){
     var content = 'MY DATA CONTENT USING MARKER';
     $('.result').append(content);  
   }

In this code the value of cat is not defined, you need to either change it to

this.cat or json[i].cat

to access the json values.