Question

对于ajax调用，我得到以下格式的响应

 var response = {
    "wwc_tag_names": [{
        "waterbreak_free": "waterbreak_free"
    }],
    "wwc_details": [{
        "generatedid": "2",
        "timer_value": "5",
        "warmup_value": "00:01:00",
        "cooldown_value": "00:01:00",
        "waterbreak_value": "00:02:30"
    }]
};

有时wwc_tag_names数组包含1-4个键值对

我试图以这种方式提取它

var ajaxresfortags = response.wwc_tag_names;
    if(ajaxresfortags.length>0)
    {
var warmuptag = ajaxresfortags[0].warmuptag;
var cooldowntag = ajaxresfortags[1].cooldowntag;
var waterbreaktag = ajaxresfortags[2].waterbreak_free;
var waterbreaktagpaid = ajaxresfortags[3].waterbreak_paid;
}

这是我的小提琴

https://jsfiddle.net/o2gxgz9r/9003/

如果存在，那么只提取warmuptag，cooldowntag，waterbreak_free，waterbreak_paid

Answer 1

您正在访问错误的变量以获取warmuptag，cooldowntag，waterbreak_free，waterbreak_paid。你应该使用ajaxres而不是ajaxresfortags：

if(ajaxresfortags.length > 0)
{
 var ajaxResDetailsData = ajaxres[0];
 var warmuptag = ajaxResDetailsData.warmuptag;
 var cooldowntag = ajaxResDetailsData.cooldowntag;
 var waterbreaktag = ajaxResDetailsData.waterbreak_free;
 var waterbreaktagpaid = ajaxResDetailsData.waterbreak_paid;
}

<强> Working Demo

如何检查是否存在然后只提取值？

1 个答案: