我如何从 Ajax 的对象中获取价值 [关闭]答案

【问题标题】：how I get Value from object from Ajax [closed]我如何从 Ajax 的对象中获取价值 [关闭]
【发布时间】：2022-01-18 14:13:33
【问题描述】：

我在 java 脚本中有这个 Ajax 调用：

function getCamInfo(value) {
  var body = "TakeNewPic=" + value;
  var req = new XMLHttpRequest();

  req.onreadystatechange = function() {
    if (this.readyState === 4 && this.status === 200) {
      var res = document.getElementById("parg");
      var outpt = JSON.parse(this.responseText);
      res.innerHTML = outpt;

      //console.log(outpt);
      /* var link = outpt.state[0];  
       var tex = document.getElementById("linkP");
       tex.value = link;*/
    }
  }

  req.open(
    "POST",
    "post_req.php",
    true,
  );

  req.setRequestHeader(
    "content-type",
    "application/x-www-form-urlencoded"
  );

  req.send(body);
}

我在响应中得到了这个 JSON 对象：

{"fingerprint":"FIG_0163","state":{"_apiVersion":2,"batteryLevel":0.78,"_batteryState":"disconnect","_cameraError":[],"_captureStatus":"idle","_capturedPictures":0,"_compositeShootingElapsedTime":0,"_function":"mySetting","_latestFileUrl":"http://192.168.254.124/files/90014a68423861503e031db848764600/100RICOH/R0011105.JPG","_mySettingChanged":false,"_pluginRunning":false,"_pluginWebServer":true,"_recordableTime":0,"_recordedTime":0,"storageUri":"http://192.168.254.124/files/90014a68423861503e031db848764600/"}}

例如，我如何从这个对象中获取一个值：只有 Url 图片：

"_latestFileUrl":"http://192.168.254.124/files/90014a68423861503e031db848764600/100RICOH/R0011105.JPG"

非常感谢

【问题讨论】：

我投票结束这个问题，因为没有尝试解决。

标签： javascript jquery json ajax object

【解决方案1】：

试试这个

var latestFileUrl = jsonObject.state._latestFileUrl;

【讨论】：

【解决方案2】：

用下面的修改你的 res.innerHTML。

res.innerHTML = outpt.state._latestFileUrl;

【讨论】：

【解决方案3】：

对于一个简单的版本：

const _fetch = () => Promise.resolve({
  "fingerprint": "FIG_0163",
  "state": {
    "_apiVersion": 2,
    "batteryLevel": 0.78,
    "_batteryState": "disconnect",
    "_cameraError": [],
    "_captureStatus": "idle",
    "_capturedPictures": 0,
    "_compositeShootingElapsedTime": 0,
    "_function": "mySetting",
    "_latestFileUrl": "http://192.168.254.124/files/90014a68423861503e031db848764600/100RICOH/R0011105.JPG",
    "_mySettingChanged": false,
    "_pluginRunning": false,
    "_pluginWebServer": true,
    "_recordableTime": 0,
    "_recordedTime": 0,
    "storageUri": "http://192.168.254.124/files/90014a68423861503e031db848764600/"
  }
})

_fetch()
  .then(x => x.state["_latestFileUrl"])
  .then(console.log);

对于一个过于复杂（但对我来说很有趣）的版本：

const _fetch = () => Promise.resolve({
  "fingerprint": "FIG_0163",
  "state": {
    "_apiVersion": 2,
    "batteryLevel": 0.78,
    "_batteryState": "disconnect",
    "_cameraError": [],
    "_captureStatus": "idle",
    "_capturedPictures": 0,
    "_compositeShootingElapsedTime": 0,
    "_function": "mySetting",
    "_latestFileUrl": "http://192.168.254.124/files/90014a68423861503e031db848764600/100RICOH/R0011105.JPG",
    "_mySettingChanged": false,
    "_pluginRunning": false,
    "_pluginWebServer": true,
    "_recordableTime": 0,
    "_recordedTime": 0,
    "storageUri": "http://192.168.254.124/files/90014a68423861503e031db848764600/"
  }
})

const compose = (...fns) =>
  fns.reduceRight((prevFn, nextFn) =>
    (...args) => nextFn(prevFn(...args)),
    value => value
  );

const prop = p => o => o[p];

const getFileUrlFromResponse = compose(prop("_latestFileUrl"), prop("state"));

_fetch()
  .then(getFileUrlFromResponse)
  .then(console.log);

【讨论】：