好的，所以这很困难。我想。
我有两个对象数组，其中包含有关Pod的一些信息。我希望提供环境的并排比较，以显示吊舱中包含完全相同的图像；诀窍在于，我也需要捕捉图像不匹配的豆荚。结果将呈现在一个看起来像这样的表上（至少是第一次迭代）：
|---------------------|------------------|------------------|
| Pod Name | Image from Env1 | Image from Env2 |
|---------------------|------------------|------------------|
| foo | foo:1.0.0 | foo:1.0.0 | <---- images match
|---------------------|------------------|------------------|
| foo | foobar:1.0.0 | | <---- No match for image tag; image name different from the pod name as the pods may contain multiple images
|---------------------|------------------|------------------|
| foo | | foobar:2.0.0 | <---- As above
|---------------------|------------------|------------------|
| bar | bar:2.0.0 | bar:2.0.0 | <---- images match
|---------------------|------------------|------------------|
| baz | baz:1.0.0 | | <----
|---------------------|------------------|------------------| Note the 'no match'; so own row for now
| baz | | baz:2.0.0 | <----
|---------------------|------------------|------------------|
稍后我将使用CSS突出显示匹配与差异。
我最努力的事情是：
foo窗格）[{ "Pod Name": "foo", "Image_Env_1": foo:1.0.0", "Image_Env_2": foo:1.0.0"}...，并在不匹配的情况下为值保留空白字符串（如表中的空白单元格）。但这可能会与foo窗格中具有其他图像标签的其他图像发生冲突。因此Env1数组类似于以下代码片段：
[
{
"Image": {
"S": "foo:1.0.0"
},
"Pod Name": {
"S": "foo"
}
},
{
"Image": {
"S": "foobar:0.2.0"
},
"Pod Name": {
"S": "foo"
}
},
{
"Image": {
"S": "bar:1.0.0"
},
"Pod Name": {
"S": "bar"
}
},
{
"Image": {
"S": "baz:1.0.0"
},
"Pod Name": {
"S": "baz"
}
},
{
"Image": {
"S": "qux:1.0.0"
},
"Pod Name": {
"S": "foo"
}
}
]
Env2数组：
[
{
"Image": {
"S": "foo:2.0.0"
},
"Pod Name": {
"S": "foo"
}
},
{
"Image": {
"S": "foobar:0.2.0"
},
"Pod Name": {
"S": "foo"
}
},
{
"Image": {
"S": "bar:1.0.0"
},
"Pod Name": {
"S": "bar"
}
},
{
"Image": {
"S": "baz:3.0.0"
},
"Pod Name": {
"S": "baz"
}
}
]
答案 0 :(得分：0)
您可以编写一个小助手方法，该方法汇总来自您的环境文件的信息，即首先按pod分组，然后汇总使用的图像。这样，在多个环境中使用图像的过滤条目就不那么重要了。
类似的东西（请注意，这适用于任意数量的环境）
function getAggregatedPodInfos(...environments) {
const aggregatedInfos = {pods: {}, environments: []};
environments.forEach(env => addEnvironmentInfos(env, aggregatedInfos))
return aggregatedInfos;
}
function addImageInfoByEnvironment(pod, environment, imageName) {
if (!pod.imageInfo[environment.envName]) {
pod.imageInfo[environment.envName] = [];
}
pod.imageInfo[environment.envName].push(imageName)
}
function addEnvironmentInfos(environment, envInfo) {
envInfo.environments.push(environment.envName);
for (const envEntry of environment.data) {
const podName = envEntry["Pod Name"]["S"];
const imageName = envEntry["Image"]["S"];
if (!envInfo.pods[podName]) {
envInfo.pods[podName] = {imageInfo: {}};
}
addImageInfoByEnvironment(envInfo.pods[podName], environment, imageName);
}
}
const res = getAggregatedPodInfos({envName: 'env1', data: env1}, {envName: 'env2', data: env2});
console.log(JSON.stringify(res))
这将打印：
{
"pods": {
"foo": {
"imageInfo": {
"env1": ["foo:1.0.0", "foobar:0.2.0", "qux:1.0.0"],
"env2": ["foo:2.0.0", "foobar:0.2.0"]
}
},
"bar": {
"imageInfo": {
"env1": ["bar:1.0.0"],
"env2": ["bar:1.0.0"]
}
},
"baz": {
"imageInfo": {
"env1": ["baz:1.0.0"],
"env2": ["baz:3.0.0"]
}
}
},
"environments": ["env1", "env2"]
}
答案 1 :(得分：0)
这很难！但是我认为该过程可以分为两个主要阶段：
第一步可以进一步分解为处理原始数据，然后对其进行处理。
在处理诸如Pod Name之类的类别时，我想做的是创建包含Set对象的Map对象。在这种情况下，它可能看起来像这样：
// Let's assume Env1 is loaded already.
const podNameMap1 = new Map();
Env1.forEach( (value, index) => {
// Here 'value' is element in Env1
const podName = value['Pod Name'].S;
const podValue = podNameMap.get(podName);
// returns undefined if it does not exist
if (!podValue) {
podNameMap.set(podName, new Set(value['Image'].S) );
// If there is no podValue, set podName to a new Set with the images in it.
} else {
// Here there is a podValue, and we can add a new member to the Set.
podValue.add(value['Images'].S);
// podValue is a Set object which means the map value needs to be set one time.
}
});
该podNameMap1的末尾包含所有吊舱及其所有图像。下一步是对Env2做同样的事情，这时有两个地图，显示了各种容器名称和图像。
之后，可以进行比较，这比创建地图要复杂得多：
// Let's assume podNameMap1 and podNameMap2 are both loaded.
// I don't know if it's possible for pods to be mismatched,
// I'm going to assume they are cannot be.
// Images in both
const inBothEnvsMap = new Map();
// Images in env1 but not env2
const inEnv1Map = new Map();
// Images in env2 but not env1
const inEnv2Map = new Map();
podNameMap1.forEach( (imageSet1, podName) => {
// Here, imageSet1 and podName are self explanitory
const imageSet2 = podNameMap1.get(podName); // assuming this is always the case
// Computing the set intersections n stuff
const inEnv1 = new Set();
const inEnv2 = new Set();
const inBothEnvs = new Set();
// Note: I wish JS had built in set operations because there's probably
// a better way to do this but alas...
imageSet1.forEach( image => {
if (imageSet2.has(image))
inBothEnvs.add(image);
else // in imageSet1 only
inEnv1.add(image);
});
// now inEnv1 AND inBothEnvs are complete
imageSet2.forEach( image => {
if (!imageSet1.has(image))
inEnv2.add(image);
});
// now this pod is done
inBothEnvsMap.set(podName, inBothEnvs);
inEnv1Map.set(podName, inEnv1);
inEnv2Map.set(podname, inEnv2);
});
此时，这是“查找匹配和差异”步骤。
继续进行下一步“将信息转换为有用的东西”的部分不是简单的设置操作所特有的，因此我无法真正给出合理的建议。