[英]javascript dynamically create multidimensional array from string
<span id="local->ethernet->port3->rx_flow">q4234</span>
<span id="local->ethernet->port3->rx">q345</span>
<span id="local->ethernet->port1->rx_flow">128</span>
<span id="remote->id">128</span>
and I need to make multidimensional array from them by ID example from element <span id="local->ethernet->port3->rx_flow">q4234</span>
并且我需要通过元素<span id="local->ethernet->port3->rx_flow">q4234</span>
ID示例从它们组成多维数组
array I need is array["local"]["ethernet"]["port3"]["rx_flow"]="q4234"
我需要的array["local"]["ethernet"]["port3"]["rx_flow"]="q4234"
是array["local"]["ethernet"]["port3"]["rx_flow"]="q4234"
function I created is: 我创建的函数是:
function make_cfg(){
var result=new Array();
var x=document.getElementById(*);
var len=x.length;
var arr;
for (var i=0; i<=len; i++;){
if(x[i].id){
if(x[i].id.indexOf("->") != -1) {
arr=x[i].id.split("->");
result=make_obj(result,arr);
}
}
}
return result;
}
And I have no idea how to make function make_obj()
而且我不知道如何制作函数make_obj()
I won't write the whole thing for you, I just help with the hard part a bit. 我不会为您写全部内容,我只是为难的部分提供帮助。
This snippet will take the two strings (basically id
and innerHTML
, here s
and s2
) and construct a nested object (there are no associative arrays in Javascript) out of it. 此代码段将使用两个字符串(基本上是id
和innerHTML
,这里s
和s2
),并从中构造出一个嵌套对象(Javascript中没有关联数组)。
var s='local->ethernet->port3->rx_flow',
s2='q4234',
a=s.split('->'),
obj=constructObject(a, s2);
function constructObject(a, final) {
var val=a.shift();
var obj={};
if (a.length>0) {
obj[val]=constructObject(a, final);
} else {
obj[val]=final;
}
return obj;
}
It uses recursion to achieve its goal. 它使用递归来实现其目标。 If you have any questions about the code, please ask. 如果您对代码有任何疑问,请询问。
Here you can try it out . 在这里您可以尝试一下 。
What is left to do? 还剩下什么呢?
I guess you want to collect these things from the spans into ONE object, my example will create one object for every s
/ s2
. 我猜您想将这些内容从跨度收集到一个对象中,我的示例将为每个s
/ s2
创建一个对象。 If you have any further questions, I am happy to help. 如果您还有其他疑问,我们很乐意为您提供帮助。
this almost worked (not so elegant as a recursive function) 这几乎可以工作(不像递归函数那么优雅)
http://jsfiddle.net/mplungjan/3zhwv/ http://jsfiddle.net/mplungjan/3zhwv/
Missing the 128 from the remote/id but the rest works. 缺少来自远程/ ID的128,其余的工作正常。 I would like to figure out what to do to get the 128 from the node that is shorter than 4 我想弄清楚如何从小于4的节点中获取128
I agree it is not flexible like a recursive function, but I wanted to see if I could make a "brute force" first and then par it down to something more clever. 我同意它不像递归函数那样灵活,但是我想看看是否可以先制造“蛮力”,然后再将其分解为更聪明的东西。
<span id="local->ethernet->port3->rx_flow">q4234</span>
<span id="local->ethernet->port3->rx">q345</span>
<span id="local->ethernet->port1->rx_flow">128</span>
<span id="remote->id">128</span>
<hr/>
<pre>
myObject = {
"local":{
"ethernet":{
"port3": {
"rx_flow":"q4234",
"rx":"q345"
}
"port1": {
"rx_flow":"128"
}
}
},
"remote":{
"id":"128"
}
}
</pre>
<script>
var spans = document.getElementsByTagName("span");
var myObject = {};
for (var i=0;i < spans.length;i++) {
var id = spans[i].id;
var parts = id.split('->');
var val = spans[i].innerHTML
if (parts[0]) { // Local or remote
if (myObject[parts[0]] == null) myObject[parts[0]]={};
if (parts[1]) { // ethernet or id
if (myObject[parts[0]][parts[1]] == null) myObject[parts[0]][parts[1]]=(parts.length==1)?val:{};
if (parts[2]) { // port3 or port1
if (myObject[parts[0]][parts[1]][parts[2]] == null) myObject[parts[0]][parts[1]][parts[2]]=(parts.length==2)?val:{};
if (parts[3]) { // rx_flow or rx
myObject[parts[0]][parts[1]][parts[2]][parts[3]]=val;
}
}
}
}
}
for (var o in myObject) { // local or remote
document.write(o+'/');
for (var p in myObject[o]) { // ethernet or id
document.write(p+'/');
for (var q in myObject[o][p]) { // ports
document.write(q+':/');
for (var r in myObject[o][p][q]) { // rx_flow or rx
document.write(r+' - '+myObject[o][p][q][r]+'<br/>');
}
}
}
}
</script>
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