'Call a Function inside a innerHTML
.then(function renderTree (data){
Object.entries(data).map((item, index) => {
let container = document.querySelector(".container")
let li = document.createElement("li")
li.id = `${item[1].id}`
container.appendChild(li)
let span = document.createElement("span")
span.innerHTML = `${item[1].name}`
li.appendChild(span)
let ul = document.createElement("ul")
ul.innerHTML = renderTree(item[1].children)
li.appendChild(ul)
// gerarLi(item[1].id, item[1].name)
})
})
I trying to continue the childs inside a <ul>, the continuation is in:
ul.innerHTML = renderTree(item[1].children)
But this code doesn't work, my index.html return like:
<ul>undefined</ul>
What can I do for this?
Solution 1:[1]
I made up my own testing data and played around with your code a bit. These were the points I found:
- Your recursive function
renderTree()always placed elements directly into the.containerdiv. This should only happen for the outer call. - You did not
returnanything from yourrenderTree()function.
I changed the function such that it now returns an HTML-string to be placed inside a <ul> structure. This function can now safely be called recursively.
const data={abc:{id:"first",name:"number one"}, def:{id:"second",name:"el segundo"},
ghi:{id:"third",name:"der dritte",
children:{jkl:{id:"fourth",name:"child one"},mno:{id:"fifth",name:"child two"}}}};
function renderTree(data){ // returns an HTML string for a <UL> DOM element
return Object.values(data).map(({id,name,children}) => {
let li = document.createElement("li");
li.id = id;
let span = document.createElement("span");
span.innerHTML = name;
li.appendChild(span);
let ul = document.createElement("ul");
if (children){
ul.innerHTML = renderTree(children);
li.appendChild(ul);
}
return li.outerHTML;
}).join("\n");
}
document.querySelector(".container").innerHTML=renderTree(data);
span {color:blue}
<ul class="container"></ul>
I also changed the Object.entries() to Object.values() since you never use the keys of the processed objects.
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
| Solution | Source |
|---|---|
| Solution 1 |
