I have an array with another nested array I need to map over. I think I almost got it but I am not getting that data back, can only see its an object. Looking at the new index I can see it is looping of the amount of objects in the object array.
Here is what I have currently:
class Menu extends Component {
constructor(props) {
super(props);
const arrayItems = [this.props.items.edges]
arrayItems.map((currentItem, index) => {
console.log("The current iteration is: " + index)
console.log("The current item is: " + currentItem)
return currentItem.map((newCurrentItem, newIndex) => {
console.log("The NEW current iteration is: " + newIndex)
console.log("The NEW current item is: " + newCurrentItem)
return newCurrentItem
})
})
...
}
Here is screenshot of what I can see in my console, which looks promising:
Can someone please point me in correct direction?
Actually, currentItem
is equal to this.props.items.edges
in your code, so you could just map this.props.items.edges
.
newCurrentItem
display as [object Object]
is because What does [object Object] mean? (JavaScript) .
So you could get data out of CurrentItem
as normal.
Your code could be simplified to:
class Menu extends Component {
constructor(props) {
super(props);
const arrayItems = this.props.items.edges
return arrayItems.map((currentItem) => {
return currentItem.id // or other attributes you what
})
...
}
After I've seen your command with sort items in state alphabetically , I believe map the node value should be what you want.
const newCurrentItem = [{ node: "abc" }, { node: "def" }];
// [Object, Object]
console.log(newCurrentItem);
// ["abc", "def"]
console.log(
newCurrentItem.map((data) => {
return data.node;
})
);
const newCurrentItem2 = [{ node: { aaa: "bbb" } }, { node: { aaa: "yyy" } }];
// [Object, Object]
console.log(newCurrentItem2);
// ["bbb", "yyy"]
console.log(
newCurrentItem2.map((data) => {
return data.node.aaa;
})
);
try arrayItems.forEach(currentItem => { //do something })
You can use .map
too bt in that case you need to extract the value of any property like @eux said
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