moveTreeNode.js ➔ ??? - Code Metrics - bencripps/react-redux-grid - Measure and Improve Code Quality continuously with Scrutinizer

moveTreeNode.js ➔ ??? A
last analyzed 2018-02-15 14:17 UTC

↳ Parent: src/util/moveTreeNode.js

Complexity

Conditions	4
Paths	4

Size

Total Lines

Duplication

Lines	0
Ratio	0 %

Importance

Changes

Metric	Value
cc	4
nc	4
dl	0
loc	54
rs	9.0306
c	0
b	0
f	0
nop	7

How to fix Long Method

import { List } from 'immutable';

import { findTreeNode } from './findTreeNode';

export const moveTreeNode = (
    treeData,
    currentIndex,
    currentPath,
    nextIndex,
    nextPath,
    childIdentifier = 'children',
    rootIdentifier = 'root'
) => {

    const originalTreeData = treeData;

    const { node: currentParent, indexPath: currentIndexPath } = findTreeNode(
        treeData,
        currentPath,
        childIdentifier,
        rootIdentifier
    );

    if (!currentParent) {

        return originalTreeData;
    }

    currentIndexPath.push(...[childIdentifier, currentIndex]);


    let node = treeData.getIn(currentIndexPath);

    treeData = treeData.deleteIn(currentIndexPath);

    const { node: nextParent, indexPath: nextIndexPath } = findTreeNode(
        treeData,
        nextPath,
        childIdentifier,
        rootIdentifier
    );

    if (!nextParent) {

        return originalTreeData;
    }

    nextIndexPath.push(childIdentifier);


    if (!List.isList(treeData.getIn(nextIndexPath))) {
        treeData = treeData.setIn(nextIndexPath, List());
    }

    node = node.set('parentId', nextPath.last());

    treeData = treeData.setIn(
        nextIndexPath,
        treeData.getIn(nextIndexPath).insert(nextIndex, node)
    );

    return treeData;
};


1			import { List } from 'immutable';
2
3			import { findTreeNode } from './findTreeNode';
4
5			export const moveTreeNode = (
6			treeData,
7			currentIndex,
8			currentPath,
9			nextIndex,
10			nextPath,
11			childIdentifier = 'children',
12			rootIdentifier = 'root'
13			) => {
14
15			const originalTreeData = treeData;
16
17			const { node: currentParent, indexPath: currentIndexPath } = findTreeNode(
18			treeData,
19			currentPath,
20			childIdentifier,
21			rootIdentifier
22			);
23
24			if (!currentParent) {
			0 ignored issues – show Best Practice introduced 2016-09-30 16:20 UTC by Report Bug Copy Issue Report If you intend to check if the variable `currentParent` is declared in the current environment, consider using `typeof currentParent === "undefined"` instead. This is safe if the variable is not actually declared. Loading history...
25			return originalTreeData;
26			}
27
28			currentIndexPath.push(...[childIdentifier, currentIndex]);
			0 ignored issues – show Bug introduced 2016-09-30 16:20 UTC by Report Bug Copy Issue Report The variable `currentIndexPath` seems to be never declared. If this is a global, consider adding a `/** global: currentIndexPath */` comment. This checks looks for references to variables that have not been declared. This is most likey a typographical error or a variable has been renamed. To learn more about declaring variables in Javascript, see the MDN. Loading history...
29
30			let node = treeData.getIn(currentIndexPath);
			0 ignored issues – show introduced 2016-09-30 16:20 UTC by Report Bug Copy Issue Report You seem to be assigning a new value to the const `node`. Consider using a var instead. Loading history... Comprehensibility Naming Best Practice introduced 2016-09-30 16:20 UTC by Report Bug Copy Issue Report The variable `node` already seems to be declared on line `17`. Consider using another variable name or omitting the `var` keyword. This check looks for variables that are declared in multiple lines. There may be several reasons for this. In the simplest case the variable name was reused by mistake. This may lead to very hard to locate bugs. If you want to reuse a variable for another purpose, consider declaring it at or near the top of your function and just assigning to it subsequently so it is always declared. Loading history...
31			treeData = treeData.deleteIn(currentIndexPath);
32
33			const { node: nextParent, indexPath: nextIndexPath } = findTreeNode(
34			treeData,
35			nextPath,
36			childIdentifier,
37			rootIdentifier
38			);
39
40			if (!nextParent) {
			0 ignored issues – show Best Practice introduced 2016-09-30 16:20 UTC by Report Bug Copy Issue Report If you intend to check if the variable `nextParent` is declared in the current environment, consider using `typeof nextParent === "undefined"` instead. This is safe if the variable is not actually declared. Loading history...
41			return originalTreeData;
42			}
43
44			nextIndexPath.push(childIdentifier);
			0 ignored issues – show Bug introduced 2016-09-30 16:20 UTC by Report Bug Copy Issue Report The variable `nextIndexPath` seems to be never declared. If this is a global, consider adding a `/** global: nextIndexPath */` comment. This checks looks for references to variables that have not been declared. This is most likey a typographical error or a variable has been renamed. To learn more about declaring variables in Javascript, see the MDN. Loading history...
45
46			if (!List.isList(treeData.getIn(nextIndexPath))) {
47			treeData = treeData.setIn(nextIndexPath, List());
48			}
49
50			node = node.set('parentId', nextPath.last());
51
52			treeData = treeData.setIn(
53			nextIndexPath,
54			treeData.getIn(nextIndexPath).insert(nextIndex, node)
55			);
56
57			return treeData;
58			};
59

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moveTreeNode.js ➔ ??? A last analyzed 2018-02-15 14:17 UTC

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moveTreeNode.js ➔ ??? A
last analyzed 2018-02-15 14:17 UTC