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2

$begingroup$

So I have the following data structures

export class BudgetGroupInfo {

    name: string;

    bcInfo: BudgetCatInfo[] = [];

}



export class BudgetCatInfo {

    name: string;

    description: string;

    bcAccounts: BCAccountInfo[] = [];

}



export class BCAccountInfo {

  name: string;

  description: string;

}

What I need to be able to do is find which BudgetGroupInfo contains both BudgetCatInfo.name and BCAccountInfo.name

This is what I have so far and it works, but I feel like this isn't the most efficient want to do this.

getGroupInfo(budCat: string, account: string, budgetGroupInfo: BudgetGroupInfo[]): BudgetGroupInfo {

    let groupInfo = null;



    budgetGroupInfo.forEach(infoGroup => {

      infoGroup.bcInfo.filter(bcInfo => bcInfo.name == budCat)

      .map(bcInfo => bcInfo.bcAccounts)

        .forEach(accounts => {

            const acctInfo: BCAccountInfo = accounts.find(acct => acct.name == account);

            if(acctInfo) {

              groupInfo = infoGroup;

            }

          });

      })



    return groupInfo;

  }

1. First I go through the groups and find which ones have the specific budCat
   


2. I take the valid BudgetCatInfo and map the bcAccount array


3. I search that array and look for the specified account
   


4. If I find it then I set it for return.

Just a note that there won't always be a result and there will never be multiple matches.

javascript typescript

share|improve this question

asked 2 days ago

Raymond HolguinRaymond Holguin

1111

New contributor

Raymond Holguin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

2

$begingroup$

So I have the following data structures

export class BudgetGroupInfo {

    name: string;

    bcInfo: BudgetCatInfo[] = [];

}



export class BudgetCatInfo {

    name: string;

    description: string;

    bcAccounts: BCAccountInfo[] = [];

}



export class BCAccountInfo {

  name: string;

  description: string;

}

What I need to be able to do is find which BudgetGroupInfo contains both BudgetCatInfo.name and BCAccountInfo.name

This is what I have so far and it works, but I feel like this isn't the most efficient want to do this.

getGroupInfo(budCat: string, account: string, budgetGroupInfo: BudgetGroupInfo[]): BudgetGroupInfo {

    let groupInfo = null;



    budgetGroupInfo.forEach(infoGroup => {

      infoGroup.bcInfo.filter(bcInfo => bcInfo.name == budCat)

      .map(bcInfo => bcInfo.bcAccounts)

        .forEach(accounts => {

            const acctInfo: BCAccountInfo = accounts.find(acct => acct.name == account);

            if(acctInfo) {

              groupInfo = infoGroup;

            }

          });

      })



    return groupInfo;

  }

1. First I go through the groups and find which ones have the specific budCat
   


2. I take the valid BudgetCatInfo and map the bcAccount array


3. I search that array and look for the specified account
   


4. If I find it then I set it for return.

Just a note that there won't always be a result and there will never be multiple matches.

javascript typescript

share|improve this question

asked 2 days ago

Raymond HolguinRaymond Holguin

1111

New contributor

Raymond Holguin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

$begingroup$

So I have the following data structures

export class BudgetGroupInfo {

    name: string;

    bcInfo: BudgetCatInfo[] = [];

}



export class BudgetCatInfo {

    name: string;

    description: string;

    bcAccounts: BCAccountInfo[] = [];

}



export class BCAccountInfo {

  name: string;

  description: string;

}

What I need to be able to do is find which BudgetGroupInfo contains both BudgetCatInfo.name and BCAccountInfo.name

This is what I have so far and it works, but I feel like this isn't the most efficient want to do this.

getGroupInfo(budCat: string, account: string, budgetGroupInfo: BudgetGroupInfo[]): BudgetGroupInfo {

    let groupInfo = null;



    budgetGroupInfo.forEach(infoGroup => {

      infoGroup.bcInfo.filter(bcInfo => bcInfo.name == budCat)

      .map(bcInfo => bcInfo.bcAccounts)

        .forEach(accounts => {

            const acctInfo: BCAccountInfo = accounts.find(acct => acct.name == account);

            if(acctInfo) {

              groupInfo = infoGroup;

            }

          });

      })



    return groupInfo;

  }

1. First I go through the groups and find which ones have the specific budCat
   


2. I take the valid BudgetCatInfo and map the bcAccount array


3. I search that array and look for the specified account
   


4. If I find it then I set it for return.

Just a note that there won't always be a result and there will never be multiple matches.

javascript typescript

share|improve this question

asked 2 days ago

Raymond HolguinRaymond Holguin

1111

New contributor

Raymond Holguin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

export class BudgetGroupInfo {

    name: string;

    bcInfo: BudgetCatInfo[] = [];

}



export class BudgetCatInfo {

    name: string;

    description: string;

    bcAccounts: BCAccountInfo[] = [];

}



export class BCAccountInfo {

  name: string;

  description: string;

}

getGroupInfo(budCat: string, account: string, budgetGroupInfo: BudgetGroupInfo[]): BudgetGroupInfo {

    let groupInfo = null;



    budgetGroupInfo.forEach(infoGroup => {

      infoGroup.bcInfo.filter(bcInfo => bcInfo.name == budCat)

      .map(bcInfo => bcInfo.bcAccounts)

        .forEach(accounts => {

            const acctInfo: BCAccountInfo = accounts.find(acct => acct.name == account);

            if(acctInfo) {

              groupInfo = infoGroup;

            }

          });

      })



    return groupInfo;

  }

share|improve this question

asked 2 days ago

Raymond HolguinRaymond Holguin

1111

New contributor

Raymond Holguin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|improve this question

asked 2 days ago

Raymond HolguinRaymond Holguin

1111

New contributor

Raymond Holguin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 2 days ago

Raymond HolguinRaymond Holguin

1111

New contributor

Raymond Holguin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 2 days ago

Raymond HolguinRaymond Holguin

1111