首页 \ 问答 \ 停止在Rails中验证的关联5(Stop Associations from Validating in Rails 5)

停止在Rails中验证的关联5(Stop Associations from Validating in Rails 5)

 使用Rails 5，我正在尝试实现Omniauth-Facebook和Clearance用于用户身份验证。 
 

 注意：我的代码与此Gist完全相同 
 

 我得到它主要工作。 但是，当使用Facebook注册（即用户从未访问过该网站）时，Rails会抛出一个错误，说Validation failed: User must exist 。 我已经把问题缩小到了Gist中的这个块： 
 

  def self.create_with_omniauth(auth_hash)
    create! do |auth|
      auth.provider = auth_hash["provider"]
      auth.uid = auth_hash["uid"]
      auth.token = auth_hash["credentials"]["token"]
    end
  end

 

 当它击中时，它试图create! 没有auth.user对象，并且失败。 以下是sessions控制器的相关代码： 
 

#-- Spectically, the line below
authentication = Authentication.find_by_provider_and_uid(auth_hash["provider"], auth_hash["uid"]) || Authentication.create_with_omniauth(auth_hash)
#-- More code, for context
if authentication.user
  user = authentication.user 
  authentication.update_token(auth_hash)
  @next = root_url
  @notice = "Signed in!"
else
  user = User.create_with_auth_and_hash(authentication,auth_hash)
  @next = edit_user_path(user)   
  @notice = "User created - confirm or edit details..."
end

 

 我在Gist中唯一遗漏的是他的Authentications表的结构。 使用我发现的上下文线索，我创建了这个表： 
 

  def change
    create_table :authentications do |t|
      t.string :provider
      t.string :uid
      t.string :token
      t.references :user, foreign_key: true

      t.timestamps
    end
  end

 

 如果需要更多信息，我会尽我所能 

Using Rails 5, I'm trying to implement Omniauth-Facebook and Clearance for user authentication.
 

Note: My code is exactly like this Gist
 

I've gotten it mostly working. However when using Facebook to register (ie a user has never visited the site), Rails throws an error saying Validation failed: User must exist. I've narrowed down the problem to this block in the Gist:
 

  def self.create_with_omniauth(auth_hash)
    create! do |auth|
      auth.provider = auth_hash["provider"]
      auth.uid = auth_hash["uid"]
      auth.token = auth_hash["credentials"]["token"]
    end
  end

 

When it hits that, it tries to create! without a auth.user object present, and fails. Here is the relevant code from the sessions controller:
 

#-- Spectically, the line below
authentication = Authentication.find_by_provider_and_uid(auth_hash["provider"], auth_hash["uid"]) || Authentication.create_with_omniauth(auth_hash)
#-- More code, for context
if authentication.user
  user = authentication.user 
  authentication.update_token(auth_hash)
  @next = root_url
  @notice = "Signed in!"
else
  user = User.create_with_auth_and_hash(authentication,auth_hash)
  @next = edit_user_path(user)   
  @notice = "User created - confirm or edit details..."
end

 

The only thing I'm missing out of the Gist is the structure of his Authentications table. Using the context clues I found, I created this table:
 

  def change
    create_table :authentications do |t|
      t.string :provider
      t.string :uid
      t.string :token
      t.references :user, foreign_key: true

      t.timestamps
    end
  end

 

If any more information is needed, I'll provide what I can

原文：https://stackoverflow.com/questions/36343793
更新时间：2022-11-20 11:11

最满意答案

 使用C99可变长度数组（VLA）功能使问题变得简单。 C11支持VLA可选，但实现必须定义__STDC_NO_VLA__以表明它不支持VLA。 
 

 这是代码的一个版本。 我重命名了Turn()函数（它将矩阵右转90°为TurnR()并添加了一个TurnL()函数，该矩阵将矩阵向左旋转90°由于代码处理非平方矩阵，因此输出矩阵是分开的从输入矩阵（如果你只想使用平方矩阵，可以稍微简化代码）。 
 

#include <stdio.h>

static void TurnR(size_t rows, size_t cols, int matrix[rows][cols], int result[cols][rows])
{
    for (size_t r = 0; r < rows; r++)
    {
        for (size_t c = 0; c < cols; c++)
            result[c][rows - 1 - r] = matrix[r][c];
    }
}

static void TurnL(size_t rows, size_t cols, int matrix[rows][cols], int result[cols][rows])
{
    for (size_t r = 0; r < rows; r++)
    {
        for (size_t c = 0; c < cols; c++)
            result[cols - 1 - c][r] = matrix[r][c];
    }
}

static void Print(const char *tag, size_t rows, size_t cols, int matrix[rows][cols])
{
    printf("%s (%zux%zu):\n", tag, rows, cols);
    for (size_t r = 0; r < rows; r++)
    {
        const char *pad = "";
        for (size_t c = 0; c < cols; c++)
        {
            printf("%s%3d", pad, matrix[r][c]);
            pad = " ";
        }
        putchar('\n');
    }
}

int main(void)
{
    int BY[4][4] = {
        {  1,  2,  3,  4, },
        {  5,  6,  7,  8, },
        {  9, 10, 11, 12, },
        { 13, 14, 15, 16, },
    };
    int out[4][4];

    Print("before", 4, 4, BY);
    TurnR(4, 4, BY, out);
    Print("right", 4, 4, out);
    TurnL(4, 4, BY, out);
    Print("left", 4, 4, out);

    int m4x6[4][6] =
    {
        {  1,  2,  3,  4,  5,  6, },
        {  7,  8,  9, 10, 11, 12, },
        { 13, 14, 15, 16, 17, 18, },
        { 19, 20, 21, 22, 23, 24, },
    };
    int m6x4[6][4];

    Print("before", 4, 6, m4x6);
    TurnR(4, 6, m4x6, m6x4);
    Print("right", 6, 4, m6x4);
    TurnL(4, 6, m4x6, m6x4);
    Print("left", 6, 4, m6x4);

    int m5x3[5][3] =
    {
        {  1,  2,  3, },
        {  4,  5,  6, },
        {  7,  8,  9, },
        { 10, 11, 12, },
        { 13, 14, 15, },
    };
    int m3x5[3][5];

    Print("before", 5, 3, m5x3);
    TurnR(5, 3, m5x3, m3x5);
    Print("right", 3, 5, m3x5);
    TurnL(5, 3, m5x3, m3x5);
    Print("left", 3, 5, m3x5);
    TurnL(3, 5, m3x5, m5x3);
    Print("doubleL", 5, 3, m5x3);

    return 0;
}

 

 并且示例输出是： 
 

before (4x4):
  1   2   3   4
  5   6   7   8
  9  10  11  12
 13  14  15  16
right (4x4):
 13   9   5   1
 14  10   6   2
 15  11   7   3
 16  12   8   4
left (4x4):
  4   8  12  16
  3   7  11  15
  2   6  10  14
  1   5   9  13
before (4x6):
  1   2   3   4   5   6
  7   8   9  10  11  12
 13  14  15  16  17  18
 19  20  21  22  23  24
right (6x4):
 19  13   7   1
 20  14   8   2
 21  15   9   3
 22  16  10   4
 23  17  11   5
 24  18  12   6
left (6x4):
  6  12  18  24
  5  11  17  23
  4  10  16  22
  3   9  15  21
  2   8  14  20
  1   7  13  19
before (5x3):
  1   2   3
  4   5   6
  7   8   9
 10  11  12
 13  14  15
right (3x5):
 13  10   7   4   1
 14  11   8   5   2
 15  12   9   6   3
left (3x5):
  3   6   9  12  15
  2   5   8  11  14
  1   4   7  10  13
doubleL (5x3):
 15  14  13
 12  11  10
  9   8   7
  6   5   4
  3   2   1

 

 我不会想用*(ptr + index)符号编写代码，尤其是使用双下标; 它太容易出错，难以阅读（确实是一场噩梦！）。 

Using the C99 variable length array (VLA) feature make the problem easy. C11 makes support for VLAs optional, but the implementation must define __STDC_NO_VLA__ to show that it doesn't support VLAs.
 

Here's one version of the code. I've renamed your Turn() function (which turns the matrix 90° right into TurnR() and added a TurnL() function which turns the matrix 90° left. Because the code handles non-square matrices, the output matrix is separate from the input matrix. (You can simplify the code slightly if you only want to work with square matrices.)
 

#include <stdio.h>

static void TurnR(size_t rows, size_t cols, int matrix[rows][cols], int result[cols][rows])
{
    for (size_t r = 0; r < rows; r++)
    {
        for (size_t c = 0; c < cols; c++)
            result[c][rows - 1 - r] = matrix[r][c];
    }
}

static void TurnL(size_t rows, size_t cols, int matrix[rows][cols], int result[cols][rows])
{
    for (size_t r = 0; r < rows; r++)
    {
        for (size_t c = 0; c < cols; c++)
            result[cols - 1 - c][r] = matrix[r][c];
    }
}

static void Print(const char *tag, size_t rows, size_t cols, int matrix[rows][cols])
{
    printf("%s (%zux%zu):\n", tag, rows, cols);
    for (size_t r = 0; r < rows; r++)
    {
        const char *pad = "";
        for (size_t c = 0; c < cols; c++)
        {
            printf("%s%3d", pad, matrix[r][c]);
            pad = " ";
        }
        putchar('\n');
    }
}

int main(void)
{
    int BY[4][4] = {
        {  1,  2,  3,  4, },
        {  5,  6,  7,  8, },
        {  9, 10, 11, 12, },
        { 13, 14, 15, 16, },
    };
    int out[4][4];

    Print("before", 4, 4, BY);
    TurnR(4, 4, BY, out);
    Print("right", 4, 4, out);
    TurnL(4, 4, BY, out);
    Print("left", 4, 4, out);

    int m4x6[4][6] =
    {
        {  1,  2,  3,  4,  5,  6, },
        {  7,  8,  9, 10, 11, 12, },
        { 13, 14, 15, 16, 17, 18, },
        { 19, 20, 21, 22, 23, 24, },
    };
    int m6x4[6][4];

    Print("before", 4, 6, m4x6);
    TurnR(4, 6, m4x6, m6x4);
    Print("right", 6, 4, m6x4);
    TurnL(4, 6, m4x6, m6x4);
    Print("left", 6, 4, m6x4);

    int m5x3[5][3] =
    {
        {  1,  2,  3, },
        {  4,  5,  6, },
        {  7,  8,  9, },
        { 10, 11, 12, },
        { 13, 14, 15, },
    };
    int m3x5[3][5];

    Print("before", 5, 3, m5x3);
    TurnR(5, 3, m5x3, m3x5);
    Print("right", 3, 5, m3x5);
    TurnL(5, 3, m5x3, m3x5);
    Print("left", 3, 5, m3x5);
    TurnL(3, 5, m3x5, m5x3);
    Print("doubleL", 5, 3, m5x3);

    return 0;
}

 

And sample output is:
 

before (4x4):
  1   2   3   4
  5   6   7   8
  9  10  11  12
 13  14  15  16
right (4x4):
 13   9   5   1
 14  10   6   2
 15  11   7   3
 16  12   8   4
left (4x4):
  4   8  12  16
  3   7  11  15
  2   6  10  14
  1   5   9  13
before (4x6):
  1   2   3   4   5   6
  7   8   9  10  11  12
 13  14  15  16  17  18
 19  20  21  22  23  24
right (6x4):
 19  13   7   1
 20  14   8   2
 21  15   9   3
 22  16  10   4
 23  17  11   5
 24  18  12   6
left (6x4):
  6  12  18  24
  5  11  17  23
  4  10  16  22
  3   9  15  21
  2   8  14  20
  1   7  13  19
before (5x3):
  1   2   3
  4   5   6
  7   8   9
 10  11  12
 13  14  15
right (3x5):
 13  10   7   4   1
 14  11   8   5   2
 15  12   9   6   3
left (3x5):
  3   6   9  12  15
  2   5   8  11  14
  1   4   7  10  13
doubleL (5x3):
 15  14  13
 12  11  10
  9   8   7
  6   5   4
  3   2   1

 

I wouldn't dream of writing the code using the *(ptr + index) notation, especially with the double subscripts; it is just too error prone and hard to read (a nightmare, indeed!).

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