Question

我正在尝试在两个表，帐户和转移之间建立关联。在转账表格中，用户应该能够选择原始账户以及转账的目的地账户。但是在页面中，第一个列表（原始帐户）为空，只有第二个（目标帐户）具有现有帐户列表。

我需要的是原始帐户列表也显示数据。

我已经阅读了“关联 - 将表连接在一起”文档的页面并在堆栈溢出中搜索了这些信息，但我遗漏了一些东西。你能帮忙吗？提前谢谢。

我的表帐户和转帐构建如下：

CREATE TABLE accounts (
    id INT UNSIGNED AUTO_INCREMENT PRIMARY KEY,
    name VARCHAR(255) NOT NULL,
    account_types_id INT UNSIGNED,
    user_id INT UNSIGNED NOT NULL,
    description TEXT,
    created DATETIME,
    modified DATETIME,
    FOREIGN KEY user_key (user_id) REFERENCES users(id),
    FOREIGN KEY account_types_key (account_types_id) REFERENCES account_types(id),
    UNIQUE KEY name_user(name, user_id)
);

和

CREATE TABLE transfers (
    id INT UNSIGNED AUTO_INCREMENT PRIMARY KEY,
    user_id INT UNSIGNED NOT NULL,
    origin_account_id INT UNSIGNED NOT NULL,
    destination_account_id INT UNSIGNED NOT NULL,
    value decimal(10,2) NOT NULL,
    description TEXT,
    created DATETIME,
    modified DATETIME,
    FOREIGN KEY user_key (user_id) REFERENCES users(id),
    FOREIGN KEY origin_account_key (origin_account_id) REFERENCES accounts(id),
    FOREIGN KEY destination_account_key (destination_account_id) REFERENCES accounts(id)
);

我的模特如下：

TransfersTable.php：

$this->belongsTo('Accounts', [
    'className' => 'Accounts',
    'foreignKey' => 'origin_account_id',
    'propertyName' => 'account',
    'joinType' => 'INNER'
]);

$this->belongsTo('Accounts', [
    'className' => 'Accounts',
    'foreignKey' => 'destination_account_id',
    'propertyName' => 'account',
    'joinType' => 'INNER'
]);

AccountsTable.php：

$this->hasMany('Transfers', [
    'className' => 'Accounts', 
    'foreignKey' => 'account_id',
    'bindingKey' => 'origin_account_id'
]);

$this->hasMany('Transfers', [
    'className' => 'Accounts', 
    'foreignKey' => 'account_id',
    'bindingKey' => 'destination_account_id'
]);

Answer 1

解决方案是在Template / Transfers / add.ctp中。就像这样：

echo $this->Form->input('origin_account_id');
echo $this->Form->input('destination_account_id', ['options' => $accounts]);

应该是这样的：

echo $this->Form->input('origin_account_id', ['options' => $accounts]);
echo $this->Form->input('destination_account_id', ['options' => $accounts]);

Cakephp 3.0 - 与同一模型的多重关系

1 个答案: