Question

请有人帮我解决这个问题。我是laravel的新手，现在我需要将一个数组从select标签存储到mysql，我有这个erorr。 SQLSTATE [42S02]：未找到基表或视图：1146表'cmsystem.account_code_project'不存在（SQL：从var actOnSelectCount = 2; var selectCount = 0; $scope.onSelectedFetchOther = function() { incrementAndActIfReady(); } $scope.onSelectedFetchDD = function(item) { incrementAndActIfReady(); } function incrementAndActIfReady = function() { selectCount++; if (selectCount === actOnSelectCount) { console.log("Doing something, 2 were selected!"); } }中选择var actOnSelectCount = 2; var selected = []; $scope.onSelectedFetchOther = function() { incrementAndActIfReady('A'); } $scope.onSelectedFetchDD = function(item) { incrementAndActIfReady('B'); } function incrementAndActIfReady = function(selector) { // Push selection to selected if not // already there. else it was a change in same // select. if (selected.indexOf(selector) != -1) { selected.push(selector); } if (selection.length === actOnSelectCount) { console.log("Doing something, 2 were selected!"); } }，其中account_code_ac_id = 1）即可。似乎它与使用belongsToMany（）函数有关。我这里应该做的是我的代码。

这是project_table

account_code_project

这是帐户代码表

project_pj_id

这是两个表的中间表

 //project table
       Schema::create('projects', function(Blueprint $table)
    {
        $table->increments('pj_id');
        $table->string('pj_title',100);
        $table->string('pj_description');
    });

项目控制器

 //codes
       Schema::create('accountcodes', function(Blueprint $table)
    {
        $table->increments('ac_id');
        $table->string('ac_code',100)->nullable();
        $table->string('ac_description')->nullable();
    });

模型

//intermediate table
    Schema::create('code_project', function(Blueprint $table){
        $table->increments('id');
        $table->integer('project_id')->unsigned();
        $table->foreign('project_id')->references('pj_id')->on('projects');

        $table->integer('account_id')->unsigned();
        $table->foreign('account_id')->references('ac_id')->on('accountcodes');
    });

谢谢。

Answer 1

您可以在belongsToMany

中提示表名

return $this->belongsToMany('App\AccountCode',"code_project");

这是因为按照惯例，您的数据透视表应该已经命名为：accountcode_project（它也不能反映您的模型名称应该是Accountcode）

Answer 2

这是因为您必须在模型关系中指定数据透视表名称：

return $this->belongsToMany('App\Project', 'code_project');

否则Eloquent会尝试按照惯例检测默认表名，我建议您遵循这些名称。

Answer 3

好吧，我通过声明两个模型的代表来解决它，所以它看起来像这个

//AccountCode
      public function projects()
{
    return $this->belongsToMany('App\Project', 'accountcode_project', 'ac_id', 'pj_id');
}


// Project
    public function accountcodes()
{
    return $this->belongsToMany('App\AccountCode', 'accountcode_project', 'pj_id', 'ac_id');
}

SQLSTATE [42S02]：找不到基表或视图（belongsToMany）Laravel 5.4

3 个答案: