Question

我的sql代码是：

Select * from table where cond(condition)1 and con2 ...... and(conX or conY).

我如何用原始表达式编写它。我不知道解决这个问题的关键工作。

更新：我的代码是：

//$builder->to sql =   Select * from table where cond(condition)1 and con2 ...... ;
    if(something != true) $builder->whereRaw(conX);  
else $builder->whereRaw((conX or ConY));
    //$builder->to sql =   Select * from table where cond(condition)1 and con2 ...... and (conX or conY);

好的，它有效，但我需要写2次conX，虽然它太长了。所以现在我想签约。

//$builder->to sql =   Select * from table where cond(condition)1 and con2 ...... and conX;
if(something = true) $builder->raw or do something;   (@)
//$builder->to sql =   Select * from table where cond(condition)1 and con2 ...... and (conX or conY);

问题在于：我应该在（@）行做什么来制作我期望的结果

Answer 1

尝试：

if (something == true) {
    $result = DB::table('table')
        ->selectRaw('*')
        ->where(condition1)
        ->where(condition2)
        ->get();
}

或者：

if (something == true) {
    $result = DB::table('table')
        ->select(DB::raw(YOUR SQL STATEMENT))
        ->get();
}

进一步阅读： https://laravel.com/docs/5.5/queries

Laravel中的原始表达？

1 个答案: