php - Laravel 5.2 选择数据未知变量

Question

我在控制器中的函数用于在用户搜索 ajax 时获取数据。

这是我的代码：

$receiptNum = Request::get('receiptNum');
// echo $receiptNum = 123456
DB::connection()->enableQueryLog();
$q = DB::table('tbl_receipt AS r')
            ->join('company AS com', 'r.com_id', '=', 'com.com_id')
            ->join('branches AS b', 'com.b_id', '=', 'b.b_id')
            ->join('employee AS e', 'r.created_by', '=', 'e.e_id')
            ->where('r.receipt_code','=',$receiptNum)
            ->get();
$query = DB::getQueryLog();
var_dump($query);

我不知道我的代码中缺少什么，它显示的原始 sql 如下所示：

SELECT *
FROM       "tbl_receipt" AS "r"
INNER JOIN "company"     AS "com" ON "r"."com_id" = "com"."com_id"
INNER JOIN "branches"    AS "b"   ON "com"."b_id" = "b"."b_id"
INNER JOIN "employee"    AS "e"   ON "r"."created_by" = "r"."e_id"
WHERE "r"."receipt_code" = ?

我试图$receiptNum用 num替换123进行测试，它也显示?在 中raw sql，请帮我解决它。

Answer 1

这 ”？” 因为它是一个声明。如果你想要一个字符串，你必须替换。

我受到https://gist.github.com/JesseObrien/7418983的启发

例如我做这个。

    $q = DB::table('tbl_receipt AS r') 
->join('company AS com', 'r.com_id', '=', 'com.com_id') 
->join('branches AS b', 'com.b_id', '=', 'b.b_id') 
->join('employee AS e', 'r.created_by', '=', 'e.e_id') 
->where('r.receipt_code','=','123') ; 

$sql = $q->toSql(); 
// $bindings = $q->getBindings(); 
foreach($q->getBindings() as $binding) 
{ 
$value = is_numeric($binding) ? $binding : "'".$binding."'"; 
$sql = preg_replace('/\?/', $value, $sql, 1); 
} 
var_dump($sql);