Laravel 5.2选择数据未知变量



当用户搜索ajax时在控制器中获取数据的函数

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);

我不知道我的代码遗漏了什么,它是这样显示原始sql:

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" = ?

我尝试用num 123代替$receiptNum进行测试,raw sql中也显示?,请帮我解决

这里的"?"是因为它是一个语句。如果你想要一个字符串,你必须替换。

我的灵感来自https://gist.github.com/JesseObrien/7418983

,例如我做了这个

    $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);

最新更新