通过php向数据库添加细节

Question

我有一个代码，员工可以通过该代码申请休假，休假的详细信息应存储在数据库中。细节还包括2张图片，即empsign和empdate，其余的细节都是文本。

<?php
    require_once('config.php'); //connection to database
    $empname = $_REQUEST['empname']; //no default
    $empid = $_REQUEST['empid']; //no default
    $deptt = $_REQUEST['deptt']; //no default
    $noofdays = $_REQUEST['noofdays']; //no default
    $from = $_REQUEST['from']; //no default
    $to = $_REQUEST['to']; //no default
    $morning = $_REQUEST['morning']; //no default
    $afternoon = $_REQUEST['afternoon']; //no default
    $reasonemp = $_REQUEST['reasonemp']; //no default
    $contactaddr= $_REQUEST['contactaddr']; //no default
    $phoneno = $_REQUEST['phoneno']; //no default
    $empsign = $_REQUEST['empsign']; //no default
    $empdate= $_REQUEST['empdate']; //no default

    $filename = $_REQUEST['empsign']; //no default
    $buffer=base64_decode($filename);
    $path = "empsign/image".$date.".jpg";
    $path1 = "http://example.com/employee/empsign/image".$date.".jpg";
    $handle = fopen($path, 'wb');
    $numbytes = fwrite($handle, $buffer);
    fclose($handle);

    $filename= $_REQUEST['empdate']; //no default
    $buffer=base64_decode($filename);
    $path = "empdate/image".$date.".jpg";
    $path1 = "http://example.com/employee/empdate/image".$date.".jpg";
    $handle = fopen($path, 'wb');
    $numbytes = fwrite($handle, $buffer);
    fclose($handle);

    $inserts = mysqli_query(
        "insert into 
        approveleave (empname,empid,deptt,noofdays,from,to,morning,afternoon,reasonemp,contactaddr,phoneno,empsign,empdate)
        values ('".$empname."','".$empid."','".$deptt."','".$noofdays."','".$from."','".$to."','".$morning."','".$afternoon."','".$reasonemp."','".$contactaddr."','".$phoneno."','".$empsign."','".$empdate."')"
        );

    $posts[0]['message'] = 'leave applied';
    $idd = mysqli_insert_id();
    $selectt = mysqli_query("select * from approveleave where id = '".$idd."'");
    $results =  mysqli_fetch_assoc($selectt);
    $posts[0]['detail'] = $results;
    header('Content-type: application/json');
    echo json_encode($posts);
?>

问题是，当我运行此代码时，数据库中没有存储任何值，而我得到的错误是

  

[{“message”：“employee added”，“detail”：false}]

可以请任何人帮助我吗？