Question

我有这个查询：

    $ManagerName = $mysqli->real_escape_string($_POST['ManagerName']);
    $Password = encryptIt($_POST['Password']);
    $Address = $mysqli->real_escape_string($_POST['Address']);
    $Email = $mysqli->real_escape_string($_POST['Email']);
    $PhoneNo = $mysqli->real_escape_string($_POST['PhoneNo']);
    $OfficeName = $mysqli->real_escape_string($_POST['OfficeName']);

    $ConsignmentNo = $_POST['percel'];
    $status = 'In Transit';

    $det = explode(",",$ConsignmentNo);

$sql = "INSERT INTO tbl_courier_officers (officer_name, off_pwd, address, email, ph_no, office, consignment, status, reg_date)
        VALUES ('$ManagerName', '$Password', '$Address', '$Email', '$PhoneNo', '$OfficeName','$ConsignmentNo','$status', NOW())";   

    $done = mysqli_query($mysqli, $sql);

    if($done){

    echo "added";

    foreach($det as $valuez)
    {
    $m = $Email;
    mysqli_query($mysqli,"INSERT INTO courier_track (email, percel_num) values('$m', '$valuez')");

    }

    mysqli_query($mysqli, "insert into user_log (username,name,action,time, user_id, mydate, mtime)values('$uname','$fullname','Added $ManagerName to courier officers table', '$tv', '$id', '$t', '$tv')");

    }else{
        echo 'Error occured: '.$mysqli->error;
    }

第一次插入为真后，我要提交第二个查询，这是一个 foreach循环。我希望它提交出现值的次数，但第一次插入后不提交。

除 foreach 中的查询外，所有查询均有效。

有人可以告诉我该怎么做吗？

Answer 1

更改这段代码

 mysqli_query($mysqli,"INSERT INTO courier_track (email, percel_num) values('$m', '$valuez')");

具有如下错误通知

 mysqli_query($mysqli,"INSERT INTO courier_track (email, percel_num) values('$m', '$valuez')") or die(mysqli_error($mysqli));

然后您将了解insert语句中的错误。

如果不起作用，请输入print_r（$ det）;并将其添加到OP中，以便我们可以正确地找出问题

在foreach中插入查询未插入

1 个答案: