Question

我使用MySql创建了以下代码，但出于某种原因，它并没有运行。

<main>
    <?php
    $conn = mysqli_connect("localhost","root"," ","dole") or die("Error" . mysqli_error($conn));
    if(isset($_POST['voucher_details']))
    {
    $tran_no = isset($_POST['tran_no']);
    $check_no = isset($_POST['check_no']);
    $payee = isset($_POST['payee']);
    $add = isset($_POST['add']);
    $amount = isset($_POST['amount']);
    $amountw = isset($_POST['amountw']);
    $obj_code = isset($_POST['obj_code']);
    $payment = isset($_POST['payment']);
    $div_no = isset($_POST['div_no']);
       $sql="INSERT INTO tbl_gip  VALUES ('tran_no','obj_code','check_no','div_no','payee','payment','add','amount','amountw')     ('{$tran_no}','{$obj_code}','{$check_no}','{$div_no}','{$payee}','{$payment}','{$add}','{$amount}','{$amountw}')";
  }
  ?>
        <section>
      <form method="POST" action="<?php echo $_SERVER['PHP_SELF']; ?>" name="voucher_details">
            <div id="form">
    <input class="form-control" name="tran_no" type="number" placeholder="Transcation No." readonly><br>
    <input class="form-control" name="obj_code" type="text" placeholder="Object-Code"><br>
    <input class="form-control" name="check_no" type="text" placeholder="Check No."><br>
    <input class="form-control" name="div_no" type="text" placeholder="DIV-no"><br>
    <input class="form-control" name="payee" type="text" placeholder="Payee"><br>
    <input class="form-control" name="payment" type="text" placeholder="Nature of Payment">br>
    <input class="form-control" name="add" type="text" placeholder="Address"><br>
    <input class="form-control" name="amount" type="text" placeholder="Amount"><br>
    <input class="form-control" name="amountw" type="text" placeholder="Amount in Words" readonly><br>
        <button type="submit" class="btn btn-success" name="save" id="save">Save</button>&nbsp;
      </div>
      </form>

它不会返回任何错误，但我认为此代码不起作用。有人可以解释我犯的错误，怎么没有错误出现？

Answer 1

您已在数据库中创建了插入查询，但您没有运行它。

你必须改为：

$sql="INSERT INTO tbl_gip  VALUES ('tran_no','obj_code','check_no','div_no','payee','payment','add','amount','amountw')     ('{$tran_no}','{$obj_code}','{$check_no}','{$div_no}','{$payee}','{$payment}','{$add}','{$amount}','{$amountw}')";

mysqli_query($conn, $sql);

Answer 2

最有可能的候选人是无效的缩进。 <?php结束标记未正确缩进。因此，这将使执行程序将整个事物假设为PHP（我不确定如此，但这是JavaScript中的情况所发生的情况）。

<?php
    //Content here
?>

Answer 3

您的变量是错误的，您应该过滤它们，并且您需要执行查询。将其更改为：

<?php
$conn = mysqli_connect("localhost","root"," ","dole") or die("Error" . mysqli_error($conn));
if(isset($_POST['voucher_details'])){
    //instead of $POST['var'], use filter_input(INPUT_POST,'var',{filter_type}),
    //same goes for $_GET['var']...use filter_input(INPUT_GET,'var',{filter_type})
    $tran_no = filter_input(INPUT_POST,'tran_no',FILTER_VALIDATE_INT);
    //...continue to filter each $var based on type;FILTER_VALIDATE_INT for ints, FILTER_SANITIZE_STRING for strings, etc.

    //Also note that your 'values' clause was in the wrong spot as well
    $sql="INSERT INTO tbl_gip (tran_no,obj_code,check_no,div_no,payee,payment,add,amount,amountw) VALUES ('{$tran_no}','{$obj_code}','{$check_no}','{$div_no}','{$payee}','{$payment}','{$add}','{$amount}','{$amountw}')";
    mysqli_query($conn, $sql);
}
?>

MySql查询不会运行，也不会返回任何错误消息

3 个答案: