Question

您好我的问题是我创建了一个表单并将所有html POST数据存储到php变量中，并希望运行查询以将该数据插入到我的本地数据库中。我知道我的数据库是连接的。当我填写表单时，它打印出错误，因为查询不起作用，但实际上并没有打印出确切的查询。我无法弄清楚出了什么问题或如何从表单中更新我的数据库。任何帮助都会非常感谢！

代码：

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd">

<html>

<!--<?php
  require_once('dbConnection.php');
?>-->

<head>

  <title>Create Account</title>

  <link rel="stylesheet" type="text/css" href="style.css">

</head>


<body>

  <div class="header">Our Really Cool Banking App</div>

  <div id="leftcolumn"> 
      <!-- Creating Buttons here -->
      <div id="nav">
        <ul>
          <li><a href="banking.php">Home</a></li>
          <li><a href="checking.php">Checking</a></li>
          <li><a href="savings.php">Savings</a></li>
          <li><a href="createaccount.php">Create Account</a></li>
          <li><a href="createloan.php">Create Loan</a></li>
        </ul>
      </div>
  </div>

<div class="inputBox">
  <form method="post" action="<?php echo htmlspecialchars($_SERVER["PHP_SELF"]);?>">
    Member ID: <input type="text" name="memberId"><br>
    Balance: <input type="text" name="Balance"><br>
    PIN: <input type="text" name="pin"><br>
    Created Date: <input type="text" name="createdDate"><br>
    Initial Balance: <input type="text" name="initBalance"><br>
    Employee ID: <input type="text" name="employeeId"><br>
    Type: <input type="radio" name="accounttype" value="Checking">Checking
    <input type="radio" name="accounttype" value="Savings">Savings<br>
    <input type="submit" value="Submit">
  </form>
</div>

<?php

  $properties = parse_ini_file("properties.ini");
  if ( !( $database = mysql_connect( $properties["dbUrl"], $properties["username"], $properties["password"] ) ) ) 
  {
    echo "<p> Be sure to fill out information in the properties.ini file </p>";
    die( "Could not connect to database </body></html>" ); 
  } 

  // open our database 
  if ( !mysql_select_db( $properties["dbName"], $database ) ) 
    die( "Could not open the database </body></html>" ); 

  $memId = mysqli_real_escape_string($database, $_POST["memberId"]);
  $balance = mysqli_real_escape_string($database, $_POST["Balance"]);
  $pin = mysqli_real_escape_string($database, $_POST["pin"]);
  $createdDate = mysqli_real_escape_string($database, $_POST["createdDate"]);
  $initBalance = mysqli_real_escape_string($database, $_POST["initBalance"]);
  $employeeId = mysqli_real_escape_string($database, $_POST["employeeId"]);
  $type = mysqli_real_escape_string($database, $_POST["accounttype"]);

  $sql="INSERT INTO Account (MemberId, Balance, PIN, creationDate, InitialBalance, CreatedByEmployee, type)
  VALUES ('$memId', '$balance', '$pin', '$createdDate', '$initBalance', '$employeeId', '$type')";

  if (!mysqli_query($database,$sql)) {
    die('Error: ' . mysqli_error($database));
  }

  echo "1 record added";

  mysql_close( $database ); 

?>

</body>

</html>

SQL插入到给出错误但不打印错误

0 个答案: