Question

<form name="applyform" action="applyform.php" method="post">
    <fieldset>
        <legend>Application Details</legend>
        <p>Name :<?php echo $row ["Emp_Fname"]; ?></p>
        <p>ID number :<?php echo $row['Emp_ID']; ?></p>
        <p>Email :<?php echo $row['Emp_Email']; ?></p>
        <p>Address :<?php echo $row['Emp_Address']; ?></p>
        <p>Handphone Number :<?php echo $row['ContactNo_HP']; ?></p>
        <p>Phone Number :<?php echo $row['ContactNo_Home']; ?></p>
        <p>Date of application :<?php echo $row['Leave_RequestDate']; ?></p>
        <p>Type of leave:
           <select name="leave type">
              <option selected>Annual leave</option>
              <option>Sick leave</option>
              <option>Emergency leave</option>
              <option>Maternity leave</option>
            </select>
        </p>
        <p>Leave duration:<input type="date" name="leave_start">to<input type="date" name="leave_end"></p>
        <p>Reason:<textarea rows="4" cols="50" name="reason"></textarea></p>
        <p><input type="submit" name="submitbtn" value="Submit"/>

这是我表格的代码。有什么不对吗？

<?php

if(isset($_POST['submitbtn'])) {

  if(!$con)  {
    die("cannot connect : " .mysql_error());
  }
  $sql = ("INSERT INTO leave(Leave_Start,Leave_End,Leave_Reason)    VALUES('$_POST[leave_start]','$_POST[leave_end]','$_POST[reason]')");
  mysql_query($sql,$con);

  mysql_close($con);
}
?>

以上是我的PHP代码。当我尝试提交表单时数据库不会更新，有人可以帮助我吗？

Answer 1

您是否忘记在PHP代码的第5行附近包含mysql_connect？

$con = mysql_connect('mysql_host', 'mysql_user', 'mysql_password');

Answer 2

而不是

$sql = ("INSERT INTO leave(Leave_Start,Leave_End,Leave_Reason) VALUES('$_POST[leave_start]','$_POST[leave_end]','$_POST[reason]')");

试

$sql = ("INSERT INTO leave(Leave_Start,Leave_End,Leave_Reason) VALUES('".$_POST["leave_start"]."','".$_POST["leave_end"]."','".$_POST["reason"]."')");

注意：执行此类查询并不安全，因为$_POST未在

之前检查

这样的事情应该更好

$sql = ("INSERT INTO leave(Leave_Start,Leave_End,Leave_Reason) VALUES('".mysqli_real_escape_string($con,$_POST["leave_start"])."','".mysqli_real_escape_string($con,$_POST["leave_end"])."','".mysqli_real_escape_string($con,$_POST["reason"])."')");

Answer 3

您需要了解一下PHP变量传递和SQL注入。您的最终查询必须看起来像

 $sql = ("INSERT INTO leave(Leave_Start,Leave_End,Leave_Reason) VALUES('{".mysqli_real_escape_string($_POST['leave_start'])."}','{".mysqli_real_escape_string($_POST['leave_end'])."}','{".mysqli_real_escape_string($_POST['reason'])."}')");

Answer 4

您可以尝试 MySQLi ，而不是谓词 MySQL 。

<?php

if(isset($_POST['submitbtn']))
{

$connection=mysqli_connect("Host","Username","Password","Database");

if(mysqli_connect_errno()){

echo "Error".mysqli_connect_error();
}

mysqli_query($connection,"INSERT INTO leave(Leave_Start, Leave_End, Leave_Reason)    VALUES('$_POST[leave_start]','$_POST[leave_end]','$_POST[reason]')");

}
?>

Answer 5

您的插入内容不正确：

$sql = ("INSERT INTO leave(Leave_Start,Leave_End,Leave_Reason)   VALUES('$_POST[leave_start]','$_POST[leave_end]','$_POST[reason]')");

应该是这样的：

$sql = ("INSERT INTO leave(Leave_Start,Leave_End,Leave_Reason)    VALUES('{$_POST["leave_start"]}','{$_POST["leave_end"]}','{$_POST["reason"]}')");

您应始终转义字符串中的特殊字符。否则他们会被误解。

http://php.net/manual/en/language.types.string.php

顺便说一句，我假设$ con是一个有效的连接对象。

为什么我提交表单时无法更新数据库？

5 个答案: