Question

我正在尝试更新管理员mysql表上的记录。但是当我尝试这种方法时仍然没有变化..我无法看到这个代码的问题..

info.php的

<?php
$query = "SELECT * from admin where username = '{$_SESSION['login_user']}'";
$result = mysqli_query($conn,$query);

while($row=mysqli_fetch_assoc($result))
    {
          echo '<input type="text" class="form-control" name="user_id" disabled value='.$row['user_id'].'>';
          echo '<input type="text" class = "form-control" name = "lastname"  value='.$row['lname'].'>';
          echo '<input type="text" class = "form-control" name = "firstname"  value='.$row['fname'].'>';
          echo '<input type="text" class = "form-control" name = "middlename"  value='.$row['mname'].'>';
          echo '<input type="text" class = "form-control" name = "address"  value='.$row['address'].'>';
          echo '<input type="text" class = "form-control" name = "contact"  value='.$row['contact'].'>';
          echo '<input type="text" class = "form-control" name = "email_add"  value='.$row['email_add'].'>';
          echo '<input type="text" class = "form-control" name = "username"  value='.$row['username'].'>';
          echo '<input type="password" class = "form-control" name = "password"  value='.$row['password'].'>';
          echo '<input type="password" class = "form-control" name = "confirmPassword" value='.$row['conf_pass'].'>';
    }?>

          <button type="submit" class="btn btn-primary" name="update">Submit</button>

update.php ----我不知道它有什么问题。

<?php
include('pgconnect.php');
if(isset($_GET['update'])){
    $acctid = $_GET['user_id'];
    $lname = $_GET['lastname'];
    $fname = $_GET['firstname'];
    $mname = $_GET['middlename'];
    $address = $_GET['address'];
    $contact = $_GET['contact'];
    $email = $_GET['email_add'];
    $user = mysql_real_escape_string($_GET['username']);    # use whatever escaping function your db requires this is very important.
    $pass = mysql_real_escape_string($_GET['password']);
    $confpass = mysql_real_escape_string($_GET['confirmPassword']);
        $sql = "UPDATE admin SET lname='$lname', fname='$fname', mname ='$mname', address='$address' contact = $contact, email_add = '$email', username ='$user',password = '$pass', conf_pass = '$confpass' where user_id = '$acctid'";
        $result = mysqli_query($conn,$sql);
            echo "$sql";
                if($result){
                    echo"Succesully Updated!";
                }
                    else
                        {
                        echo"Cannot be Updated!";
                    }
        mysqli_close($conn);
    }?>

Answer 1

了解表格背后的机制：

的index.html：

<form action="update.php?param=234" method="post">
  <input type="text" name="firstfield" />
  <button name="update">Submit</button>
</form>

update.php：

<?php

echo $_GET['param']; //should output 234
echo $_POST['firstfield']; //should output whatever you put in your textfield
var_dump($_POST['update']); //should return something

你注意到了，参数之后？在您的操作中，作为GET值传递，所有其他值都是POST值。

无法在mysql数据库php中更新值

1 个答案: