Question

由于某种原因它不会更新，而是会添加新数据。我是初学者，我很难找到错误。

读取功能非常正常，插入工作也很好。它只是更新无法正常工作或插入而不是更新查询或数据。

<?php include "db.php"; ?>
<?php include "functions.php" ?>
<?php
  if(isset($_POST['submit']))

  {$username = $_POST['username'];

      $password = $_POST['password'];

      $id = $_POST['id'];

      $querys = "UPDATE users SET ";

  $querys .= "username = '$username', ";

  $querys .= "password = '$password' ";

  $querys .= "WHERE id = $id ";


  $result = mysqli_query($connection, $querys);

  if(!$result)
      {
          die('Query FAILED'. mysqli_error($connection));
      }
  }

  ?>
<!DOCTYPE html>
<html lang="en">
  <head>
    <meta charset="UTF-8">
    <title>Document</title>
    <link rel="stylesheet" 
      href="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.6/css/bootstrap.min.css
      ">
  </head>
  <body>
    <div class="container">
      <div class="col-xs-6">
        <form action="login_create.php" method="post">
          <div class="form-group">
            <label for="username">Username</label>
            <input type="text" name="username" class="form-control">
          </div>
          <div class="form-group">
            <label for="password">Password</label>
            <input type="password" name="password" class="form-control">
          </div>
          <div class="form-group">
            <select name="id" id="">
            <?php
              showAllData();
              ?>
            </select>
          </div>
          <input class="btn btn-primary" 
            type="submit" name="submit" value="Update">
        </form>
      </div>
    </div>
  </body>
</html>

感谢您的帮助。

Answer 1

您好，这是更新的一个小例子

if(isset($_POST["submit"])) {

$link=Conect(); 
$id=$_POST["id"];
$value1=$_POST["value1"];
$value2=$_POST["value2"];
$value3=$_POST["value3"];
$query = "UPDATE table_name SET field1='".$value1."',
 field2='".$value2."',
 "."field3='".$value3."' WHERE id='".$id."'";
$action =mysql_query($query,$link);

    $result=mysql_query($query,$link) or die("Error: ".mysql_error());
    echo "was update the code: ".$id;

}

祝你好运

SQL Update错误。而不是更新，它的添加

1 个答案: