无法使用php更新mysql数据库

问题描述：

I am trying to update a record on my customer mysql table. However I get the following errors:

Notice: Undefined index: username in E:\EasyPHP-12.1\www\Register1.php on line 313

Notice: Undefined variable: db_password in E:\EasyPHP-12.1\www\Register1.php on line 315 error updating record

I can't see anything wrong with my code. And have been looking at it for a few hours with no luck.

<div class="footer">
  <img src="images/gardening5.jpg" alt="Image">
  <div>
    <h1>Tips of the Week</h1>
    <form method= "get">
      <form method="get">
      <p>Username:
        <input name="username" type="text">
        </p>
      </form>
      <p>
      <form method="get">
        <p>Password:
          <input name="password" type="password">
        </p>
        <p>First Name:
          <input name="first name" type="first name">
        </p>
        <p>Last Name:
           <input name="last name" type="last name">
        </p>
        <p>1st Line of Address:
          <input name="1st line of address" type="1st line of address">
        </p>
        <p>2nd Line of Address:
          <input name="2nd line of address" type"2nd line of address">
        </p>
        <p>Town:
           <input name="town" type"town">
        </p>
        <p>PostCode:
          <input name="postcode" type="postcode">
        </p>
        <p>Phone Number
          <input name="phone_number" type="phone_number">
        </p>
        <p>
          <input name="submit3" type="submit" value="update customer record">
        </p>
       </form>
     </form>
<?php
  $host="localhost"; // Host name
  $tbl_name="customer"; // Table name
  $db_user="root";
  $db_pass="";

  $connect = mysql_connect("$host", "$db_user", "$db_pass");
  $db_name="the_shop"; // Database name
  mysql_select_db("$db_name");

  if(isset($_GET['submit3'])){

    $db_username = $_GET['username'];

    $sql3 = "UPDATE `customer` SET `Password`='.$db_password.' WHERE `Username`='.$db_username.";
    $result3 = mysql_query($sql3);

    mysql_query($sql3) or die('error updating record');
    echo $sql3;
  }
?>

答

There are several problems with your code:

You're not getting username parameter with GET method. html markup should be corrected to properly define one form instead of having several.

You don't have $db_password variable defined. It's probably meant to be

$db_password = $_GET['password'];

There is invalid concatenation in $sql3.Therefore change

$sql3 = "UPDATE `customer` SET `Password`='.$db_password.' WHERE `Username`='.$db_username.";

$sql3 = "UPDATE `customer` SET `Password`='$db_password' WHERE `Username`='$db_username'";

答

your query should be like below :

$sql3 = "UPDATE `customer` SET `Password`='".$db_password."' 
WHERE `Username`='".$db_username."'";

答

The username field in your HTML is in a different form than the submit button, so it's not being sent along with the rest of the fields.
You have no $db_password variable; you do have a $db_pass variable. Is that what you meant?
For requests that modify data, you should really use POST, not GET requests.

As noted in the comments:

You should use $db_password = $_GET['password'], and correct the quotes as in peterm and Mehdi's answers.

无法使用php更新mysql数据库

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