title: Show phpmyadmin cell in html textbox value
body: <p>I have a database in phpmyadmin called <code>fleet hire motors</code>, and in that database is a table called <code>customer</code>. </p>
  
  <p>In that table are columns called <code>customerID</code> and <code>Surname</code>. I have already done some coding on one page that lets the user select the <code>customerID</code> to edit the <code>Surname</code>. </p>
  
  <p>On the next page I want a textbox. in that textbox, the default value should be what the current <code>Surname</code> is. </p>
  
  <p>So, if i was to edit customer with <code>customerID</code> 1 (of which surname is currently <code>Brown</code> and I want to change to <code>Green</code>) the second page would show <code>Surname: [Brown]</code>, where [] encloses a textbox. </p>
  
  <p>I currently do not have any code, and would like to keep it primarily php. The first page is called <code>editcustomer.php</code>, and the second is called <code>editcustomer2.php</code>. </p>
  
  <p>Any help is appreciated.</p>
  
  <p>My current code is:</p>
  
  <pre><code>&lt;html&gt; &lt;head&gt; &lt;title&gt;Edit Customer&lt;/title&gt; &lt;/head&gt;&lt;body&gt;
  &lt;?php mysql_connect("localhost","username","password") or die(mysql_error()); 
  mysql_select_db("fleet hire motors") or die(mysql_error()); ?&gt; 
  &lt;?php 
  $CustomerID = $_GET["CustomerID"]; 
  $query=mysql_query(" SELECT * FROM customer WHERE CustomerID = '$CustomerID' ") or die(mysql_error()); 
  while ($row = mysql_fetch_array($query)) { 
      b$CustomerID = $row["CustomerID"]; 
  } ?&gt; 
  First Name: &lt;input name="FirstName" type="text" value="
  &lt;?php  
      $FirstName = $_GET["CustomerID"]; 
      include 'db.php'; 
      $query=mysql_query(" SELECT FirstName  FROM customer  WHERE CustomerID = '$CustomerID' ") or die(mysql_error()); 
  ?&gt; "&gt;
  &lt;br&gt; &lt;input name="submitbtn" type="submit" value="Save"&gt; &lt;input name="resubmitbtn" type="submit" value="Reset"&gt; &lt;/form&gt; &lt;/body&gt; &lt;/html&gt;
  </code></pre>
  
  <p>Sorry for all the edits, as I am new to stackoverflow and just learning how to do it.</p>
  
  <p>I have now updated my coding thanks to a response, but it is still not working. My most current coding is:</p>
  
  <pre><code>&lt;html&gt;
  &lt;head&gt;
  &lt;title&gt;Edit Customer&lt;/title&gt;
  &lt;/head&gt;
  &lt;body&gt;
  &lt;?php
  mysql_connect("localhost","username","password") or die(mysql_error());
  mysql_select_db("fleet hire motors") or die(mysql_error());
  ?&gt;
  &lt;?php 
  $CustomerID = $_GET["CustomerID"]; 
  $query=mysql_query(" SELECT * FROM customer WHERE CustomerID = '$CustomerID' ") or         die(mysql_error()); 
  $row = mysql_fetch_array($query);
  if (!$row || !is_array($row)){
  $CustomerID = 0;
  $CustomerFirstName = '';
  }
  else {
  $CustomerID = $row["CustomerID"]; 
  $CustomerFirstName = $row['FirstName'];
  }
  ?&gt; 
  First Name: &lt;input name="FirstName" type="text" value="&lt;?php echo $CustomerFirstName; ?  &gt;"&gt;
  &lt;input name="submitbtn" type="submit" value="Save"&gt;
  &lt;input name="resubmitbtn" type="submit" value="Reset"&gt;
  &lt;/form&gt;
  &lt;/body&gt;
  &lt;/html&gt;
  </code></pre>
  
  <p>This does not give me anything in the textbox, and my submit button does not work.</p>