How to populate php/html form with MySQL data

**peterv** · 06-12-2009, 10:13 PM

I am learning PHP/MySQL on my own. I've got a MySQL database and a few php/html forms that work fine. What I want to do, but haven't been able to figure out, is to display a form so I can enter a key value to retrieve a record from the database and display that data in the form. I then want to be able to modify 1 or more fields and then update the record in the database. I think I can figure out how to do the update, but I can't figure out how to populate the form with the selected record's data. Anyone have any examples of how to do this? I'd appreciate the help.

**n1tr0b** · 06-13-2009, 02:15 AM

This is the most simplest i know..

PHP Code:


<?php

$con = mysql_connect("localhost","user","pass");

if (!$con)

  {

  die('Could not connect: ' . mysql_error());

  }



mysql_select_db("my_db", $con);



$result = mysql_query("SELECT * FROM field");



while($row = mysql_fetch_array($result))

  {

  echo '<form action="">';

  echo '<select name="nameofname :D">';

  echo '<option value=". $row['nameoffield'] .">". $row['nameoffield']. "</option>';

  echo '</form>';

  }



mysql_close($con);

?>

**heavensgate15** · 06-13-2009, 03:08 AM

Hmmm.. What you mean is that you display the data from the database in a form manner? If that so, then try something like this:

Code:

<?php

  include('dbconnect.php');

  $data = "select * from table_name where ID = 100";

  $query = mysql_query($data);
  
  $data2 = mysql_fetch_array($query);
  
  
?>

<html>
<head>
      <title></title>
 </head>

<body>

// form to display record from database
<form>
  Name: <input type="text" value="<?php echo $data2[name]?>"/> </br>
  age: <input type="text" value="<?php echo $data2[age]?>"/> </br>
  hobby: <input type="text" value="<?php echo $data2[hobby]?>"/>
</form>

</body>

</html>

**forum_amnesiac** · 06-13-2009, 01:41 PM

This is very simple to do with 3 separate scripts:

The first 1 here is called Form.htm and asks for what record to find:

Form.htm

HTML Code:

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN"
"http://www.w3.org/TR/html4/loose.dtd">
<html>
<head>
      <title></title>
 </head>

<body>

<!-- form to get key detail of record in database -->
<form name="form" method="POST" action="form1.php">
  Keyfield <input type="text" name="search"> <br><br>
  <input type="submit"  value="submit">
</form>

</body>

</html>

The second script is called Form1.php, it is called by Form.htm, and displays the relevant record and allows changes to be made.

Form1.php

PHP Code:


<?php



$connection = mysql_connect('XXXXX','XXXXXXX','XXXXXXX') or die ("Couldn't connect to server."); 

$db = mysql_select_db('XXXXXXXX', $connection) or die ("Couldn't select database."); 



$search=$_POST['search'];



$data = 'SELECT * FROM `table_name` WHERE `ID` = "'.$search.'"';

  $query = mysql_query($data) or die("Couldn't execute query. ". mysql_error());

  $data2 = mysql_fetch_array($query);

  

  

?>



<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN"

"http://www.w3.org/TR/html4/loose.dtd">

<html>

<head>

      <title></title>

 </head>



<body>



<!-- form to display record from database -->

<form name="form" method="POST" action="form2.php">

  Name: <input type="text" name="namefield" value="<?php echo $data2[name]?>"/> <br>

  age: <input type="text" name="agefield" value="<?php echo $data2[age]?>"/> <br>

  hobby: <input type="text" name="hobbyfield" value="<?php echo $data2[hobby]?>"/><br><br>

    <input type="hidden" name="keyfield" value="<?php echo $search?>">  

  <input type="submit"  value="submit">

</form>



</body>



</html>

The third script is called Form2.php, it is called by Form1.php, it updates the record and redisplays the entered data.

Form2.php

PHP Code:


<?php



$connection = mysql_connect('XXXXX','XXXXXXX','XXXXXXX') or die ("Couldn't connect to server."); 

$db = mysql_select_db('XXXXXXXX', $connection) or die ("Couldn't select database."); 



$Key=$_POST['keyfield'];

$Name=$_POST['namefield'];

$Age=$_POST['agefield'];

$Hobby=$_POST['hobbyfield'];



$data = "UPDATE `table_name` SET name='$Name', age='$Age', hobby='Hobby' WHERE ID=".'"'.$Key.'"';

  $query = mysql_query($data) or die("Couldn't execute query. ". mysql_error());

  

?>



<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN"

"http://www.w3.org/TR/html4/loose.dtd">

<html>

<head>

      <title></title>

 </head>



<body>



<!--  display the changed record from database -->

  Name: <?php echo $Name?><br>

  Age: <?php echo $Age?> <br>

  Hobby: <?php echo $Hobby?><br><br>



</body>



</html>

I haven't tested this code but it is hopefully ok and will give you an idea of how you could do what you have requested.

I have used a hidden input field to pass the "key" field between Form1.php and Form2.php because sometimes this may not a field you want to alter, just a field you want to search by, eg an ID number

**heavensgate15** · 06-14-2009, 01:34 PM

Ya, that was it, ehehhe... hey forum_amnesiac, why do you use $_POST['value'] instead of $_POST['search'] in Form1.php?

**forum_amnesiac** · 06-14-2009, 01:58 PM

typo, as I said I hadn't tested it.

I'll correct it now

**heavensgate15** · 06-16-2009, 11:19 AM

ahw.. I thought it was another knowledge to earn.. haahahhaha

**mukeshgulia** · 07-16-2009, 09:28 AM

hi all,
I want to do the similar operation as described above, but in a different manner.
My form comprises of 1000's html text boxes. User will enter 2 or 3 fields and then on a button/submit a php script will find thr record in the database using these fields entered by the user.
Record reterived by the user will populate all the other fields in the form.
I hope u understand what I mean? I do not want to jump to any other page.

Moreover, my form already have a submit button which saves the data in mysql. Do I need to create a different form for search and give the search textboxes there only. Or I can use the same form.

Please suggest. I am struck for two days now because of this.

Thanks in advance.

**vinaykr** · 08-26-2015, 04:21 PM

Originally Posted by mukeshgulia

hi all,
I want to do the similar operation as described above, but in a different manner.
My form comprises of 1000's html text boxes. User will enter 2 or 3 fields and then on a button/submit a php script will find thr record in the database using these fields entered by the user.
Record reterived by the user will populate all the other fields in the form.
I hope u understand what I mean? I do not want to jump to any other page.

Moreover, my form already have a submit button which saves the data in mysql. Do I need to create a different form for search and give the search textboxes there only. Or I can use the same form.

Please suggest. I am struck for two days now because of this.

Thanks in advance.

How can this be acheived using PDO

**Beverleyh** · 08-26-2015, 06:16 PM

vinaykr, Your question isn't clear - neither is what you are referring to in the post from 6 years ago.

Please start your own thread, clearly stating your question and clearly describing what you're trying to achieve. Sample code/data would be helpful.

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