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How do i display records of an item into textbox based on drop down list

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Join Date: Jul 2009

Posts: 15

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karin21

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Newbie Poster

How do i display records of an item into textbox based on drop down list

Jul 5th, 2009

hi, i have a script here which display an item name from database into textfield based on the combo box selection, i cn display the item name but my problem is i dont know how to display the item information into the textbox? i'm required to use php but i think it can handle by javascript, hope my simple explanation works n_n

ex. item color
apple red
car blue

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 php Syntax (Toggle Plain Text) 
<script type="text/javascript" language="javascript">
 
function populate(oSelect) {
 
var opt, opt2, a=0, i = 0, textarea = document.getElementById('notes');
 
while (opt = oSelect.options[i++])
 
if (opt.selected && textarea.value.indexOf(opt.value) == -1)
 
textarea.value += opt.value + '\n';
 
return true;
 
}
 
</script>
 
<body>
 
<?php
 
$db_host = 'localhost';
$db_user = 'root';
$db_pass = '';
$db_db = 'dbname';
 
$db_link = mysql_connect($db_host, $db_user, $db_pass) or die('MySQL Connection Error:'.mysql_error());
mysql_select_db($db_db) or die('MySQL Error: Cannot select table');
 
$sql = "SELECT id,item,colorinfo FROM table ORDER by id";
$result = mysql_query($sql);
 
echo "<form name=f1 method=post action='' onSubmit='return false;'>";
echo "<select name=m1 size='8' onDblClick='populate(this)'>";
while ($row=mysql_fetch_assoc($result)) {
$id=$row['id'];
$display_name=$row['item'];
$display_color=$row['color'];
echo "<option value='$display_name'>$display_name</option>";
 
}
 
echo "</select>";
echo "<br>";
 
echo "<th>Color:<input type='text' name='color' id='color'>";
echo "</form>";
echo "</td>";
echo "<td valign='top' align='left'>";
echo "<form name=f2 method=post action='' onSubmit='return false;' >";
echo "<p align='left'>";
 
echo "</form>";
 
?>
 
</table>
<form>
<TEXTAREA class=formfield2 name="notes" id="notes" rows=12 cols=16 wrap="virtual"></TEXTAREA>
<input type="button" value="clear" onClick="if (confirm('Clear the field?'))notes.value='',">
</form>
</body>
</html>
<script type="text/javascript" language="javascript">

function populate(oSelect) {

var opt, opt2, a=0, i = 0, textarea = document.getElementById('notes');

while (opt = oSelect.options[i++])

if (opt.selected && textarea.value.indexOf(opt.value) == -1)

textarea.value += opt.value + '\n';

return true;

}

</script>

<body>

<?php

$db_host = 'localhost';
$db_user = 'root';
$db_pass = '';
$db_db = 'dbname';

$db_link = mysql_connect($db_host, $db_user, $db_pass) or die('MySQL Connection Error:'.mysql_error());
mysql_select_db($db_db) or die('MySQL Error: Cannot select table');

$sql = "SELECT id,item,colorinfo FROM table ORDER by id";
$result = mysql_query($sql);

echo "<form name=f1 method=post action='' onSubmit='return false;'>";
echo "<select name=m1 size='8' onDblClick='populate(this)'>";
while ($row=mysql_fetch_assoc($result)) {
$id=$row['id'];
$display_name=$row['item'];
$display_color=$row['color'];
echo "<option value='$display_name'>$display_name</option>";

}

echo "</select>";
echo "<br>";

echo "<th>Color:<input type='text' name='color' id='color'>";
echo "</form>";
echo "</td>";
echo "<td valign='top' align='left'>";
echo "<form name=f2 method=post action='' onSubmit='return false;' >";
echo "<p align='left'>";

echo "</form>";

?>

</table>
<form>
<TEXTAREA class=formfield2 name="notes" id="notes" rows=12 cols=16 wrap="virtual"></TEXTAREA>
<input type="button" value="clear" onClick="if (confirm('Clear the field?'))notes.value='',">
</form>
</body>
</html>

Last edited by peter_budo; Jul 6th, 2009 at 5:15 am. Reason: Keep It Organized - For easy readability, always wrap programming code within posts in [code] (code blocks) and [icode] (inline code) tags.

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Join Date: Oct 2008

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adatapost

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Re: How do i display records of an item into textbox based on drop down list

Jul 6th, 2009

Your thread is belongs to PHP forum.

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Join Date: Jul 2009

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karin21

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Re: How do i display records of an item into textbox based on drop down list

Jul 22nd, 2009

i think this is a PHP forum

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Join Date: Oct 2006

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ardav

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Re: How do i display records of an item into textbox based on drop down list

Jul 22nd, 2009

You will definitely need a server-side language to get info from a database. pHp with MySQL is great. If you have a dropdown (select) you can use AJAX to retrieve info into a textarea.

I suggest that you use something like the Prototype library to deal with the ajax implementation.

1. Download this from http://www.prototypejs.org (currently version 1.6).
2. Place this file into a folder, e.g. "/js/"
3. Create a new js file, e.g. custom.js and place it in the same folder.
4. Link the two js files to the php file's head area.
5. Create the form select widget via php or use static html.

e.g. for php:

 Help with Code Tags 
 PHP Syntax (Toggle Plain Text) 
<select name="m1" id="m1" onchange="popMe();return false">
<?php
$q = "SELECT * FROM table...";
$r = mysql_query($q);
while($d = mysql_fetch_array($r)){
   output .= "\n\t<option id=\"{d['id']}\">{$d['type']}</option>";
}
echo $output;
?>
</select>
 
...
 
<textarea name="notes" id="notes"></textarea>
<select name="m1" id="m1" onchange="popMe();return false">
<?php
$q = "SELECT * FROM table...";
$r = mysql_query($q);
while($d = mysql_fetch_array($r)){
   output .= "\n\t<option id=\"{d['id']}\">{$d['type']}</option>";
}
echo $output;
?>
</select>

...

<textarea name="notes" id="notes"></textarea>

6. Write the popMe() function in your custom.js file:

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function popMe(){
  var op = $F('m1');
  var url = "/includes/getrecs.php";
  var param = "id=" + op;
  var oGetInfo = new Ajax.Updater("notes", url,{method: 'post',parameters: param});
}
function popMe(){
  var op = $F('m1');
  var url = "/includes/getrecs.php";
  var param = "id=" + op;
  var oGetInfo = new Ajax.Updater("notes", url,{method: 'post',parameters: param});
}

7. Write the php code for getting the data and call the file getrecs.php.

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 PHP Syntax (Toggle Plain Text) 
...DB connection details...
 
$id = $_POST['id'];
$q = "SELECT info FROM table WHERE id='{$id}'";
$r = mysql_query($r);
$d = mysql_fetch_array($r);
$info = stripslashes($d['info']);
echo $info;
...DB connection details...

$id = $_POST['id'];
$q = "SELECT info FROM table WHERE id='{$id}'";
$r = mysql_query($r);
$d = mysql_fetch_array($r);
$info = stripslashes($d['info']);
echo $info;

That's it. I haven't checked the code, it's off the top of my head, so there may be typos or something, but it's pretty close.

Happy Humbugging Christmas

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Join Date: Sep 2007

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spthorn

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Re: How do i display records of an item into textbox based on drop down list

Jul 22nd, 2009

Karin,

I see a couple problems.

First, you're not accessing the right column name for the color. The SELECT statement says it's 'colorinfo', but when you access it, it's as 'color'. Change the assignment line to:
$display_color=$row['colorinfo']; and the following line to
echo "<option value='$display_color'>$display_name</option>";
In that JavaScript function, use the following line to get both the color and name into the textarea:
textarea.value += opt.innerHTML + ' ' + opt.value + '\n';
-steve

Last edited by peter_budo; Jul 22nd, 2009 at 3:02 pm. Reason: Keep It Organized - For easy readability, always wrap programming code within posts in [code] (code blocks) and [icode] (inline code) tags.