[RESOLVED] Filling a form on load using PHP and javascript.

Galuf

I have a php page that has a form on it. I want to fill the form with data from an SQL database if data for that combination is available. I currently am unable to post the script, but will post it this evening if it will help. The problem is that I am unable to fill the form with the data. I don't know if the database is finding data or not. Is there an easy way to auto fill a form if the data's there. I'm using the javascript

document.getElement.By.Name("name".value=<?php echo $variable; ?>";

is that the proper way to do this, or am I going about it the wrong way. Thanks for any help that you can give.

Galuf

bretticus

document.getElement.By.Name("name".value=<?php echo $variable; ?>";

Ummm...that should be giving an syntax error.

Should probably be...

document.getElementById("name").value='<?php echo $variable; ?>'";

However, you don't have to use javascript to fill in the values. Something like this works just fine (example to illustrate.)

<?php
mysql_connect("localhost", "mysql_user", "mysql_password") or
    die("Could not connect: " . mysql_error());
mysql_select_db("mydb");
$query = "SELECT id, name FROM mytable WHERE id = '" . mysql_real_escape_string($_GET['id']) . "'";
$result = mysql_query($query);

$id = ""; 
$name = ""; 

if ($row =  mysql_fetch_assoc($result)) {
    $id = $row['id']; 
    $name = $row['name']; 
}

mysql_free_result($result);
?>
<html>
<body>
<form>
ID: <input type="text" name="id" value="<?php echo $id?>">
Name: <input type="text" name="name" value="<?php echo $name?>">
</form>
</body>
</html>

Galuf

<?php
ob_start();
$host="...";
$username="...";
$password="...";
$submit=$_POST['submit'];
$cname=$_POST['cname'];
$snumber=$_POST['snumber'];
$db_name="...";
if ($submit=="Check In")
{
$table_name="checkin";
mysql_connect("$host", "$username", "$password")or die("cannot connect");
mysql_select_db("$db_name")or die("cannot select DB");
$sql="SELECT * FROM $table_name WHERE storename='$cname' and storenum='snumber'";
$result=mysql_query($sql);
$field1=$snumber;
$field2=mysql_result($result,0,"instdate");
$field3=mysql_result($result,0,"techname");
$field4=$cname;
$field5=mysql_result($result,0,"city");
$field6=mysql_result($result,0,"state");
$field7=mysql_result($result,0,"timeonsite");
$field8=mysql_result($result,0,"serveronsite");
$field9=mysql_result($result,0,"serverlocation");
$field10=mysql_result($result,0,"port1");
$field11=mysql_result($result,0,"port2");
$field12=mysql_result($result,0,"port3");
$field13=mysql_result($result,0,"port4");
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
<html>
<head>
<link rel="stylesheet" type="text/css" href="/include/css.css" />
<link rel="stylesheet" type="text/css" href="/include/print.css" media="print" />
<title>...</title>
<script type="text/javascript">
function updateData()
{
	document.getElementByName("storenumber").value="<?php echo $field1; ?>";
	document.getElementByName("instdate").value="<?php echo $field2; ?>";
	document.getElementByName("techname").value="<?php echo $field3; ?>";
	document.getElementByName("storename").value="<?php echo $field4; ?>";
	document.getElementByName("city").value="<?php echo $field5; ?>";
	document.getElementByName("state").value="<?php echo $field6; ?>";
	document.getElementByName("timeonsite").value="<?php echo $field7; ?>";
	document.getElementByName("serveronsite").value="<?php echo $field8; ?>";
	document.getElementByName("serverlocation").value="<?php echo $field9; ?>";
	document.getElementByName("port1").value="<?php echo $field10; ?>";
	document.getElementByName("port2").value="<?php echo $field11; ?>";
	document.getElementByName("port3").value="<?php echo $field12; ?>";
	document.getElementByName("port4").value="<?php echo $field13; ?>";
}
</script>
</head>
<body>
<form action="submit.php" method="post">
	<table align="center" width="80%">
			<Caption>...</caption>
		<tr>
			<td>Store Number:</td><td> <input type="text" name="storenumber" size="4" maxlength="4"></td>
			<td>Date of Install:</td><td> <input type="text" name="instdate"></td>
			<td>Tech Name:</td><td> <input type="text" name='techname'></td>
		</tr>
		<tr>
			<td>Store Chain:</td><td> <select name="storename">
      <option value="">1</option>
      <option value="">2</option>
      <option value="">3</option>
      <option value="">4</option>
      </select></td>
			<td>City:</td><td><input type="text" name='city'></td>
			<td>State:</td><td><input type="text" name="'state'" size="2" maxlength="2"></td>
		</tr>
		<tr></tr>
			<td>Tech Time on Site:</td><td><input type="text" name="timeonsite" size="5" maxlength="5"></td>
			<td>Server On Site:</td><td> <input type="checkbox" name="serveronsite"></td>
			<td>Server Location:</td><td> <input type="text" name="serverlocation"></td>
		</tr>
	</table>
	<table align="center" width="80%">
			<tr>
			<td>Server Port 1:</td><td> <input type="text" name="'port1'" size="4" maxlength="4"></td>
			<td>Server Port 2:</td><td> <input type="text" name="'port2'" size="4" maxlength="4"></td>
			<td>Server Port 3:</td><td> <input type="text" name="'port3'" size="4" maxlength="4"></td>
			<td>Server Port 4:</td><td> <input type="text" name="'port4'" size="4" maxlength="4"></td>
		</tr>
	</table>
	<table align="center" width="80%">
		<tr>
			<td align="right"><input type="submit" name="checkin" value="Submit"></td><td><input type="reset" name="reset" value="Reset"></td>
		</tr>
	</table>
	<script type="text/javascript">
	updateData();
	</script>
</form>

</body>
</html>
<?php
}
elseif ($submit=='Check Out')

From the elseif the same xhtml page is created, but with a diffrent set of options.

Galuf

Now I have the Values for field1 and field4 showing up, but the rest are not showing up. I updated my code using the information provided.

bretticus

Galuf wrote:

<?php
$sql="SELECT * FROM $table_name WHERE storename='$cname' and storenum='snumber'";
?>

$field1 comes from $snumber which comes from $_POST['snumber']. Are you sure you are posting data to this script?

Also, in the code above, did you really mean to not inject the store name from your POST variable ($snumber) or did you just miss the '$'?

bretticus

Galuf wrote:
Now I have the Values for field1 and field4 showing up, but the rest are not showing up. I updated my code using the information provided.

Oh whoops. I got that backward, but the solution is still the same: you are missing a '$' in from of snumber in your query.

bretticus

Oh, and one last thing. Don't forget to filter your variables that get set outside the script to avoid xsite script attacks. For example...

$cname=$_POST['cname'];
$snumber=$_POST['snumber'];

...should be...

$cname=mysql_real_escape_string($_POST['cname']);
$snumber=mysql_real_escape_string($_POST['snumber']);

Galuf

Here is the most recent version of the script. I found the problem you were right it was on the preceding page. I didn't have a value in the drop down box. Now the only problem is that fields10-13 and field 8 do not show up. I thank you for all of the help with this. I only hope that I can learn enough to be able to help someone else out in return.

<?php
ob_start();
$host="...";
$username="...";
$password="abc123";
$submit=$_POST['submit'];
$cname=mysql_real_escape_string($_POST['cname']);
$snumber=mysql_real_escape_string($_POST['snumber']);
$db_name="...";
if ($submit=="Check In")
{
$table_name="...";
mysql_connect("$host", "$username", "$password")or die("cannot connect");
mysql_select_db("$db_name")or die("cannot select DB");
$sql="SELECT * FROM $table_name WHERE storename='$cname' AND storenum='$snumber'";
$result=mysql_query($sql);
$field1=$snumber;
$field2=mysql_result($result,0,"instdate");
$field3=mysql_result($result,0,"techname");
$field4=$cname;
$field5=mysql_result($result,0,"city");
$field6=mysql_result($result,0,"state");
$field7=mysql_result($result,0,"timeonsite");
$field8=mysql_result($result,0,"serveronsite");
$field9=mysql_result($result,0,"serverlocation");
$field10=mysql_result($result,0,"port1");
$field11=mysql_result($result,0,"port2");
$field12=mysql_result($result,0,"port3");
$field13=mysql_result($result,0,"port4");
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
<html>
<head>
<link rel="stylesheet" type="text/css" href="/include/css.css" />
<link rel="stylesheet" type="text/css" href="/include/print.css" media="print" />
<title>...</title>
</head>
<body>
<form action="submit.php" method="post">
	<table align="center" width="80%">
			<Caption>...</caption>
		<tr>
			<td>Store Number:</td><td> <input type="text" name="storenumber" size="4" maxlength="4" value="<?php echo $field1?>"></td>
			<td>Date of Install:</td><td> <input type="text" name="instdate" value="<?php echo $field2?>"></td>
			<td>Tech Name:</td><td> <input type="text" name='techname' value="<?php echo $field3?>"></td>
		</tr>
		<tr>
			<td>Store Chain:</td><td> <select name="storename" value="<?php echo $field4?>">
      <option value="">1</option>
      <option value="">2</option>
      <option value="">3</option>
      <option value="">4</option>
      </select></td>
			<td>City:</td><td><input type="text" name="city" value="<?php echo $field5?>"></td>
			<td>State:</td><td><input type="text" name="state" size="2" maxlength="2" value="<?php echo $field6?>"></td>
		</tr>
		<tr></tr>
			<td>Tech Time on Site:</td><td><input type="text" name="timeonsite" size="5" maxlength="5" value="<?php echo $field7?>"></td>
			<td>Server On Site:</td><td> <input type="checkbox" name="serveronsite" value="<?php echo $field8?>"></td>
			<td>Server Location:</td><td> <input type="text" name="serverlocation" value="<?php echo $field9?>"></td>
		</tr>
	</table>
	<table align="center" width="80%">
			<tr>
			<td>Server Port 1:</td><td> <input type="text" name="'port1'" size="4" maxlength="4" value="<?php echo $field10?>"></td>
			<td>Server Port 2:</td><td> <input type="text" name="'port2'" size="4" maxlength="4" value="<?php echo $field11?>"></td>
			<td>Server Port 3:</td><td> <input type="text" name="'port3'" size="4" maxlength="4" value="<?php echo $field12?>"></td>
			<td>Server Port 4:</td><td> <input type="text" name="'port4'" size="4" maxlength="4" value="<?php echo $field13?>"></td>
		</tr>
	</table>
	<table align="center" width="80%">
		<tr>
			<td align="right"><input type="submit" name="checkin" value="Submit"></td><td><input type="reset" name="reset" value="Reset"></td>
		</tr>
	</table>
</form>

</body>
</html>

Thanks,

Galuf

bretticus

You may need to run your query from phpMyadmin / Mysql Query Browser / command line to see if you are using the right column names or if they are even accessible from your query.

Also, just for practise sake, mysql_result is used pretty much for just one result from a query. Optimally, you'll want to use one of the mysql fetch functions instead:

$result=mysql_query($sql);
if ( $row = mysql_fetch_assoc($result) ) {
	$field1=$snumber;
	$field2=$row["instdate"];
	$field3=$row["techname"];
	$field4=$cname;
	$field5=$row["city"];
	$field6=$row["state"];
	$field7=$row["timeonsite"];
	$field8=$row["serveronsite"];
	$field9=$row["serverlocation"];
	$field10=$row["port1"];
	$field11=$row["port2"];
	$field12=$row["port3"];
	$field13=$row["port4"];
}

Finally, inserting the serveronsite variable value for the checkbox value will not check it. You need an if statement of some kind.

//your code
<td>Server On Site:</td><td> <input type="checkbox" name="serveronsite" value="<?php echo $field8?>"></td>

//my suggestion
<td>Server On Site:</td><td> <input type="checkbox" name="serveronsite" value="1"<?php echo ($field8 == 1) ? " checked" : ""?>></td>

Galuf

Thank you so much for all of your help. It is working perfectly.