PHP - display the values as well as get values in same form

[rosy]

Hi,
I would like to retrieve values from table and display in the form . The form has more than 15 elements having textbox,textarea, select boxes and submit button.
When i open the form(by clicking client name), it should display the corresponding value based on the client from the database to my form. It should make use to understand what are the fields already entered and what should be entered in the form.

<?php
$salesrecordsql = mysql_query("SELECT * FROM sales_records WHERE clientname='$contact_company' AND rep_sid = '$rep_id'");
$salesrecordrow = mysql_fetch_array($salesrecordsql);
extract($salesrecordrow);
?>

<INPUT TYPE="text" NAME="contactname" VALUE="<?php print $contactname; ?>" SIZE="25" id="contactname" CLASS="text" tabindex="19">
<td align="right" rowspan="4"><b>Address:</b></td>
<td rowspan="4"><textarea cols="20" rows="4" name="contactaddress" class="text" tabindex="21"><?php print $contactaddress; ?></textarea></td>
</tr>

$leads = mysql_query("SELECT source FROM leadsource WHERE lead_id=$LeadSource");
      $leadsrow = mysql_fetch_array($leads);
      extract($leadsrow);

print "<td>\n<select name=\"LeadSource\" id=\"Leadsource\" size=\"1\" class=\"text\" tabindex=\"4\"\n";
$leadsql = mysql_query("SELECT * FROM leadsource WHERE leadstatus='0' ORDER BY source");

print "<option value=\"\" SELECTED> </option>\n";
while ($leadrow = mysql_fetch_array($leadsql)) {
extract($leadrow);
print "<option value=\"$lead_id\">$source</option>\n";
}
print "</select>\n</td>\n";

}  
<INPUT TYPE="submit" NAME="submit" VALUE="Add Sales Record" CLASS="button" >

How can i make it both like getting input as well as display the already entered values on the same page?
Any help would be appreciated.
Thanks.

[webnick]

Try something like this-

<?
$sql = "SELECT * FROM sales_records WHERE clientname='$contact_company' AND rep_sid = '$rep_id'"
$result = mysql_query($sql);
$row = mysql_fetch_assoc($result);
?>
<INPUT TYPE="text" NAME="contactname" VALUE="<?= $row['contactname']; ?>" SIZE="25" id="contactname" CLASS="text" tabindex="19">
<td align="right" rowspan="4"><b>Address:</b></td>
<td rowspan="4"><textarea cols="20" rows="4" name="contactaddress" class="text" tabindex="21"><?= $row['contactaddress']; ?></textarea></td>
</tr>

<select name="LeadSource" id="Leadsource" size="1" class="text" tabindex="4">
<?
$sql = "SELECT * FROM leadsource";
$result = mysql_query($sql);
while ($row = mysql_fetch_assoc($result)) {

echo "\n<option value=\"$row['lead_id']\"";
if($_GET['lead_id']==$LeadSource){ echo " selected=\"selected\"";
echo ">$row['source']</option>\n";
	}
}
?>
</select>
<INPUT TYPE="submit" NAME="submit" VALUE="Add Sales Record" CLASS="button" >

[rosy]

Hi ,
when i tried using Value=<?$row['contactname']?>, i dint work.
i would like to check if it already has values in db, if it has values display to user else display the input field to get input from user.
How can do this dynamically using php.
I tried using ,

<INPUT TYPE="text" NAME="contactname" VALUE="<?php print $contactname; ?>" SIZE="25" id="contactname" CLASS="text" tabindex="19"> 

i could able to retieve value in text box. which is in editable format. I dont want to display it as editable but like readonly format and should insert value once, even if it is submitted more than once.
please help.
Thanks

[webnick]

<?$row['contactname']?> will only work if the table rows have a column named 'contactname', like your 'clientname'.

You have a couple options for testing if a value is set. Try either of these-

<INPUT TYPE="text" NAME="contactname" VALUE="<?php echo $row['contactname']; ?>" id="contactname" <? if($row['contactname'] != "") { echo "disabled=\"disabled\""; }?> /> 

...

 if(isset($row['contactname']) && $row['contactname']!= "") { 
echo $row['contactname'];
} else {
echo "<INPUT TYPE=\"text\" NAME=\"contactname\" VALUE=\"". $row['contactname'] ."\" id=\"contactname\" />"; 
}

The idea is to query the database for your client and if it exists, echo their values in the form. Then use a second query to find all the possible categories they can be added to. Those should be two different tables (normalized). I can't test the script without setting up a similar table but you could emulate some tutorial to get going. http://www.devshed.com/c/a/MySQL/Displaying-Multiple-Records-Per-Row-in-a-MySQL-Query-Result-Set/

http://www.devshed.com/c/a/MySQL/Building-a-Simple-Affiliate-System-in-PHPMySQL/

Nick

[rosy]

I tried that one too it doesnt work,
Is there any other way i can get this done?
please help.

[ppowell]

When you do your query, are you retrieving all of the values based upon some external criteria? If so, then you will have one row with each field in that row for each form element.

Standard rule of thumb (and coding requirements that I set up) is that the form element name should be identical to the resultset row:

class MySQLQuery {

/**
	 * Retrieve resultset as an array of objects set up by mysql_fetch_object() PHP function
	 *
	 * @access public
	 * @return object $result
	 */
	function getResult() {										// OBJECT METHOD
		$runquery = $this->runQuery();
		$count = 0;
		while ($row = mysql_fetch_object($runquery)) {
			$result[$count] = $row;
			$count++;
		}
		return $result;
	}

}

$query = new MySQLQuery('SELECT * FROM blah', $this->dbAP->getDBConn());
$result = $query->getResult();
// NOW YOU HAVE A SINGLE RESULTSET ROW, A SINGLE-ELEMENT ARRAY OF OBJECTS

?>
<input name="first_name" size="30" maxlength="75" value="<?= str_replace('"', '&quot;', str_replace('\\', '', $result[0]->first_name)) ?>">

That would be a sample of how you could handle this.

HTH
Phil