passing selected data to database

ataloss · October 11th, 2013, 02:03 PM

Someone please tell me what I'm doing wrong. It doesn't work.

HTML Code:

<HTML>
<body bgcolor="#ccffff">
<FORM name=Form action="rateselect.php" method=post>
<B><center>
Set tax rate <SELECT name=taxrate>
<OPTION value=0.0000 selected>0.0%
<OPTION value=0.02900>2.9%
<OPTION value=0.04000>4.0%
<OPTION value=0.04225>4.225%
<OPTION value=0.04500>4.5%
<OPTION value=0.04700>4.7%
<OPTION value=0.05000>5.0%
<OPTION value=0.05300>5.3%
<OPTION value=0.05500>5.5%
<OPTION value=0.05600>5.6%
<OPTION value=0.05750>5.75%
<OPTION value=0.06000>6.0%
<OPTION value=0.06250>6.25%
<OPTION value=0.06500>6.5%
<OPTION value=0.06850>6.85%
<OPTION value=0.06875>6.875%
<OPTION value=0.07000>7.0%
<OPTION value=0.08250>8.25%
</OPTION></SELECT><p>
<CENTER>
<INPUT type=image height=24 alt="submit button" width=129 src="rollsubmit.gif" 
        border=0>
</CENTER>
</FORM></B></BODY></HTML>

PHP Code:


			
<?php

if (isset( $_POST['taxrate']) ) 

{

$taxrate=$_POST['taxrate'];

}

mysql_connect("localhost","root","");

mysql_select_db('numbersdb') or die( "Unable to select database");

print $_POST['taxrate'];

$query = "

INSERT INTO numbdata (taxrate)

VALUES('$taxrate')";

echo "data inserted</font><br /><br />"; 

$stat = mysql_query($query) or die('Query failed: ' . mysql_error()); 

mysql_close();

?

Adebayo · October 17th, 2013, 12:47 AM

Hello,

What exactly do you want to do? If you want to update the table, you have to use update query or you will have to use insert to insert new record.
Regard

Adebayo · October 17th, 2013, 12:54 AM

Oh sorry, I didn't read the code to the end. What problem do you have with the code? I think you should use update instead of insert and since you are define a particular column in a table, you should just update. Or what do you think?

ataloss · October 17th, 2013, 04:17 PM

I can't seem to get the code right

PHP Code:


			
<?php 

// error_reporting(0);

error_reporting(E_ALL ^ E_NOTICE); 

mysql_connect("localhost","root",""); 

mysql_select_db("numbersdb") or die( "Unable to select database"); 

if(!empty($_POST["submit"])) 

{ 

 $taxrate = $_POST['taxrate']; 

 $query="SELECT * FROM numbdata Where id='$id'"; 

 $result=mysql_query($query); 

 if(mysql_num_rows($result)) 

 {  

}else{echo "No listing for taxrate $taxrate .<br />Select another? .<br />";} 

} 

if(!empty($_POST["update"])) 

{ 

$sql = "UPDATE numbdata SET 

taxrate = '" .mysql_real_escape_string($_POST['taxrate']) . "', 

WHERE taxrate='".$_POST['taxrate']."'"; 

mysql_query($sql) or die(mysql_error());

 echo "Record for taxrate ".$_POST["taxrate"]." has been updated"; 

 } 

?>

ataloss · October 18th, 2013, 11:25 AM

I have a one record database table (numbdata)with fields such as taxrate.
I'm trying to select one option using HTML, pass it to PHP and
update the one field (taxrate)in numbdata.

ataloss · October 22nd, 2013, 06:47 PM

hello all, Below is my current code w/response. I hope for a solution, thanks

Quote:

GA is invalid.
Select another ?

Abbreviate; NY, CA, etc;

PHP Code:


			
<html><body bgcolor="ccffff"><center>

<?php

error_reporting(E_ALL ^ E_NOTICE);

mysql_connect("localhost","root","");

mysql_select_db("taxdb") or die( "Unable to select database");

if(!empty($_POST["submit"]))

      {

 $state = $_POST['state'];

$taxrate = $_POST['taxrate'];

 $query="SELECT * FROM taxtbl Where id='$id'";

 $result=mysql_query($query);

 if(mysql_num_rows($result))

      {

  echo "<form action='#' method='post'><b><br /><br />

    <table border='1'>

    <tr>

<th>state</th>

    </tr>";

  while($row = mysql_fetch_assoc($result))

      {

   echo "<tr>

<td><input type='text' size=2 name='state' value='" . $row['state'] . "'></td>

   </tr>";

      }

   echo "</table>

<input type='submit' name='update' value='Update Record' />

   </form>";

      } 

else{echo "$state is invalid.<br />Select another ?<br />";}

      }

if(!empty($_POST["update"]))

      {

$sql = "UPDATE taxtbl SET

 taxrate = '" . mysql_real_escape_string($_POST['taxrate']) . "', 

    WHERE state'".$_POST["state"]."'";

    mysql_query($sql) or die("Update query failed.");

    echo "tax rate".$_POST["state"]." has been set ...";

      }

?>

<form method="post" action="#"> <br />

Abbreviate; 

 <input type="text" name="state"/>NY, CA, etc;  <p>

<input type="submit" name="submit" value="make selection"/>

</form>

</center></BODY></HTML>