[SOLVED] Using POST to update Mysql db

[cleary1981]

Can anyone tell me where i have went wrong here? when I submit the database isn't updated.

 

<form id=addCompany action="addCompany.php" method="POST">
<label for="accno">Account Number</label>
<input type = "text" value = "" name = "accno" id = "accno"></br>
<label for="compName">Company Name</label>
<input type = "text" value = "" name = "compName" id = "compName"></br>
<label for="contact">Contact Name</label>
<input type = "text" value = "" name = "contact" id = "contact"></br>
<label for="add1">Address</label>
<input type = "text" value = "" name = "add1" id = "add1"></br>
<input type = "text" value = "" name = "add2" id = "add2"></br>
<label for="town">Town/City</label>
<input type = "text" value = "" name = "town" id = "town"></br>
<label for="county">County</label>
<input type = "text" value = "" name = "county" id = "county"></br>
<label for="pcode">Post Code</label>
<input type = "text" value = "" name = "pcode" id = "pcode"></br>
<label for="tel">Telephone No.</label>
<input type = "text" value = "" name = "tel" id = "tel"></br>
<input type = "submit" value = "Submit">

 

 

<?php

if($_POST['submit'])
{
   require "config.php"; // database connection details
   
   //convert all the posts to variables:
   $accno = $_POST['accno'];
   $compName = $_POST['compName'];
   $contact = $_POST['contact'];
   $add1 = $_POST['add1'];
   $add2 = $_POST['add2'];
   $town = $_POST['town'];
   $county = $_POST['county'];
   $pcode = $_POST['pcode'];
   $tel = $_POST['tel'];
   
   //Insert the values into the correct database with the right fields
   $result=MYSQL_QUERY("INSERT INTO company VALUES ('$accno', '$compName', '$contact', '$add1', '$add2', '$town', '$county', '$pcode', '$tel')"); 

}
?>

[samshel]

//Insert the values into the correct database with the right fields
   $result = mysql_query("INSERT INTO company VALUES ('$accno', '$compName', '$contact', '$add1', '$add2', '$town', '$county', '$pcode', '$tel')") or die (mysql_error()); 

 

try this.

[cleary1981]

tried that. It didnt make anydifference.

[samshel]

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

[cleary1981]

that didnt work either.

[samshel]

<?php

//try debugging here....

var_dump($_POST);

 

if($_POST['submit'])

{

  require "config.php"; // database connection details

 

  //convert all the posts to variables:

  $accno = $_POST['accno'];

  $compName = $_POST['compName'];

  $contact = $_POST['contact'];

  $add1 = $_POST['add1'];

  $add2 = $_POST['add2'];

  $town = $_POST['town'];

  $county = $_POST['county'];

  $pcode = $_POST['pcode'];

  $tel = $_POST['tel'];

 

  //Insert the values into the correct database with the right fields

  $result=MYSQL_QUERY("INSERT INTO company VALUES ('$accno', '$compName', '$contact', '$add1', '$add2', '$town', '$county', '$pcode', '$tel')");

 

}

?>

[ignace]

are you working with multiple database connections? cause if anyone would care to read a manual they would notice that mysql_query() actually have 2 parameters, so try adding the db connection resource (optional is only to be used when you in no way are able to write the parameter):

 

$query = "INSERT INTO ..";

$result = mysql_query($query, $dbConnection);

 

if (!$result) die(mysql_error());

[cleary1981]

What do you mean multiple database connections? I am querying one table in one db. I have seen examples of mysql_query used in this way.

[cleary1981]

is there a better way of doing this?

 

[samshel]

the way is correct, make sure the execution goes in the

 

if($_POST['submit'])

 

condition... and then print debug statements inside the condition.

[cleary1981]

the output to that is

 

array(9) { ["accno"]=> string(4) "uhuh" ["compName"]=> string(4) "uhuh" ["contact"]=> string(4) "uhuh" ["add1"]=> string(4) "uhuh" ["add2"]=> string(4) "uhuh" ["town"]=> string(4) "uhuh" ["county"]=> string(4) "uhuh" ["pcode"]=> string(4) "uhuh" ["tel"]=> string(4) "uhuh" }

[samshel]

as you can see the value of 'submit' is not being passed here..so if you name the button as "submit" it should come here and the code in the if condition should be executed...

[cleary1981]

yeah that worked had left " out of the code.