[thelist] php mysql update problem

[Austin Harris]

Thanks Chris,

It seems to work but I'm not sure if I have done it right, as I said in
original post very new to this :)

Thanks again, Austin

// do the update before getting records.
  if ($_POST['submit']=='Submit') {
    // do the update
    $update_sql = "UPDATE jayandmel SET Present='Car' WHERE id='1'";
    $result = mysql_query($update_sql);
    my_debug("SQL -- $update_sql");
    
  }
  
// carry on and grab the required records as normal.

mysql_select_db("austin",$db);

$result = mysql_query($update_sql);
my_debug("result -- $result SQL -- $update_sql");
$result = mysql_query("SELECT * FROM jayandmel",$db);


on 27/6/03 1:01 pm, christopher at ideadesigners.com at
christopher at ideadesigners.com wrote:

> [resent because I'm a dope and didn't include the subject the first time. :) ]
> 
> Austin Harris wrote:
>> // do the update before getting records.
>> if ($_POST['submit']=='Submit') {
>> // do the update
>> $update_sql = "UPDATE jayandmel SET Present='2' WHERE Present='House'";
>> my_debug("SQL -- $update_sql");
>> }
>> 
>> // carry on and grab the required records as normal.
> 
> Hi Austin,
> 
> You need to run the SQL statement after you have constructed it inside
> variable $update_sql.
> 
> Your debug statement outputs the SQL but you don't actually use it on
> your DB to do the update :)
> 
> ADD:
> 
>> $update_sql = "UPDATE jayandmel SET Present='2' WHERE
> Present='House'";
> 
> $result = mysql_query($update_sql);
> 
>> my_debug("SQL -- $update_sql");
> 
> then debug on:
> 
> my_debug("result -- $result SQL -- $update_sql");
> 
> to give you the status of the result......although you should see the
> table contents change.