[wdvltalk] Re: php

Fri, 13 Feb 2004 10:21:05 -0800 

I think the message you are getting is that there is a
good connection to the database server. I do not think
you have yet inserted yet. Try 
$result = mysql_query($queryNo)
   or die("Invalid query: " . mysql_error());
 somewhere before > $id= mysql_insert_id();

--- "Trusz, Andrew" <[EMAIL PROTECTED]> wrote:
> No difference. The id and '' combination works fine
> in the mysql interface.
> My main question is: is the use of mysql_insert_id()
> correct?
> 
> There may be a problem with the connection between
> the php and mysql on this
> machine. I'll try the code on one I know works
> later. But I wanted to know
> if the use of the insert)_id function was proper
> since it doesn't work on
> the mysql command line interface.
> 
> If I can't get that to work, I'll have to use an
> ugly single table or invent
> a more elaborate way of setting up the id.
> 
> drew
> 
> -----Original Message-----
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] 
> Sent: Friday, February 13, 2004 12:13 PM
> To: [EMAIL PROTECTED]
> Cc: [EMAIL PROTECTED]
> Subject: [wdvltalk] Re: php
> 
> $Query = "Insert into $TableName (id, name, date)
> values ('', 'number 
> three', '2004-12-02')";
> 
> loose all references to id.... making...
> $Query = "Insert into $TableName (name, date) values
> ('number three', 
> '2004-12-02')";
> 
> My first thought was that you're forcing a blank
> into the id field...
> Try this... no pormises though...
> (It's friday, and I'm 2 mins from leaving.. sorry)
> 
> 
> 
> 
> 
> [EMAIL PROTECTED] 
> 13/02/2004 17:09
> Please respond to
> [EMAIL PROTECTED]
> 
> 
> To
> [EMAIL PROTECTED]
> cc
> 
> Subject
> [wdvltalk] php
> 
> 
> 
> 
> 
> 
> I’m trying to do  multi-table insert using an auto
> incremented “id” from
> the main table (register) as the id in the second
> table (money). I can
> establish the connection to the database and I can
> insert into each table
> using the mysql command line. So the pieces fit. The
> script runs but
> nothing is added to the tables. As you’ll see,
> I’m trying to use
> mysql_insert_id() to get the id from the insert into
> the main table to use
> as a value in the additional table.
> 
> I get the “success” message and an id of 0 when
> it should be 3 and no data
> inserted in either table.
> 
> <?php 
> /* Set variables for database access */
> $Host = "localhost";
> $User = "root";
> $Password = "";
> $DBName = "MySQL.respite";
> $TableName = "register";
> $TableName2 = "money";
> 
> $Link= @mysql_connect($Host, $User, $Password) or
> die(mysql_error());
> if ($Link) {
>                  $msg = "success";
>                  }
> $Query = "Insert into $TableName (id, name, date)
> values ('', 'number
> three', '2004-12-02')";
> $id= mysql_insert_id();
> 
> 
> $Query2= "Insert into $TableName2 (id, cc, pay)
> values ('$id',
> '1234-9857', '3,000')";
> 
> $Query3 = "Select * from register money where id =
> '$id'";
>  
> mysql_close($Link)
> ?>
> <? echo "$msg"; ?>
> <br />
> <? echo "$id"; ?>
> 
> 
> drew
> 
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