[PHP-DB] Re: never seen this before..
I believe your assumption is correct. VARCHAR is not the best datatype
suitable for indexing. I understand why you want to utilize this type,
so that you are not storing empty characters. I am assuming that the id
column is numerical in nature. In that case, make the data type INT.
This way it increments just a number and it only stores the amount of
characters needed.
i.e. : id INT NOT NULL auto_increment,
This should solve your problem, post again if not
gl -- Seth
Jas wrote:
> Here is the db table structure...
> CREATE TABLE demo_sessions (
>id varchar(255) NOT NULL auto_increment,
>ipaddy varchar(255) NOT NULL,
>date varchar(255) NOT NULL,
>PRIMARY KEY (id)
> );
> Let me know if this is correct, I have a feeling that the varchar is my
> problem but I could be wrong.
> TIA
> Jas
>
> <[EMAIL PROTECTED]> wrote in message
> [EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
>
>>TIA --
>>
>>
>>column in your table denotes your primary key? After that, how is it
>>incremented? Is it auto or do you have some other form of it. I think
>>your code is correct so the error seems to be how the data ID is
>>
> generated.
>
>>gl -- Seth
>>
>>Jas wrote:
>>
>>
>>>I am getting an error when inserting data into a mysql database table.
>>>
> It
>
>>>is telling me that there is a duplicate entry in the database everytime
>>>
> it
>
>>>gets 10 records inserted to the table, if I flush the table and try it
>>>
> again
>
>>>it works fine. Is there some kind of limit on some tables? Here is the
>>>error php outputs once the table gets 10 records: Duplicate entry '10'
>>>
> for
>
>>>key 1
>>>Here is the code...
>>>require '/home/web/b/bignickel.net/scripts/admin/db.php';
>>> $table = "demo_sessions";
>>> $sql_insert = "INSERT INTO $table (ipaddy,date) VALUES
>>>('$ipaddy','$date')" or die('Could not insert client info');
>>> $result = @mysql_query($sql_insert,$dbh) or die(MYSQL_ERROR());
>>>
>>>Any help on this would be great
>>>TIA
>>>Jas
>>>
>>>
>>>
>>>
>
>
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RE: [PHP-DB] Re: never seen this before..
Must be an old or broken version of mysql. I'm running 3.23.50 and it won't
even allow the `id' field to be a varchar(255). Change it to int() and
watch the magic.
-Original Message-
From: Jas [mailto:[EMAIL PROTECTED]]
Sent: Monday, June 24, 2002 1:05 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] Re: never seen this before..
Here is the db table structure...
CREATE TABLE demo_sessions (
id varchar(255) NOT NULL auto_increment,
ipaddy varchar(255) NOT NULL,
date varchar(255) NOT NULL,
PRIMARY KEY (id)
);
Let me know if this is correct, I have a feeling that the varchar is my
problem but I could be wrong. TIA Jas
<[EMAIL PROTECTED]> wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
> TIA --
>
>
> column in your table denotes your primary key? After that, how is it
> incremented? Is it auto or do you have some other form of it. I think
> your code is correct so the error seems to be how the data ID is
generated.
>
> gl -- Seth
>
> Jas wrote:
>
> > I am getting an error when inserting data into a mysql database
> > table.
It
> > is telling me that there is a duplicate entry in the database
> > everytime
it
> > gets 10 records inserted to the table, if I flush the table and try
> > it
again
> > it works fine. Is there some kind of limit on some tables? Here is
> > the error php outputs once the table gets 10 records: Duplicate
> > entry '10'
for
> > key 1
> > Here is the code...
> > require '/home/web/b/bignickel.net/scripts/admin/db.php';
> > $table = "demo_sessions";
> > $sql_insert = "INSERT INTO $table (ipaddy,date) VALUES
> > ('$ipaddy','$date')" or die('Could not insert client info');
> > $result = @mysql_query($sql_insert,$dbh) or die(MYSQL_ERROR());
> >
> > Any help on this would be great
> > TIA
> > Jas
> >
> >
> >
>
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[PHP-DB] Re: never seen this before..
Here is the db table structure...
CREATE TABLE demo_sessions (
id varchar(255) NOT NULL auto_increment,
ipaddy varchar(255) NOT NULL,
date varchar(255) NOT NULL,
PRIMARY KEY (id)
);
Let me know if this is correct, I have a feeling that the varchar is my
problem but I could be wrong.
TIA
Jas
<[EMAIL PROTECTED]> wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
> TIA --
>
>
> column in your table denotes your primary key? After that, how is it
> incremented? Is it auto or do you have some other form of it. I think
> your code is correct so the error seems to be how the data ID is
generated.
>
> gl -- Seth
>
> Jas wrote:
>
> > I am getting an error when inserting data into a mysql database table.
It
> > is telling me that there is a duplicate entry in the database everytime
it
> > gets 10 records inserted to the table, if I flush the table and try it
again
> > it works fine. Is there some kind of limit on some tables? Here is the
> > error php outputs once the table gets 10 records: Duplicate entry '10'
for
> > key 1
> > Here is the code...
> > require '/home/web/b/bignickel.net/scripts/admin/db.php';
> > $table = "demo_sessions";
> > $sql_insert = "INSERT INTO $table (ipaddy,date) VALUES
> > ('$ipaddy','$date')" or die('Could not insert client info');
> > $result = @mysql_query($sql_insert,$dbh) or die(MYSQL_ERROR());
> >
> > Any help on this would be great
> > TIA
> > Jas
> >
> >
> >
>
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[PHP-DB] Re: never seen this before..
TIA --
column in your table denotes your primary key? After that, how is it
incremented? Is it auto or do you have some other form of it. I think
your code is correct so the error seems to be how the data ID is generated.
gl -- Seth
Jas wrote:
> I am getting an error when inserting data into a mysql database table. It
> is telling me that there is a duplicate entry in the database everytime it
> gets 10 records inserted to the table, if I flush the table and try it again
> it works fine. Is there some kind of limit on some tables? Here is the
> error php outputs once the table gets 10 records: Duplicate entry '10' for
> key 1
> Here is the code...
> require '/home/web/b/bignickel.net/scripts/admin/db.php';
> $table = "demo_sessions";
> $sql_insert = "INSERT INTO $table (ipaddy,date) VALUES
> ('$ipaddy','$date')" or die('Could not insert client info');
> $result = @mysql_query($sql_insert,$dbh) or die(MYSQL_ERROR());
>
> Any help on this would be great
> TIA
> Jas
>
>
>
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