When ever i add headers in my pages i get the following errors
Can anybody tell me what's wrong?
Warning: Cannot add header information - headers already sent by (output started
at /usr/src/apache_1.3.14/web/server/apache/htdocs/searchdisplay.php:4) in /usr
/src/apache_1.3.14/web/server/apache/
On the contrary, MySQL is much better at handling table crashes and data corruption
than PostgreSQL is. What you may have heard is that due to lack of transaction
support, critical data may be lost "in transit" from your application to the database,
in the event of a system crash or a dropped c
Hi,
you can't have ANY output whatsoever before adding a
header. Even whitespaces before the opening schrieb im Newsbeitrag
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
> When ever i add headers in my pages i get the following errors
> Can anybody tell me what's wrong?
>
> Warning: Cannot add he
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>>This also requires a specific version of the Java Runtime Environment
>>(available at www.blackdown.com for Linux).
>
> Could you explain this? I don't recall having to install any specific
>Java Runtime Environment on our Linux box...
Sure, I was referring to installing the entire Oracle d
You need to put the headers before any output has been sent to the user, not
line on line 6, line 2 if possible. This includes spaces or blank lines
before the start of your php.
e.g. Here is script that creates a simple image:
Regards,
Corin Rathbone
[EMAIL PROTECTED]
P.S. If you use the scrip
HI,
If i say header("Location:http://$HTTP_HOST/$DOCROOT/say.htm");
the file displayed will be http://localhost/say.htm
but now say i have a user called php
and that user wants to display the following
http://localhost/~php/say.htm
specifying equivalent variable l
On Fri, Apr 06, 2001 at 02:15:34PM +0300, Dainius Vaskelis wrote:
> I suppose that you are using MySQL (because of LIMIT operator in SQL query)
I m using PostgreSql
> So:
> > In the second query i write a query whereby i say
> > $result=pg_Exec($database,"SELECT mem_id,name from search LIMIT $co
> As for a full comparison between the two, I think the bottom line is
> that MySQL is slightly more light-weight, but easier to use and faster
> than PostgreSQL. So if you're looking for a database for a relatively
> noncritical web application, I'd say go with MySQL, especially since
> that's wh
Hello, PHP turn me crazy, I'm triyong to undertand what happend here.
The parse error apear in line: odbc exec.
I'd try with odbc_execute and odbc_prepare too.
Thanks.
Fer
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Are encode and decode secure enough for storing credit
card numbers on a Web server?
Jeff Oien
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No
On Sat, 7 Apr 2001, Jeff Oien wrote:
> Are encode and decode secure enough for storing credit
> card numbers on a Web server?
> Jeff Oien
>
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>
You shouldn't store credit card numbers on a web server - your bank probably
prohibts it in your merchant agreement as well.
jason
> Are encode and decode secure enough for storing credit
> card numbers on a Web server?
> Jeff Oien
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Guys,
How to view the result from the table.
If I only want to display persons begin with letter A in the front.
_
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wyoy could use a SQL statement like this:
select * from myTable where LastName like "A%"
- Frank
> Guys,
>
> How to view the result from the table.
> If I only want to display persons begin with letter A in the front.
>
>
>
> _
> w
Hello boclair,
On 05-Apr-01 01:58:24, you wrote:
>I have a problem with a user input in a form required to be an
>integer, creating a variable for a mysql query.
>I have instances where integer, 0, is being typed as letter,o.
>The last discussion of validating the input was
>http://marc.theaim
Hello Sharmad,
On 06-Apr-01 08:33:54, you wrote:
>Hi,
> Thanks for firstly helping me out with the prevoius query.
> I have a database with mem_id(int) and name(text) as its fields.
> In the first form (HTML) ,I take mem_id,name and count(used for LIMIT)
> from the browse
It still not working
I want to get the result of people with the first later is "A" in my database
Here is my syntax :
$result = mysql_query("SELECT * FROM mydata WHERE name like '$begin'%",$db);
the $begin is from the URL link
www.sample.com/name?begin=a
Thanks
--- "Frank M
Put the percent symbol inside of the single quotes:
$result = mysql_query("SELECT * FROM mydata WHERE name like '$begin%'
",$db);
"Naga Sean" <[EMAIL PROTECTED]> wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
> It still not working
> I want to get the result of people with the fir
Okay it's working now..
How if I want to display all the data begin with NUMBERS. like "1,2,3,...,0"
--- "Frank M. Kromann" <[EMAIL PROTECTED]>
> wrote:
>wyoy could use a SQL statement like this:
>
>select * from myTable where LastName like "A%"
>
>- Frank
>
>> Guys,
>>
Manuel,
Manuel Lemos <[EMAIL PROTECTED]> wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
| Hello boclair,
|
| On 05-Apr-01 01:58:24, you wrote:
|
| >I have a problem with a user input in a form required to be an
| >integer, creating a variable for a mysql query.
|
| >I have instance
Manuel,
Manuel Lemos <[EMAIL PROTECTED]> wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
| Hello boclair,
|
| On 05-Apr-01 01:58:24, you wrote:
|
| >I have a problem with a user input in a form required to be an
| >integer, creating a variable for a mysql query.
|
| >I have instance
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