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PHP - Mysql_insert_id Returns 0, Connections Issue?

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Mysql_insert_id Returns 0, Connections Issue?	View Content

after inserting data i get 0 sometimes
i created a mysql_connect for it

Code: [Select]
<?php
$maincon = mysql_connect('local', "root", "root");
mysql_select_db("root");

then

//insert and

$success = mysql_query($query);
echo $ref_id = mysql_insert_id($maincon);
?>
$ref_id is returned as 0, although the connection is there
var_dumping the query returns the values

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Php Script Causing Apache Max Connections Issue

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Over the past few days I have been running into some issues with my server crashing due to apache max connections issues. I am running my site off of a hosted Cloud VPS with 200GB of storage, 8192MB Memory, 8TB of transfer, Apache, MySQL, PHP5, and CentOS. I am afraid the issue doesn't necessarily lay in the configuration of Apache, but the way I have scripted the php on my site, the reason I am reaching out here. My site isn't your average website, it is more of a web-based customer management program. There are currently only 2 pages you can actually access via the url bar (signin.php and index.php). All other content is loaded via AJAX and JQuery processes (.load and $.getScript). All AJAX requests are pointed toward a single file called functions.php where a _POST parameter contains the function name and any additional _POST data required by the function. FOR EXAMPLE: AJAX Call:

$.post('functions/functions.php',{func:'myFunctionName',ops:'whatever',a:'whatever',b:'whatever'},function(data){
         DO WHATEVER I WANT WITH THE RETURN DATA HERE
},"json");

PHP (functions.php)

require 'dbcon.php';
include 'main_class.php';
include 'f_customerdetail.php';
include 'f_listoptions.php';
include 'f_route.php';
include 'f_useractions.php';

if (isset($_POST['func'])){ 
   $userfunc = $_POST['func']; 
   $funcops = $_POST['ops'];

   if ($funcops != ''){ 
      $userfunc($funcops); 
   }else{ 
      $userfunc(); 
   } 
}

How To Use This ! Mysql_insert_id

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Hello dear friends,

say i've database table with (id,name) and want to add more informations
by insert new names within an file for example has

Code: [Select]
$q1 = "INSERT INTO `mytable` VALUES (???,name1);
mysql_query($q1) or die(mysql_error()." at row ".__LINE__);

$q2 = "INSERT INTO `mytable` VALUES (???,name2);
mysql_query($q2) or die(mysql_error()." at row ".__LINE__);

ect.....

how then it automatic detect the last id to go on after it ?

some says use mysql_insert_id but i didn't understand how to apply it in this way also the example at php.net is bad not explain much

please how can i use it
thanks

Mysql_insert_id

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How can you debug to find out why it's not getting the mysql_insert_id number? I have it echoed all the queries and all with all the correct values from my form but the only problem is that its not getting the insert id number of the id.

$query1 = "INSERT INTO `efed_bio` (charactername,username,posername,style_id,gender,status_id,division_id,alignment_id,sortorder) VALUES ('".$charactername."','".$username."','".$posername."','".$style."','".$gender."','".$status."','".$division."','".$alignment."','".$sort."')";
            mysql_query($query1);
            $query1_id = mysql_insert_id();
            echo $query1;
            echo $query1_id;
            $query2 = "INSERT INTO `efed_bio_allies` (bio_id) VALUES (".$query1_id.")";
            mysql_query($query2);
            echo $query2;
            $query3 = "INSERT INTO `efed_bio_rivals` (bio_id) VALUES (".$query1_id.")";
            mysql_query($query3);
            echo $query3;
            $query5 = "INSERT INTO `efed_bio_singles` (bio_id) VALUES (".$query1_id.")";
            mysql_query($query5);
            echo $query5;

Mysql_insert_id Problem

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Hi all,

i'm a bit more than a newbie.

Here's the scenario.

I have a page with tabbed navigation. On each tab there's a form (diving centers, teachers, and so on). Each form has a submit button with a unique name. Every submit button is processed by a series of if statements in an external included php file.

As you can see in my code below what the file does is to process the if statements based on the submit pressed. It' important to notice that a JS script checks that the first form is filled with all the infos. If not is not possible to proceed to complete all the others.

On the first form the user can register general information. Obviously there's an Id field (auto increment) that i grab with mysql_nsert_id.

Also obvious is the fact that, for query reasons, i want to store the grabbed id in an id field present in all the tables of my database. The tables get the data from the various forms displayed on the tabbed navigation.

Here's the code of the included file (notice that i use to start with a simple coding to test everything 's working fine)

<?php
   require_once('Connections/Scubadiving.php');

if (!empty($_POST['theButton'])) {
   $nome=$_POST["nome"];
$indirizzo=$_POST["indirizzo"];
$insertSQL = "INSERT INTO centrisub (nome, indirizzo) VALUES ('$nome' , '$indirizzo')";

mysql_select_db($database_Scubadiving, $Scubadiving);
$Result1 = mysql_query($insertSQL, $Scubadiving) or die(mysql_error());
$last_id = mysql_insert_id ($Scubadiving);

}

if (!empty($_POST['istruttori'])) {
   echo $_SESSION["$last_id"];
$insertSQL = "INSERT INTO istruttori (nome, data_nascita, curriulum, altre_info, idcentrisub) VALUES ('$_POST[nome]','$_POST[data_nascita]','$_POST[curriculum]','$_POST[altre_info]','$last_id')";

mysql_select_db($database_Scubadiving, $Scubadiving);
$Result1 = mysql_query($insertSQL, $Scubadiving) or die(mysql_error());

}
?>

What i don't understand is why $last_id is not passed form the first if statment (processed when the user submit the generel infos form) to the other if statement (in this case the teachers form).

I've tried also with $_SESSION (trying to assign $last_id) but no success. (Probably because i don't know exactly how to use it)

Hope everything's clear.

What i'm missing ?

Thanks in advance for your help.

Is This The Best Way To Use Mysql_insert_id() Or Do I Need To Select It?

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I have two tables. The first with an auto-increment field of id and the second with a article_id field. I want to get the value of the auto-increment field in the first table and insert it into the article_id field of the second table (so they match). I am using the mysql_insert_id() command for the first time and I am wondering if I can run a query like this and turn it into a variable or if I need to use a SELECT query from the the mysql_insert_id() field from the first table before inserting it into the second table?

Any feedback is appreciated.

Thanks,

kaiman

Here is what I have so far (untested):

// insert data into blog database
$sql1="INSERT INTO $tbl_name1(author, title, content, date)VALUES('$author', '$title', '$content', NOW()) LIMIT 1";
$result1=mysql_query($sql1) or trigger_error("A mysql error has occurred!");
$article_id = mysql_insert_id ();

// if successfully inserted data into database, redirect user
if($result1){
header( "Location: http://www.mydomain.com/blog/add/success/" );
}
else {
header( "Location: http://www.mydomain.com/blog/add/error/" );
exit;
}

// insert data into blog categories database
$sql2="INSERT INTO $tbl_name2(article_id, category)VALUES('$article_id', '$category' LIMIT 1";
$result2=mysql_query($sql2) or trigger_error("A mysql error has occurred!");

// if successfully inserted data into database, redirect user
if($result2){
header( "Location: http://www.mydomain.com/blog/add/success/" );
}
else {
header( "Location: http://www.mydomain.com/blog/add/error/" );
exit;
}

Problem With Mysql_insert_id()

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When I run this Prepared Statement...

// Build query.
$q2 = "INSERT INTO member(email, activation_code, salt, hash, first_name,
username, register_ip, register_hostname, location, created_on)
VALUES(?, ?, ?, ?, ?, ?, ?, ?, ?, NOW())";

// Prepare statement.
$stmt2 = mysqli_prepare($dbc, $q2);

// Bind variables to query.
mysqli_stmt_bind_param($stmt2, 'sssssssss', $email, $activationCode, $salt, $hash, $firstName,
$username, $ip, $hostName, $location);

// Execute query.
mysqli_stmt_execute($stmt2);

// Capture New ID.
$_SESSION['memberID'] = mysql_insert_id();

I am getting this error...
Quote

Warning: mysql_insert_id() [function.mysql-insert-id]: A link to the server could not be established in /Users/user1/Documents/DEV/++htdocs/05_Debbie/members/create_account.php on line 269

What am I doing wrong?

Debbie

Problems With Mysql_insert_id

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Hello there, well basically what this function does is look at all the entry's to find a unused port in the database but when I run the function the first time if the table is empty mysql_insert_id works fine, but then if I run it again the mysql_insert_id keeps returning 0, I am using it immediately after the query but its not working properly, can anybody shed some ideas into the matter, anything is appreciated!, thanks for your time!:

function addServerToBuildPool($ServerGame, $ServerOwner, $ServerSlots, $ServerBox)
{
$Server = mysql_fetch_array(mysql_query("SELECT * FROM xhost_boxs WHERE box_id = '".mysql_real_escape_string($ServerBox)."'"));
$AddServer = mysql_query("INSERT INTO xhost_servers (server_game, server_slots, server_owner, server_ip, server_is_setup) VALUES('".mysql_real_escape_string($ServerGame)."', '".mysql_real_escape_string($ServerSlots)."', '".mysql_real_escape_string($ServerOwner)."', '".mysql_real_escape_string($Server['box_ip'])."', 'No')");

$ServerID = mysql_insert_id();

$FindPort = mysql_query("SELECT server_port FROM xhost_servers ORDER BY server_id ASC");

$Port = 0;

while($row = mysql_fetch_assoc($FindPort))
{
$Port = ($Port == 0)? $row['server_port'] : $Port;

if($row['server_port'] != $Port)
{
break;
}

$Port++;
}

mysql_query("UPDATE xhost_servers SET server_port = '".$Port."' AND server_username = 'server".$ServerID."' AND server_password = '".rand(5000000, 900000000)."' WHERE server_id = '".$ServerID."'") or die(mysql_error());
}

Problem With Mysql_insert_id()

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hello, im asking for your help again. i have a problem with this mysql_insert_id() function im using. it's used to get the values of id's and insert it to a field of another table right?but in my case where i have 4 tables and using the function to get the unique id,it no longer gets the id on the third table which is supposed to be doing so that this id will be inserted to the field of the final table. these are dynamic textboxes im working on by the way..here's my php code:

html
Code: [Select]
<html>
<head>
<script language="JavaScript">

function AddTextBox()
{
document.getElementById('container').innerHTML+='<input type="text" size="15" maxlength="15" name=block[]><br>';
}
function AddTextBox2()
{
document.getElementById('container2').innerHTML+='<input type="text" size="15" maxlength="15" name=room[]><br>';
}

</script>
</head>
<body>
<form name="form1" method="post" action="adnew.php">
<input type="hidden" name="cid">
Course:<input type="text" name="course">
<input type="hidden" name="yid">
Year: <select name="year">
<option value="1">1</option>
<option value="2">2</option>
<option value="3">3</option>
<option value="4">4</option>
</select>
<table width="23%" border="0">
<tr>
<td width="37%" height="1" colspan="2"> <input type="hidden" name="block_id">
Block:
<input name="button" type="button" onClick="AddTextBox();" value="Add textbox"></td>
</tr>
<tr>
<td height="21"><div id="container"></div></td>
</tr>
</table>
<table width="23%" border="0">
<tr>
<td width="64%">Room:
<input name="button2" type="button" onClick="AddTextBox2();" value="Add textbox"></td>
</tr>
<tr>
<td><div id="container2"></div></td>
</tr>
</table>

<br><br><input type="Submit" name="submit" value=" Add ">
</form>
</body>
</html>

php
Code: [Select]
<?php
include("dbcon.php");
?>
<?php
$id_c=$_POST['cid'];
$block=$_POST['block'];
$block_id=$_POST['block_id'];
$room=$_POST['room'];
$intblock=0;
$introom=0;

$sql=mysql_query("INSERT INTO course VALUES ('$id_c','$_POST[course]')") or die (mysql_error());
$id_c = mysql_insert_id();

$sql=mysql_query("INSERT INTO year VALUES ('$_POST[yid]','$id_c','$_POST[year]')") or die (mysql_error());
$_POST['yid'] = mysql_insert_id();

while(count($block)>$intblock) {
if (($block[$intblock]<>"")){
$sql=mysql_query("INSERT INTO block VALUES ('$block_id', '$_POST[yid]', '".$block[$intblock]."')") or die (mysql_error());
mysql_query($sql);
}
else{
echo "Block ".($intblock+1)." is missing values and cannot be inserted.";
}
$intblock=($intblock+1);
}
$block_id = mysql_insert_id();
while (count($room)>$introom) {
if (($room[$introom]<>"")){
$sql=mysql_query("INSERT INTO room VALUES ('$block_id', '".$room[$introom]."')") or die (mysql_error()); // this is the 4th table..
mysql_query($sql);
}
else{
echo "Room ".($introom+1)." is missing values and cannot be inserted.";
}
$introom=($introom + 1);
}

echo "Successfully added.";
echo "<br><a href='index.php'>Add another</a>";
?>

Quote

$block_id = mysql_insert_id();

this is what i'm having problems with..

Quote

$sql=mysql_query("INSERT INTO room VALUES ('$block_id', '".$room[$introom]."')") or die (mysql_error());

and this is for my 4th table..

i have found out that you have to put the function after an INSERT command,but in my case, i have a switch statement and after i tried putting it inside the switch, i get a message that says "duplicate entry 1 for b_id...etc.." any suggestions?advise?

Mysql_insert_id Problem

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I get mysql_insert_id problem. It returns 0. I do not know how to fix it. Please tell me. Thank you very much.

<?php session_id(); session_start();?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />
<title>Untitled Document</title>
</head>

<body>
<?php

include("connection.php");

$subtotal=$_POST['subtotal'];
$tax=$_POST['tax'];
$total=$_POST['total'];
$today=date("Y-m-d");
$email=$_SESSION['email'];
$od=mysql_insert_id();
$number=$_POST['number'];
$name=$_POST['name'];
$month=$_POST['month'];
$year=$_POST['year'];
$code=$_POST['code'];
$method=$_POST['type'];
$add1="select * from customer where email='$email'";
$add2=mysql_query($add1);
$add3=mysql_fetch_array($add2);

extract($add3);

$qiu1="INSERT INTO test (oid) VALUES ('$email')";

$qiuzhen=mysql_query($qiu1);

$orderid=mysql_insert_id();

$query44="INSERT INTO income (subtotal,tax,total,time,email,address,credit,name,month,year,code,method,custid) VALUES ('$subtotal','$tax','$total','$today','$email','$add1',$number,'$name','$month','$year','$code','$method','$orderid')";
$result=mysql_query($query44);
echo "successful!";
$orderid=mysql_insert_id();

echo "$email";
echo "$total";
$sessid=session_id();
$qu="select * from carttemp where sess='$sessid'";

$result=mysql_query($qu);

while($w=mysql_fetch_array($result)){

extract($w);

$query7="INSERT INTO try (prodnum,quan,custnum) VALUES ('$prodnum','$quan','$custid')";

$iii=mysql_query($query7) or (mysql_error());

}

?>

</body>
</html>

Getting Mysql_insert_id() From Multiple Inserts

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If I am running a query like this:

INSERT INTO `cam_locations` (`id`, `datacenter`, `address1`, `address2`, `city`, `state`, `zip`, `country`, `phone`) VALUES
(1, 'Austin Data Center', '', '', 'Austin', 'Texas', '', 'United States', ''),
(2, 'Sunnyvale Date Center', '', '', 'Sunnvale', 'California', '', 'United States', ''),
(4, 'BoxBorough Data Center', '', '', 'Boston', 'Massachusetts', '', 'United States', '')

It it possible to use mysql_insert_id() and get all the id's inserted? Maybe like an array of them or something? Thanks

How Do I Use Mysql_insert_id() In Conjuntion With Sprintf() To Link Tables?

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Set up
Windows Vista
* XAMPP 1.7.3, including:
* Apache 2.2.14 (IPv6 enabled) + OpenSSL 0.9.8l
* MySQL 5.1.41 + PBXT engine
* PHP 5.3.1
* phpMyAdmin

Ultimate objective: I want to insert session data into multiple tables whilst ensuring that the data is in the appropriate column.

Problem: I managed to insert the data into the correct tables and column, but after reading various forums I have been given the impression that if I was to have multiple site users the data's columns could get muddled up if they execute the script at the same time (hope I'm making sense, say if I'm not). The suggested method was mysql_insert_id() but I do not know how to make this work with in conjunction with sprintf().

As I said the script worked before I added the code with the star by it in an attempt to reach my ultimate objective.

<?php

//let's start our session, so we have access to stored data
session_start();

session_register('membership_type');
session_register('terms_and_conditions');

include 'db.inc.php';

$db = mysql_connect('localhost', 'root', '') or
    die ('Unable to connect. Check your connection parameters.');
mysql_select_db('ourgallery', $db) or die(mysql_error($db));


//let's create the query
$query = sprintf("INSERT INTO subscriptions (
name, email_address, membership_type)
VALUES ('%s','%s','%s')",
mysql_real_escape_string($_SESSION['name']),
mysql_real_escape_string($_SESSION['email_address']),
mysql_real_escape_string($_SESSION['membership_type']));

//let's run the query
$result = mysql_query($query, $db) or die(mysql_error($db));

*mysql_real_escape_string($_SESSION['name']) = mysql_insert_id($db);*

$query = sprintf("INSERT INTO site_user_info (
terms_and_conditions, name_on_card,
credit_card_number )
VALUES ('%s','%s','%s')",
*mysql_real_escape_string($_SESSION['name']),*
mysql_real_escape_string($_SESSION['terms_and_conditions']),
mysql_real_escape_string($_POST['name_on_card']),
mysql_real_escape_string($_POST['credit_card_number']),
mysql_real_escape_string($_POST['credit_card_expiration_data']));

//let's run the query
$result = mysql_query($query, $db) or die(mysql_error($db));

*mysql_real_escape_string($_SESSION['credit_card_number']) = mysql_insert_id($db);*

$query = sprintf("INSERT INTO card_numbers (
credit_card_expiration_data)
VALUES ('%s')",
*mysql_real_escape_string($_POST['credit_card_number']),*
mysql_real_escape_string($_POST['credit_card_expiration_data']));

$result = mysql_query($query, $db) or die(mysql_error($db));
echo '$result';

?>

All the other facets of the script work but I get the following error message with the above script: Fatal error: Can't use function return value in write context in C:\x\xampp\htdocs\form_process.php on line 27

But it's not so much the error message its the ultimate objective. Any help appreciated.

Curl Connections

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Hi,

I would just like to know what limitations apply for opening X amount of connections to different websites with Curl? (specifically, I'm using Curl multi and have my own way of regulating how many connections occur at a given time).

I ask this because it seems I can open 100 connections and get the info from the websites quite quickly. If I were to open say 1000 would that crash my router or what would happen?

Thanks in advance!

Too Many Mysql Connections?

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On my new site on average every page that loads opens 9 MySql connections and runs 14 MySql Queries, And as I use Ajax to dynamically update the pages, so this repeats every 60 seconds per page.

Is this too many connections?
I am not having any problems at the moment but the site is getting more busy, and I don't want the site to start running slow, or the MySql server to crash.

Any comment would be much appreciated.

Help With Database Connections

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Wasn't really sure what to call this topic, and I'm also not sure how to ask this question. Here is what I have, and the problem I'm having.

I have an html form on a webpage that sends information into 2 different tables in a database I have. One table is not important to the topic, the other table is where i'm having my problem.

The user submits their information from the form and the info goes into the tables. There is another page that displays the names of the users on a "wall" that populates from the second table in the database. The second table takes information from the form they filled out and displays their name on the wall based on that (most importantly whether or not they can be e-mailed from the wall or not).

The issue I am having (and this is probably the long way of asking this) is that when some people fill out the form their info does not go into either database. I know the functionality works, because their have been a couple of users that have signed up and everything worked fine for them.

I'm thinking this may be a connection issue possibly due to a firewall or something. Any ideas on this? The website is

http://itgetsbetter.tsagl.org

the form to fill out is under the button "Take the Pledge" and the names are displayed under "View the Wall". If someone wouldn't mind to try out the functionality of the form for me (I need people that have some technical experience to do this because most of the people that I work with have no idea what I am talking about when trying to troubleshoot this with them). I assure you that if everything works, I will remove your information from my database. This site is only out to a few people at the moment, but I am planning on taking it public in the next couple of days.

I would display code here, but I'm not sure what code you would need to see in order to help with this. I just think this is a connection issue; I just need to know how to test it or how to rectify the issue.

Thanks for your help (and please don't hate too much on the design on the site; i'm more of a programmer than a designer).

Looping Socket Connections

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Hello,

I'm trying to update my database with current statistics from my gameservers, but running the loop of socket connections to all of the servers on page load takes too long. For every server in the array it takes roughly 4 seconds to complete each one.. with a massive list of say 42 it takes awhile. However, I'm running from shared hosting and was wondering if that was the reason each server takes so long to query. I was thinking about running a cron job every 1 minute to update the information so it's still relativity new and current. Will the 1 minute cron job affect anything from the shared hosting? I'm using 1and1 web-hosting on a 1&1 Business Package.

Thanks

How To Manage Database Connections

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At the moment I am writing an application that supports plugins. Every plugin implements an interface so that the main application can control the plugin with the interface methods. There could be plugins that need a database connection and others that do not. What I am doing at the moment is loading every plugin with a __construct($databaseHandler). For plugins that don't need a database connection this is a little overhead.

I could also use a Singleton or something else "staticish" so that the plugin can get the database instance whenever it wants, but then the plugin can only be used for my application where this specific Singleton is present and maybe I would like to use the plugin for other purposes later in another environment without changing it.

Do you have any better ideas or how do you normally manage your database connections?

Odbc Connections Not Closing

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Hi

Have an issue with a script that connects to multiple Access databases to extract data.

There is one master database and then numerous small databases (I take no responsibility for the design!). The master is opened and then the sub databases are looped around, opened process and closed in turn.

However after about 20 connections I get the error [ODBC Microsoft Access Driver] Too many client tasks for any new connections. It is related to the number of connections rather than the number of operations on databases (ie, if I comment out one of the pieces of SQL run on each database it makes no difference).

I am closing the connection and unsetting the variable that stores the connection. As such there shouldn't be an excess of connections open at any one time.

Any suggestions?

All the best

Keith

Two Mysql Connections Via A Php Class

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Hi, I am quite new to using php classes so it maybe something simple here, however I am trying to establish to connection to two different databases using one PHP class; The problem is that this will work for one but when I create a new connection they will both fail.

<?php
putenv("TZ=Europe/London");
 
// start database connection
$funky_db = new Database('xxxx', 'xxxx', 'xxxx', 'xxxx');
 
// start database connection
$funky_cc = new Database('xxxx', 'xxxx', 'xxxx', 'xxxx');
 
class Database {
 
    private $host;
    private $user;
    private $pass;
    private $name;
    private $link;
    private $error;
    private $errno;
    private $query;
 
    function __construct($host, $user, $pass, $name = "", $conn = 1) {
        $this -> host = $host;
        $this -> user = $user;
        $this -> pass = $pass;
        if (!empty($name)) $this -> name = $name;      
        if ($conn == 1) $this -> connect();
    }
 
    function __destruct() {
        @mysql_close($this->link);
    }
 
    public function connect() {
        if ($this -> link = mysql_connect($this -> host, $this -> user, $this -> pass, TRUE)) {
            if (!empty($this -> name)) {
                if (!mysql_select_db($this -> name)) $this -> exception("Could not connect to the database!");
            }
        } else {
            $this -> exception("Could not create database connection!");
        }
    }
 
    public function close() {
        @mysql_close($this->link);
    }
 
    public function query($sql) {
        if ($this->query = @mysql_query($sql)) {
            return $this->query;
        } else {
            $this->exception("Could not query database!".$this->name);
            return false;
        }
    }
 
    public function num_rows($qid) {
        if (empty($qid)) {         
            $this->exception("Could not get number of rows because no query id was supplied!");
            return false;
        } else {
            return mysql_num_rows($qid);
        }
    }
 
    public function fetch_array($qid) {
        if (empty($qid)) {
            $this->exception("Could not fetch array because no query id was supplied!");
            return false;
        } else {
            $data = mysql_fetch_array($qid);
        }
        return $data;
    }
 
    public function fetch_array_assoc($qid) {
        if (empty($qid)) {
            $this->exception("Could not fetch array assoc because no query id was supplied!");
            return false;
        } else {
            $data = mysql_fetch_array($qid, MYSQL_ASSOC);
        }
        return $data;
    }
 
public function fetch_object($qid) {
        if (empty($qid)) {
            $this->exception("Could not fetch object assoc because no query id was supplied!");
            return false;
        } else {
            $data = mysql_fetch_object($qid);
        }
        return $data;
    }
 
    public function fetch_all_array($sql, $assoc = true) {
        $data = array();
        if ($qid = $this->query($sql)) {
            if ($assoc) {
                while ($row = $this->fetch_array_assoc($qid)) {
                    $data[] = $row;
                }
            } else {
                while ($row = $this->fetch_array($qid)) {
                    $data[] = $row;
                }
            }
        } else {
            return false;
        }
        return $data;
    }
 
    public function last_id() {
        if ($id = mysql_insert_id()) {
            return $id;
        } else {
            return false;
        }
    }
 
    private function exception($message) {
        if ($this->link) {
            $this->error = mysql_error($this->link);
            $this->errno = mysql_errno($this->link);
        } else {
            $this->error = mysql_error();
            $this->errno = mysql_errno();
        }
        if (PHP_SAPI !== 'cli') {
        ?>
 
            <div class="alert-bad">
                <div>
                    Database Error
                </div>
                <div>
                    Message: <?php echo $message; ?>
                </div>
                <?php if (strlen($this->error) > 0): ?>
                    <div>
                        <?php echo $this->error; ?>
                    </div>
                <?php endif; ?>
                <div>
                    Script: <?php echo @$_SERVER['REQUEST_URI']; ?>
                </div>
                <?php if (strlen(@$_SERVER['HTTP_REFERER']) > 0): ?>
                    <div>
                        <?php echo @$_SERVER['HTTP_REFERER']; ?>
                    </div>
                <?php endif; ?>
            </div>
        <?php
        } else {
                    echo "MYSQL ERROR: " . ((isset($this->error) && !empty($this->error)) ? $this->error:'') . "\n";
        };
    }
 
}
?>

Number Of Connections/queries

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Hey,

I would there be a way i could create a form in which upon entering a domain name/URL it will tell you how many connections there are to the site and how many (if any) MySQL queries are running?

If I add this line to a wordpress footer it displays the number of sql queries.

Code: [Select]
<?php echo get_num_queries(); ?> queries in <?php timer_stop(1); ?> seconds.

Any help is greatly appreciated.

Thanks.

Including Functions & Db Connections

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Hi,

Just a quick query.

I call my db connection and configuration files on every script.
I also have just made a functions script where i'm storing some functions
except when i call the function, my script fails because of a missing database connection.

But on the page that fails, i call the configuration files including the db connection at the start of the script.

But when i get to where i include the function they are all on the same script so should it fail??

The page did work fine when the function was inbedded in the script. But as i'm using the function on a few pages, i thought it would be better practise and more efficent to just include the function on each page i need, rather than just copy and paste all the php in the required pages.

Or do i have to declare the db connection inside the function page as well???

Thanks