[mysql how to organize by the no. of rows?]

try

SELECT *  FROM `table` 
ORDER BY (SELECT COUNT(`id`) FROM `table` a WHERE a.threadid=table.threadid) DESC

[check how many times some houer pass between 2 diff dates]

try

<?php
function no_of_times ($srch,$start,$end) {
$start =strtotime($start);
$nstart = strtotime(date('Y-m-d ',$start ).$srch);
$out = ($start <= $nstart) ? 1 : 0;
$out += floor( (strtotime($end) - $nstart) / 86400);
return  $out;
}
$srch ='22:00:00';
$start = '2/04/07 22:00:00';
$end = '2/05/07 22:00:00';
echo no_of_times($srch, $start, $end);
?>

[[SOLVED] Help please! While loop not looping!]

you use same variable name ($result) in both while loop

try to make names diferent ($result1 and $result2)

[[SOLVED] Calculate the age of a person?]

Because it's not yet been exactly three years. Those amounts are rounded down, that's what Floor does. You know by looking at it that if tomorrow is their birthday, they are not yet 4 years old. However, since we don't know the time they were born, we assume it was midnight, and when you round out the days, it turns out to be 1460. If you put in a different date such as 2003-03-05, you'll see the days go down, because it's not quite four years yet. Their age is still 3.

 

no. the error is thet betwen this two date is 2004-02-29.

it cause wrong output

[[SOLVED] Calculate the age of a person?]

Made a little change, now this should work:

 

<?php

function calc_age($date)
{
$seconds = time() - strtotime($date);
$year = 31556926;
return floor($seconds / $year);
}

?>

 

 

Orio.

it stil output same results

[[SOLVED] Calculate the age of a person?]

3 * 365 = 1095

and 4* 365 = 1460

how years is 3?

[[SOLVED] Calculate the age of a person?]

i try Orio's way

<?php
function calc_age($date)
{
$seconds = time() - strtotime($date);
$year = 60*60*24*365.25;
return floor($seconds / $year);
}
echo date('Y-m-d'), "\n"; //output 2007-02-04
echo calc_age('2004-02-05'), "\n"; // output 3
echo calc_age('2005-02-05'); // output 1
?>

Is it OK?

[[SOLVED] Calculate the age of a person?]

I try jesirose's script

function birthdateToAge($birthdate){
$age = array();
$now = time();
$age['seconds']	= $now-$birthdate;
$age['minutes']	= floor($age['seconds']/60);
$age['hours']	= floor($age['minutes']/60);
$age['days']	= floor($age['hours']/24);
$age['years']	= floor($age['days']/365);
return $age;
}

$age = birthdateToAge(strtotime("March 20, 1980"));
print $age['years'];

it output

date born: 2003-02-05<br />

today: 2007-02-04<br />

old: 4

 

Is it OK?

 

[Convert MySQL 14 digit timestamp into Readable format]

try

<?php echo date("D M j G:i:s T Y",strtotime($row_Recordset4['date_posted']));?>

[[SOLVED] Mysql date sorting]

number 071230105050 == 71230105050

if you want 0 in front try

$a=071230105050;
echo $a,"<br />\n";
$b = sprintf('%012s',$a);
echo $b;

or use unix timestamp

$date = '07-12-30 10:50:50';
echo $x = strtotime($date),"<br />\n"; //date to int (Unix timestamp - number of sec from 1970-1-1) 
$date1 = date('y-m-d h:i:s',$x); // int to date
echo $date1;

[mysql while loop?]

try

<?php
$link = mysql_connect("**********", "*******", "*******") or die(mysql_error());
mysql_select_db("******") or die(mysql_error());

$x = 0;


mysql_query("CREATE TABLE temp(
id INT NOT NULL AUTO_INCREMENT,
PRIMARY KEY(id),
username VARCHAR(30),
warning INT)")
or die(mysql_error());

while ($x <= 101){
        $y = 0;
while ($y <= 101){
	$afs2 = 'afs'.$x.$y;
	mysql_query("ALTER TABLE temp ADD $afs2 VARCHAR(30) NULL DEFAULT NULL") or die(mysql_error());
	$y++;
	}
$x++;
}
echo 'Table Updated!';
?>

remove ' from query

 

btw

This code will stop when $x=11 and $y=1 because afs111 is duplicate ($x=1 and $y=11)

[Dynamic Tree Menu]

change line

$query2=mysql_query("select DISTINCT Custom2 from lampworld ORDER by Custom1");

to

$query2=mysql_query("select DISTINCT Custom2 AS name from lampworld WHERE Custom1='$name'");

[xml creation problem]

if you have line mysql_fetch_assoc($Recordset1); before for loop temowe it

[Arrays not outputting]

can you use include?
try[code]<?php
$test = @include_once('/config.php');
if ($test) echo 'yes'; else echo 'no';
?>[/code]

[How to handle the dates greater than "2057-12-31"]

try[code]<?php
function in_day($a){
$monts=array(0,31,28,31,30,31,30,31,31,30,31,30,31);
$a = explode('-',$a);
if (($a[0] % 400 == 0) or ($a[0] % 4 == 0 and $a[0] % 100 != 0))  $monts[2] = 29;
$y = $a[0] -1;
$out = $y *365 + (int) ($y / 4) - (int) ($y / 100) + (int) ($y / 400);
for ($i=0;$i<$a[1];$i++) $out += $monts[$i];
$out += $a[2] - 1;
return $out;
}

$date1='2045-12-31';
$date2='2047-12-31';
echo in_day($date2) - in_day($date1);

?>[/code]

[Arrays and Loops]

try[code]foreach($arr_am as $ai => $aa)
{
// foreach($arr_id as $ai)
// {
$x = 0;

while($aa > $x)
{
echo $arr_id[$ai];
$x++;
}
// }
} [/code]

[Need assistance with a script]

[code]<?php
$Sigs=array('weekend.jpeg','url for first picture','2ndpicture','3rd_picture',etc);
$times = array(0,5,9,14,19,24);
$time = date('H');
$day = date('D');
$w_days = array('Sun','Sat');
if (in_array($day,$w_days)) $i=0;
else {
$i = 0;
while ($time >= $times[$i]) $i++;
}
echo '<img src="',$Sigs[$i],'" >';
?>[code][/code][/code]

[Need assistance with a script]

try[code]<?php
$times = array(0,5,9,14,19,24);
$time = date('H');
$day = date('D');
$w_days = array('Sun','Sat');
if (in_array($day,$w_days)) echo 'Pisture for weekends';
else {
$i = 0;
while ($time > $times[$i]) $i++;
echo 'Picture no ',$i;
}
?>[/code]

[It's not working!]

if you script redirect to login_false.php $rec_num is zero
I think that in your data table columns username and userid is not same
You compare column userid with $username. Is it OK?

[Eval returns nothing?]

change
print $str.' = '.eval($str.';').'<br />';
to
print $str.' = '; eval('print '.$str.';'); print '<br />';

[Need help with separating an array.]

try[code]$name=explode("<br>", $row['resourcesneeded']);
$amountIn = count($name);

for($x = 0; $x < $amountIn; $x++){

  print "$name[$x]<p>";
  $out = explode(' ',$name[$x],2);
  print 'number is ',$out[0],' and word is ',$out[1];

}[/code]
for 5 Fibre Glass it output 5 Fibre
Add 2 in explode function

[It's not working!]

you check (userid = $username  in query). Is it OK?

[Math - Adding up values when using While]

OK
1st
you use variable $id before yuo set up it in 1st query ($query_payment)
It cause thet query returns no rows and $payment_num is zero and you never go in while loop

line $id = $_GET['id']; move before line $query_payment = ...

2nd 
Before while loop $paysubtotal = $total (not payment yet)
In while loop for each paymant you must subtract payment_amount from paysubtotal ($paysubtotal = $paysubtotal - $payment_amount;)

insert line $paysubtotal = $total; before while loop
change line $paysubtotal = $total - $payment_amount; in $paysubtotal = $paysubtotal - $payment_amount;

[Sorting MYSQL Results]

try order by (prod_size_color + 0)

[Math - Adding up values when using While]

move line[code]$paysubtotal = $total - $payment_amount;[/code] in if part
[code]if ( $payment_id == "$id" ) {

echo"$$payment_amount - $payment_date<br />";
                $paysubtotal = $total - $payment_amount;
}[/code]
btw you can add WHERE part in your SQL[code]$query_payment="SELECT id, payment_id, payment_amount, payment_date FROM customer_payments WHERE payment_id='$id'";[/code]
and you don't need if statemans at all