How is the loop algorithm?

[zac1987]

I use mysql to retrieve data from database and use php to display the data onto website directly by the following codes :

$query2 = "SELECT message FROM messages WHERE username='{$username1}'";

$result2 = mysql_query($query2,$connection) or die (mysql_error());

confirm_query($result2);

while($userinfo = mysql_fetch_array($result2)){

echo "<div class=\"messagepiece\">" . $userinfo['message'] . "</div><br>";

}

 

 

I want the div(s) arrange style like on the picture http://2aek.com/inventory/divs.jpg

display the 1st 4 divs onto 1st column, jump to 2nd column to display another 4 divs....

 

Below are the codes of how the divs arrangement : 

 

<div class="test1" style="left:0px; top:0px;">

asdawdaw

</div>

 

<div class="test1" style="left:0px; top:70px;">

asdawdaw

</div>

 

<div class="test1" style="left:0px; top:140px;">

asdawdaw

</div>

 

<div class="test1" style="left:0px; top:210px;">

asdawdaw

</div>

 

<div class="test1" style="left:200px; top:35px;">

asdawdaw

</div>

 

<div class="test1" style="left:200px; top:105px;">

asdawdaw

</div>

 

<div class="test1" style="left:200px; top:175px;">

asdawdaw

</div>

 

<div class="test1" style="left:200px; top:245px;">

asdawdaw

</div>

 

<div class="test1" style="left:400px; top:0px;">

asdawdaw

</div>

 

<div class="test1" style="left:400px; top:70px;">

asdawdaw

</div>

 

<div class="test1" style="left:400px; top:140px;">

asdawdaw

</div>

 

<div class="test1" style="left:400px; top:210px;">

asdawdaw

</div>

 

So the looping styles should be something like :

loop left 0 for four times > 200 for four times > 400 for four times.

loop top 0 > 70 > 140 > 210 then 35 > 105 > 175 > 245 then again 0 > 70 > 140 > 210.

Seriously I don't know how to write the algorithm of looping.

 

I guess the looping algorithm for left should be something like :

i = 0 ; i=>4 ; i++{

k=0; k=>20000; k+200{}

}

But it is so wrong, it will stop at 4th loop and each loop will be 0, 200, 400, 800. That's not what I want.

 

Seriously I can't figure out the algorithm, my brain is stupid, so can you guys guide me please?

 

[attachment deleted by admin]

[AbraCadaver]

I'm no hero with CSS, especially positioning, but this seems to work:

 

$top  = 0;
$left = 0;

while($userinfo = mysql_fetch_array($result2)){
    echo "<div style=\"left:{$left}px; top:{$top}px; position:absolute;\">{$userinfo['message']}</div>\n";
    
    switch($left) {
        case 0:
            $left = 400;
            break;
        case 200:
            $left = 0;
            $top +=35;
            break;
        case 400:
            $left = 200;
            $top +=35;
            break;
    }
}

[zac1987]

I'm no hero with CSS, especially positioning, but this seems to work:

 

$top  = 0;
$left = 0;

while($userinfo = mysql_fetch_array($result2)){
    echo "<div style=\"left:{$left}px; top:{$top}px; position:absolute;\">{$userinfo['message']}</div>\n";
    
    switch($left) {
        case 0:
            $left = 400;
            break;
        case 200:
            $left = 0;
            $top +=35;
            break;
        case 400:
            $left = 200;
            $top +=35;
            break;
    }
}

 

I did test your switch case codes, it is not working. Sad.

[zac1987]

Oh, sorry, I forget to put position:absolute just now. Your switch case codes are working now!! Thank you so much. 

[zac1987]

Please see the picture for reference

 

I want to display all the divs as the picture (below section of picture). Here is my current codes :

$top  = 30;

$left = 10;

 

switch($top) {    

        case $top < 100:

            for ($i=1; $i<=4; $i++)

  {

$left = 10;

$top  += 30;

          break;

  }

 

        case $top > 100:

for ($i=1; $i<=4; $i++)

  {

$left = 200;

$top += 60;

            break;

}

 

        case $top > 600:

for ($i=1; $i<=4; $i++)

  {

$left = 600;

$top += 60;

            break;

  }

}

 

The result of code display the 1st div = $left = 10; $top  = 30; . And all the rest of divs are having the same $left = 600; but different $top  = 180; $top  = 300; $top  = 420; $top  = 540; $top  = 660;

 

I don't know what's wrong with it. Can you give me some guide? Thank you.

[zac1987]

ok, i know the problem already, i cannot use for loop because it will loop the top value +60 for 3 times = +180, then pass the +180 to the mysql while loop.

[sasa]

try

<?php
$left = 0;
$top = 0;
$num = mysql_num_rows($result2);
$len1 = ceil($num/3);
$len2 = $len1 + floor($num/3);
while($userinfo = mysql_fetch_array($result2)){    
    if($count==$len1){$top=30;  $left=100;}
    if($count==$len2){$top=0;  $left=200;}
    echo "<div style=\"left:{$left}px; top:{$top}px; position:absolute;\">{$userinfo['message']}</div>\n";
    $top +=60;
    $count++;
}
?> 

[zac1987]

try

<?php
$left = 0;
$top = 0;
$num = mysql_num_rows($result2);
$len1 = ceil($num/3);
$len2 = $len1 + floor($num/3);
while($userinfo = mysql_fetch_array($result2)){    
    if($count==$len1){$top=30;  $left=100;}
    if($count==$len2){$top=0;  $left=200;}
    echo "<div style=\"left:{$left}px; top:{$top}px; position:absolute;\">{$userinfo['message']}</div>\n";
    $top +=60;
    $count++;
}
?> 

 

WOW. It is so hard to understand your code algorithm. Let's say $num=9, $len1=3, $len2=6.

You didn't assign a value for $count, so $count=0, how can $count equal to $len1 or $len2? I have tested your code, it works for displaying the divs for 2 column, how if I want 50 column?

[sasa]

all div in 1st column have same left property and each next top is 2 row biger

i just calculate numberd of data in 1st column ($len1)

when 1st column is full go to 2nd column

all div in 2nd column have same left property and each next top is 2 row biger

i calculate number of data in 2nd column and add to $len1 ($len2)

when fierst two columns is full go to 3rd..

[zac1987]

all div in 1st column have same left property and each next top is 2 row biger

i just calculate numberd of data in 1st column ($len1)

when 1st column is full go to 2nd column

all div in 2nd column have same left property and each next top is 2 row biger

i calculate number of data in 2nd column and add to $len1 ($len2)

when fierst two columns is full go to 3rd..

 

How you do the "when 1st column is full go to 2nd column" ? How you know when the 1st column is full?

 

I have 16 rows of records on database.

When I use your codes, first 5 records are not displayed. It starts display the result from 6th record.

I echo out the value of count, len1 and len2 of every process and the output is :

count=1, len1=6, len2=11.

count=2, len1=6, len2=11.

count=3, len1=6, len2=11.

count=4, len1=6, len2=11.

....

count=16, len1=6, len2=11.

 

my database has 16 records, so len1=ceil(16/3)=6

len2=6+floor(16/3)=11

 

if($count==$len1){$top=30;  $left=10;}

if($count==$len2){$top=80;  $left=200;}

So it mean if $count==6, then only execute the code. That's why it fail to display the first 5 records on my database.

 

Why you use num/3? but not num/2 or num/5?

[zac1987]

I used firebug to detect the 1st div, then i just realize the 1st div is at the back of the 5th div, which mean the 5th div is cover on top of 1st div, so I can't see the 1st div, which mean they both divs are at the same left and same top.

[zac1987]

I just change the value of left :

if($count==$len1){$top=80;  $left=200;}

if($count==$len2){$top=30;  $left=400;}

Then I saw the 3 columns already!!! Thank you!!! So if I want to add another new column (4th column), I just need to write something like this?

$len3 = $len2 + floor($num/3);

if($count==$len3){$top=30;  $left=600;}

 

It is okay, I go try it now, thanks anyway Thank you guru Sasa. 

 

[zac1987]

Finally I know why you do num/3, because u though I just want 3 columns. Haha. Actually I want 1000 columns.

 

I have edit your codes to :

$len1 = 3;

$len2 = 6;

$len3 = 9;

$len4 = 12;

$len5 = 15;

while($userinfo = mysql_fetch_array($result2)){

$lastid=$userinfo['id'];

  if($count==$len1){$top=80;  $left=200;}

    if($count==$len2){$top=30;  $left=400;}

if($count==$len3){$top=80;  $left=600;}

if($count==$len4){$top=30;  $left=800;}

if($count==$len5){$top=80;  $left=1000;}

 

It works perfect for me!!! Thanks again for your guidance, I really appreciate it. Thanks.

[zac1987]

Is it I need to write $len1 until $len1000 manually?

$len1 = 3;

$len2 = 6;

$len3 = 9;

$len4 = 12;

$len5 = 15;

.....

$len1000 = 3000;

since all data below are +3, +3, +3

Can I use for loop to assign the value for $len1 to $len1000?

for ($i=3; $i<=3000; $i+3)

{

echo $len . $i .... er, I don't know how to write it. 

[sasa]

<?php
$left = 0;
$top = 30;
$num = mysql_num_rows($result2);
$num_col = 5;
$len1 = floor($num/$num_col);
for($i=0;$i<$num_col; $i++)$len[$i]=$len1;
$j = 0;
$r = $num % $num_col;
for($i=0; $i< $r; $i++){
    if($j >= $num_col) $j = 1;
    $len[$j]++;
    $j +=2;
}
$count = 0;
$i = 0;
while($userinfo = mysql_fetch_array($result2)){
    if($count==$len[$i]){$top=($top%100==30) ? 80 : 30; $i++; $left=200 * $i; $count=0;}
    echo "<div style=\"left:{$left}px; top:{$top}px; position:absolute;\">{$userinfo['message']}</div>\n";
    $top +=100;
    $count++;
}
?> 

[zac1987]

<?php
$left = 0;
$top = 30;
$num = mysql_num_rows($result2);
$num_col = 5;
$len1 = floor($num/$num_col);
for($i=0;$i<$num_col; $i++)$len[$i]=$len1;
$j = 0;
$r = $num % $num_col;
for($i=0; $i< $r; $i++){
    if($j >= $num_col) $j = 1;
    $len[$j]++;
    $j +=2;
}
$count = 0;
$i = 0;
while($userinfo = mysql_fetch_array($result2)){
    if($count==$len[$i]){$top=($top%100==30) ? 80 : 30; $i++; $left=200 * $i; $count=0;}
    echo "<div style=\"left:{$left}px; top:{$top}px; position:absolute;\">{$userinfo['message']}</div>\n";
    $top +=100;
    $count++;
}
?> 

 

My god, I took many hours to understand your codes. Finally I understand that. It works perfectly for me now. Thank you. You are the smartest person that I ever seen. Thank you so much.