PHP: changing row color showing mysql results

emilioestevezz · 09-08-2004, 07:17 PM

Hi, i have this little script that shows results from a mysql query, it works but i want to change something. i´ve got 4 records at the databse and i want to show results but one record with a white background color then the second with a black background color then the third again with white background color an so on how can i modify this script to show results in that way???

And here is the script:

// Muestra todos los apellidos encontrados en la tabla
if(mysql_num_rows($result)) {
$rank = 1;
while($row = mysql_fetch_row($result))

print("<td width=\"6%\" bgcolor=\"$color\"><HR>");
print("Id:<a href=\"borra.php?id=$row[0]\"><img src=\"images/tacho.gif\" alt=\"Borrar\" width=\"15\" height=\"22\" border=\"0\" align=\"right\">$row[0]</a> ");
print("Apellido:$row[1] ");
print("Nombre:$row[2] ");
print("Teléfono:$row[4] ");
print("Dni:$row[9] ");
print("Fecha:$row[5] ");
print("Hora:$row[6] ");
print("Abodado:$row[7] ");
print("Asunto:$row[8] ");
print("Donde nos conoció:$row[3] ");
print("<HR></td>");

}else{
echo ("No hay registros para mostrar!");
}

Thanks.
Emilio

rjcrews · 09-08-2004, 07:55 PM

make a counter and do even odds?

if(counter==1)
make it white
counter--
else
make it black
counter++

or,

if white
make it black
else
make it white

lots of ways...

CroMagnon · 09-08-2004, 08:11 PM

Code:

$colours = array( 0 => "#000000", 1 => "#ffffff" );
$rank = 1;
while($row = mysql_fetch_row($result)) {
    $color = $colours[ ($rank % 2) ];
    $rank++;

rjcrews · 09-08-2004, 08:12 PM

foolish mod

emilioestevezz · 09-09-2004, 12:57 PM

Quote:

Originally posted by CroMagnon

Code:

$colours = array( 0 => "#000000", 1 => "#ffffff" ); $rank = 1; while($row = mysql_fetch_row($result)) { $color = $colours[ ($rank % 2) ]; $rank++;

Thanks CroMagnon, but maybe theres something wrong on the adaptation that i´ve made, cos it showes results as before all with white background.

I also would like to put a afunction that when there is no data to show tell the user that .

heres is how it is after modifications:

// Muestra todos los apellidos encontrados en la tabla
$colours = array( 0 => "#CCDFED", 1 => "#E1DDF3" );
$rank = 1;

while($row = mysql_fetch_row($result)) {
$color = $colours[ ($rank % 2) ];
$rank++;

print("</tr><tr>");

print("<td width=\"6%\" bgcolor=\"$color\"><HR>");
print("Id:<a href=\"borra.php?id=$row[0]\"><img src=\"images/tacho.gif\" alt=\"Borrar\" width=\"15\" height=\"22\" border=\"0\" align=\"right\">$row[0]</a> ");
print("Apellido:$row[1] ");
print("Nombre:$row[2] ");
print("Teléfono:$row[4] ");
print("Dni:$row[9] ");
print("Fecha:$row[5] ");
print("Hora:$row[6] ");
print("Abodado:$row[7] ");
print("Asunto:$row[8] ");
print("Donde nos conoció:$row[3] ");
print("<HR></td>");
}

?>

Is this ok???

CroMagnon · 09-09-2004, 05:41 PM

The <TD> tag looks good, and the logic seems right (I have copied the top portion of your code and tried it out on my server, and it seems to work).

First step I would say is to check the source code of the page from your browser - see if the TD tags are getting the right colour values printed out.

If they are there, then this is very odd - perhaps you have some CSS being included that is overriding your specifications? If they are not, it should give you a hint as to what's really happening. If you'd like, you can paste the HTML here and I'll have a look too. Also, make sure your table is closed off properly - </tr> and </table>. I don't think this would affect the colours, but web browsers are strange beasts.

To tell the user when there is no data, simply put a check just after the while loop - if there is one row of data, $rank will be incremented:

[code]print("<HR></td>");
}
if ($rank == 1) {
print( "No data!" );
}

emilioestevezz · 09-10-2004, 01:12 PM

Quote:

Originally posted by CroMagnon
The <TD> tag looks good, and the logic seems right (I have copied the top portion of your code and tried it out on my server, and it seems to work).

First step I would say is to check the source code of the page from your browser - see if the TD tags are getting the right colour values printed out.

If they are there, then this is very odd - perhaps you have some CSS being included that is overriding your specifications? If they are not, it should give you a hint as to what's really happening. If you'd like, you can paste the HTML here and I'll have a look too. Also, make sure your table is closed off properly - </tr> and </table>. I don't think this would affect the colours, but web browsers are strange beasts.

To tell the user when there is no data, simply put a check just after the while loop - if there is one row of data, $rank will be incremented:

[code]print("<HR></td>");
}
if ($rank == 1) {
print( "No data!" );
}

Okay, here is the entire showdata.php:

<?php

$conn = @mysql_connect("localhost", "ssfceo101", "ssfceo23arm99");

if (!$conn) {
echo( "No se pudo conectar " .
"al servidor MySQL." );
exit();
}

if (! @mysql_select_db("clientes") ) {
echo( "No se puede encontrar " .
"la base de datos clientes!" );
exit();
}

// Request all data
$result = mysql_query(
"SELECT * FROM newclients");
if (!$result) {
echo("Error realizando el query: " .
mysql_error() . "");
exit();
}

// Muestra todos los apellidos encontrados en la tabla
$colours = array( 0 => "#CCDFED", 1 => "#E1DDF3" );
$rank = 1;

[code]print("<HR></td>");
}
if ($rank == 1) {
print( "No data!" );
}

while($row = mysql_fetch_row($result)) {
$color = $colours[ ($rank % 2) ];
$rank++;

print("</tr><tr>");

print("<td width=\"6%\" bgcolor=\"$color\"><HR>");
print("Id:<a href=\"borra.php?id=$row[0]\"><img src=\"images/tacho.gif\" alt=\"Borrar\" width=\"15\" height=\"22\" border=\"0\" align=\"right\">$row[0]</a> ");
print("Apellido:$row[1] ");
print("Nombre:$row[2] ");
print("Teléfono:$row[4] ");
print("Dni:$row[9] ");
print("Fecha:$row[5] ");
print("Hora:$row[6] ");
print("Abodado:$row[7] ");
print("Asunto:$row[8] ");
print("Donde nos conoció:$row[3] ");
print("<HR></td>");
}

?>

i also added the no data function but its complaining about some parse error. Checkit out.

Thanks.

CroMagnon · 09-11-2004, 03:46 AM

Sorry, I forgot to close my code tag, so it appeared in the text. Also, you've pasted it in the wrong place - I put last two lines of your code so you could see where to paste it. The other problem was a missing <table> tag (!) - this is why you didn't see any colour changes... the browser wasn't rendering the data inside a table.
The whole codefile should read like this:

Code:

<?php

$conn = @mysql_connect("localhost", "ssfceo101", "ssfceo23arm99");

if (!$conn) {
echo( "<P>No se pudo conectar " .
"al servidor MySQL.</P>" );
exit();
}

if (! @mysql_select_db("clientes") ) {
echo( "<P>No se puede encontrar " .
"la base de datos clientes!</P>" );
exit();
}

// Request all data
$result = mysql_query(
"SELECT * FROM newclients");
if (!$result) {
echo("<P>Error realizando el query: " .
mysql_error() . "</P>");
exit();
}

// Muestra todos los apellidos encontrados en la tabla
$colours = array( 0 => "#CCDFED", 1 => "#E1DDF3" );
$rank = 1;
print("<table>")
while($row = mysql_fetch_row($result)) {
$color = $colours[ ($rank % 2) ];
$rank++;

print("</tr><tr>");

print("<td width=\"6%\" bgcolor=\"$color\"><small><HR>");
print("<font face=\"Arial\ size=\"2\"><strong>Id:</strong><a href=\"borra.php?id=$row[0]\"><img src=\"images/tacho.gif\" alt=\"Borrar\" width=\"15\" height=\"22\" border=\"0\" align=\"right\">$row[0]</a></font></small><br>");
print("<font face=\"Arial\ size=\"2\"><strong>Apellido:</strong>$row[1]</font><br>");
print("<font face=\"Arial\ size=\"2\"><strong>Nombre:</strong>$row[2]</font><br>");
print("<font face=\"Arial\" color=\"#CC0000\ size=\"2\"><strong>Teléfono:</strong>$row[4]</font><br>");
print("<font face=\"Arial\" color=\"#CC6600\ size=\"2\"><strong>Dni:</strong>$row[9]</font><br>");
print("<font face=\"Arial\ size=\"2\"><strong>Fecha:</strong>$row[5]</font><br>");
print("<font face=\"Arial\ size=\"2\"><strong>Hora:</strong>$row[6]</font><br>");
print("<font face=\"Arial\ size=\"2\"><strong>Abodado:</strong>$row[7]</font><br>");
print("<font face=\"Arial\ size=\"2\"><strong>Asunto:</strong>$row[8]</font><br>");
print("<font face=\"Arial\ size=\"2\"><strong>Donde nos conoció:</strong>$row[3]</font><br>");
print("<HR></td>");
}

if ($rank == 1) {
print( "No data!" );
}

?>

Cedrik · 09-11-2004, 04:52 AM

Instead of :

PHP Code:



$colours = array( 0 => "#CCDFED", 1 => "#E1DDF3" ); 
$rank = 1; 
print("<table>") 
while($row = mysql_fetch_row($result)) { 
    $color = $colours[ ($rank % 2) ]; 
    $rank++;

I would do :

PHP Code:



$rank = 1; 
$flag = FALSE; 
print("<table>") 
while($row = mysql_fetch_row($result)) { 
    $color = ($flag) ? "#E1DDF3" :  "#CCDFED"; 
    $flag = !$flag; 
    $rank++;

Cedrik · 09-11-2004, 05:01 AM

Also instead of :

PHP Code:



print("</tr><tr>"); 
 
print("<td width=\"6%\" bgcolor=\"$color\"><small><HR>"); 
print("<font face=\"Arial\ size=\"2\"><strong>Id:</strong><a href=\"borra.php?id=$row[0]\"><img src=\"images/tacho.gif\" alt=\"Borrar\" width=\"15\" height=\"22\" border=\"0\" align=\"right\">$row[0]</a></font></small><br>"); 
print("<font face=\"Arial\ size=\"2\"><strong>Apellido:</strong>$row[1]</font><br>"); 
print("<font face=\"Arial\ size=\"2\"><strong>Nombre:</strong>$row[2]</font><br>"); 
print("<font face=\"Arial\" color=\"#CC0000\ size=\"2\"><strong>Teléfono:</strong>$row[4]</font><br>"); 
print("<font face=\"Arial\" color=\"#CC6600\ size=\"2\"><strong>Dni:</strong>$row[9]</font><br>"); 
print("<font face=\"Arial\ size=\"2\"><strong>Fecha:</strong>$row[5]</font><br>"); 
print("<font face=\"Arial\ size=\"2\"><strong>Hora:</strong>$row[6]</font><br>"); 
print("<font face=\"Arial\ size=\"2\"><strong>Abodado:</strong>$row[7]</font><br>"); 
print("<font face=\"Arial\ size=\"2\"><strong>Asunto:</strong>$row[8]</font><br>"); 
print("<font face=\"Arial\ size=\"2\"><strong>Donde nos conoció:</strong>$row[3]</font><br>"); 
print("<HR></td>");

I would prefer :

PHP Code:



echo <<<END 
</tr><tr> 
<td width="6%" bgcolor="$color"><small><HR> 
<font face="Arial" size="2"><strong>Id:</strong><a href="borra.php?id=$row[0]"> 
<img src="images/tacho.gif" alt="Borrar" width="15" height="22" border="0" align="right"> 
</a></font></small><br> 
<font face="Arial" size="2"><strong>Apellido:</strong>$row[1]</font><br> 
<font face="Arial" size="2"><strong>Nombre:</strong>$row[2]</font><br> 
<font face="Arial" color="#CC0000" size="2"><strong>Teléfono:</strong>$row[4]</font><br> 
<font face="Arial" color="#CC6600" size="2"><strong>Dni:</strong>$row[9]</font><br> 
<font face="Arial" size="2"><strong>Fecha:</strong>$row[5]</font><br> 
<font face="Arial" size="2"><strong>Hora:</strong>$row[6]</font><br> 
<font face="Arial" size="2"><strong>Abodado:</strong>$row[7]</font> 
<font face="Arial" size="2"><strong>Asunto:</strong>$row[8]</font><br> 
<font face="Arial" size="2"><strong>Donde nos conoció:</strong>$row[3]</font><br> 
<HR></td> 
END;

CroMagnon · 09-11-2004, 05:17 AM

Heh, I didn't know PHP supported the ? operator.

In that case, you could just do:

Code:

 $color = ($rank % 2) ? "#E1DDF3" :  "#CCDFED";

and dispense with $flag altogether. Of course, it's just a matter of personal preference, since they all do the same thing.

Cedrik · 09-11-2004, 05:20 AM

it's just a matter of personal preference
yep

I dislike the % operator

emilioestevezz · 09-13-2004, 11:04 AM

Quote:

Originally posted by CroMagnon
Heh, I didn't know PHP supported the ? operator.

In that case, you could just do:

Code:

$color = ($rank % 2) ? "#E1DDF3" : "#CCDFED";

and dispense with $flag altogether. Of course, it's just a matter of personal preference, since they all do the same thing.

Ok i´ve modified the ode but the little bugger still complains : Parse error on line 30 (while section)

// Muestra todos los apellidos encontrados en la tabla
$color = ($rank % 2) ? "#E1DDF3" : "#CCDFED";
$rank = 1;
print("<table>")
while($row = mysql_fetch_row($result)) {
$color = $colours[ ($rank % 2) ];
$rank++;

print("</tr><tr>");

print("<td width=\"6%\" bgcolor=\"$color\"><HR>");
print("Id:<a href=\"borra.php?id=$row[0]\"><img src=\"images/tacho.gif\" alt=\"Borrar\" width=\"15\" height=\"22\" border=\"0\" align=\"right\">$row[0]</a> ");
print("Apellido:$row[1] ");
print("Nombre:$row[2] ");
print("Teléfono:$row[4] ");
print("Dni:$row[9] ");
print("Fecha:$row[5] ");
print("Hora:$row[6] ");
print("Abodado:$row[7] ");
print("Asunto:$row[8] ");
print("Donde nos conoció:$row[3] ");
print("<HR></td>");
}

if ($rank == 1) {
print( "No data!" );
}

?>

this is supposed to e line 30:

while($row = mysql_fetch_row($result)) {

I don´t see any error here???

Thanks.
Emilio

emilioestevezz · 09-13-2004, 11:19 AM

Quote:

Originally posted by emilioestevezz
Ok i´ve modified the ode but the little bugger still complains : Parse error on line 30 (while section)

// Muestra todos los apellidos encontrados en la tabla
$color = ($rank % 2) ? "#E1DDF3" : "#CCDFED";
$rank = 1;
print("<table>")
while($row = mysql_fetch_row($result)) {
$color = $colours[ ($rank % 2) ];
$rank++;

print("</tr><tr>");

print("<td width=\"6%\" bgcolor=\"$color\"><HR>");
print("Id:<a href=\"borra.php?id=$row[0]\"><img src=\"images/tacho.gif\" alt=\"Borrar\" width=\"15\" height=\"22\" border=\"0\" align=\"right\">$row[0]</a> ");
print("Apellido:$row[1] ");
print("Nombre:$row[2] ");
print("Teléfono:$row[4] ");
print("Dni:$row[9] ");
print("Fecha:$row[5] ");
print("Hora:$row[6] ");
print("Abodado:$row[7] ");
print("Asunto:$row[8] ");
print("Donde nos conoció:$row[3] ");
print("<HR></td>");
}

if ($rank == 1) {
print( "No data!" );
}

?>

this is supposed to e line 30:

while($row = mysql_fetch_row($result)) {

I don´t see any error here???

Thanks.
Emilio

I´ve got it!!!!!!

gotta replace print("<table>") with print("<table>");

but i just have other detail to fix, i´ve recently changed the print part as Cedrik suggested it works ok, but it doesn´t show me the id number it shows an "_" instead of the id number any hints???

My boss like flash very much (darn!!!!!#%&$#%$"), so he ask me to convert this code so i could show data on a flash animation so if you can give me any pointers to do this. So far i´ve been reading some tuts but i can´t find a good one, i think its the reverse thing but i just can´t figure it out.

Emilio.

CroMagnon · 09-13-2004, 05:21 PM

Quote:

gotta replace print("<table>") with print("<table>");

Good catch. I've always hated that about PHP - it tells you there's a problem with line 30, and you look at line 30 until your eyes bleed, but you can't see any problem... then you realise you left the semicolon off line 29.

One other thing you might want to fix - you're assigning $color twice, but you don't need to. You do, however, need to do it inside the while loop or it will never change. Move line 28 to line 31, and remove the current line 31 (also remove the line that sets the $colours array since you won't need it with the new method).

If you copied Cedrik's code exactly as written, then you are missing the $row[0] part that needs to be after the <img> tag.