Displaying data horizontally using php and mysql

Question

Hi I'm attempting to display data retrieved from a mysql table horizontally in an html table using php. The code below works well except for the fact that it leaves out the first record (starts at the second record) in my database. I'm sure it has something to do with the counter but I can't seem to figure out how to get it to stop doing this. If anyone can point out my error I'd really appreciate it!

$items = 5;
$query = "SELECT * FROM members ";

$result = mysql_query($query) 
  or die(mysql_error());

$row = mysql_fetch_array($result);

if (mysql_num_rows($result) > 0) {

echo '<table border="1">';

$i = 0;

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

     $first_name = $row['first_name'];

           if ($i==0) {
                       echo "<tr>\n";
                      }

           echo "\t<td align=\center\">$first_name</td>\n";

           $i++;

           if ($i == $items) {

                              echo "</tr>\n";
                              $i = 0;
                              }
 }//end while loop

 if ($i > 0) {

                 for (;$i < $items; $i++) {
                      echo "<td>&nbsp;</td>\n";
                  }

              echo '</tr>';

              }//end ($i>0) if 

 echo '</table>';

 }else {

   echo 'no records found';
 }

You shouldn't use mysql_ function, and your indentation is cumbersome — samayo
– samayo, Commented Jul 14, 2013 at 18:52

srakl · Accepted Answer · 2013-07-14 18:55:09Z

1

try and remove the 1st

$row = mysql_fetch_array($result);

you are calling it twice, that's why it skips 1 row in your while loop

answered Jul 14, 2013 at 18:55

srakl

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1 Comment

matt tuman Over a year ago

That's it thanks! I would have marked it as the correct answer but apparently I can't after only nine minutes.

echo_Me · Accepted Answer · 2013-07-14 20:10:32Z

1

try this simpler.

       $items = 5;
  $query = "SELECT * FROM members ";

 $result = mysql_query($query) or die(mysql_error());
 if (mysql_num_rows($result) > 0) {

 echo '<table border="1">';
 while($row = mysql_fetch_array($result)){

 $first_name = $row['first_name'];
        echo "<tr>";
       for ($i=0 ; $i <= $items ;$i++) {

                    echo "<td align='center'>".$first_name."</td>";
                    }
}//end while loop
     echo "</tr>";
    echo '</table>';
}else{ echo 'no records found'; }

edited Jul 14, 2013 at 20:10

answered Jul 14, 2013 at 18:59

echo_Me

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1 Comment

echo_Me Over a year ago

@matttuman because i have removed \n and \t .try add them again and see

brasofilo · Accepted Answer · 2014-05-30 16:46:42Z

0

I have run into this issue before. Try the do while loop instead. Example

do {

// code

 } while($row = mysql_fetch_array($result)); //end while loop

edited May 30, 2014 at 16:46

brasofilo

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answered May 30, 2014 at 16:40

James

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Comments

DavidMc · Accepted Answer · 2019-03-12 13:17:38Z

0

$row = mysql_fetch_array($result); 1) remove this line of code from ur scripts 2) only use while loop code instead.

answered Mar 12, 2019 at 13:17

DavidMc

1

1 Comment

Nico Haase Over a year ago

Please explain that further such that others can learn from your answer

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