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I tried this code to display MySQL data into coded HTML table via PHP:

<?php
    // Get all the data from the "login" table
    $result = mysql_query("SELECT * FROM login") or die(mysql_error());

    echo '<table class="table">
          <thead>
              <tr>
                  <th>FirstName</th> 
                  <th>LastName</th> 
                  <th>user</th> 
              </tr>
          </thead>
          <tbody>';

    // keeps getting the next row until there are no more to get
    while($row = mysql_fetch_array( $result )) {
        **if( $i % 2 == 0 )** {  //line 20
            $class = " class='odd'";
        }
        else {
            $class = "";
        }
        // Print out the contents of each row into a table
        echo "<tr" . $class . "><td>"; 
        echo $row['FirstName'];
        echo "</td><td>"; 
        echo $row['LastName'];
        echo "</td><td>"; 
        echo $row['user'];
        echo "</td><td>";  
    }
    echo "</tbody></table>";
?>

But it shows me this error: Undefined variable: i in line 20 The line 20 error text is marked as bold in above code.

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closed as too localized by cryptic ツ, SztupY, PeeHaa, tereško, Baba Feb 25 at 22:18

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up vote 1 down vote accepted

Try this

 $i=0;
 while($row = mysql_fetch_array( $result )) {
    if( $i % 2 == 0 ) {
        $class = " class='odd'";
    } else {
        $class = "";

    }
    // Print out the contents of each row into a table
    echo "<tr" . $class . "><td>"; 
    echo $row['FirstName'];
    echo "</td><td>"; 
    echo $row['LastName'];
    echo "</td><td>"; 
    echo $row['user'];
    echo "</td><td>";  
  $i++;
}
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1  
Thanks Vims, thank you very very very much – Entire Ict Feb 24 at 7:41
up vote 1 down vote

add $i=0; before the while loop starts, and also increment it $i++ before while loop ends.

$i = 0;

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

//your code

  $i++;
}
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Thank you Vora, thanks very very much – Entire Ict Feb 24 at 7:38
Accept an answer that solved your problem by ticking the mark under the score – codingbiz Feb 24 at 7:57
up vote 0 down vote

Add 

$i = 0;

before the while where you are getting the error and use $i++ on you evaluation

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Thanks a lot ares – Entire Ict Feb 24 at 7:40
up vote 0 down vote

You were never incrementing your $i

$i = 0; 

while($row = mysql_fetch_array( $result )) {
  if($i % 2 == 0)
  ....

  $i++;  //You need to increment your $i
}
share|improve this answer
Thanks for answer, thanks a lot – Entire Ict Feb 24 at 7:39

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