Display in each table row by using the same database table data Asked Feb 03 '16 at 11:56 Active Feb 03 '16 at 12:39 Viewed 918 times 0 I'm going to construct a table to let my user enter their SPM result then my system will filter and show them which course is eligible to apply. I've tried to make 10 rows in a table and each row consisted of 2 <select>, one is SPM subject and another one is the grade they obtained. The options of Subject and Grade were fetch from my database. Here is my coding for my table: <?php $result = mysql_query("SELECT * FROM spm_subject"); $result2 = mysql_query("SELECT * FROM spm_grade"); ?> <table class="p1" bgcolor="#FFFFCC" bordercolor="#000000" align="center" width="771" border="2"> <tr> <td><div align="center"><strong>No.</strong></div></td> <td><div align="center"><strong>Subject Name</strong></div></td> <td><div align="center"><strong>Grade</strong></div></td> </tr> <form action="checkresult2.php"> <?php for($i=1; $i<=10; $i++) {?> <tr> <td width="44"><div align="center"><?php echo $i; ?></div></td> <td width="601"> <select> <option value="">--- Please choose a subject ---</option> <?php while($s = mysql_fetch_assoc($result)) {?> <option name="subj"><?php echo $s["name"]; ?></option> <?php } ?> </select> </td> <td width="104"><div align="center"> <select> <option value="">&nbsp;</option> <?php while($g = mysql_fetch_assoc($result2)) {?> <option name="grad"><?php echo $g["grade"]; ?></option> <?php } ?> </select> </div> </td> </tr> <?php } ?> <tr> <td colspan="3"> <div align="center"> <input type="submit" value="Submit"> </div> </td> </tr> </form> </table> The table was constructed but only the FIRST ROW able to display data from the both database tables inside the both <select>. Any solution to solve this problem? php mysql database edited Feb 03 '16 at 12:05 fusion3k 11,568 4 25 47 asked Feb 03 '16 at 11:56 Wong Juin Hao 51 1 12 *Any solution to solve this problem?* What problem. ? You have not mention your problem. – Nana Partykar Feb 03 '16 at 12:09 each row must have a `select` with ALL `name` fields and another with ALL `grade` fields? – fusion3k Feb 03 '16 at 12:10 What error you are getting @Wong ? – Nana Partykar Feb 03 '16 at 12:12 @NanaPartykar First row of the table is able to display the table data inside only, but others cannot – Wong Juin Hao Feb 03 '16 at 12:16

Answer 1

1) Remove <form> inside <table>. Because, A form is not allowed to be a child element of a table, tbody or tr. You can have an entire table inside a form. You can have a form inside a table cell. You cannot have part of a table inside a form. For more info, click this

2) <option> don't have name attribute. Check here

3) Give name to <select></select> . Check here

4) Since, multiple values are being submitted to checkresult2.php page, you have to define name as name="subj[]" and name="grad[]" as array type.

5) An Alternative way for showing data in rest 9 rows is : before while loop give that query there itself, like below.

Updated Code.

<form action="checkresult2.php">
    <table class="p1" bgcolor="#FFFFCC" bordercolor="#000000" align="center" width="771" border="2">
          <tr>
              <td><div align="center"><strong>No.</strong></div></td>
              <td><div align="center"><strong>Subject Name</strong></div></td>
              <td><div align="center"><strong>Grade</strong></div></td>
          </tr>

            <?php 
          for($i=1; $i<=10; $i++)
          {?>
          <tr>
              <td width="44"><div align="center"><?php echo $i; ?></div></td>
              <td width="601">
                  <select name="subj[]">
                      <option value="">--- Please choose a subject ---</option>
                          <?php
                          $result = mysql_query("SELECT * FROM spm_subject");
                          while($s = mysql_fetch_assoc($result))
                          {?>
                              <option value="<?php echo $s['name']; ?>"><?php echo $s["name"]; ?></option>
                      <?php } ?>
                  </select>
              </td>
              <td width="104"><div align="center">
                  <select name="grad[]">
                      <option value="">&nbsp;</option>
                          <?php
                          $result2 = mysql_query("SELECT * FROM spm_grade");
                          while($g = mysql_fetch_assoc($result2))
                          {?>
                              <option value="<?php echo $g['grade']; ?>"><?php echo $g["grade"]; ?></option>
                      <?php } ?>
                  </select>
              </div>
              </td>
          </tr>
            <?php }?>
          <tr>
              <td colspan="3">
                  <div align="center">
                      <input type="submit" value="Submit">
                  </div>
              </td>
          </tr>
    </table>
</form>

[NOTE: mysql_ is deprecated in PHP 5.5.0, and it was removed in PHP 7.0.0. Use MySQLi or PDO_MySQL extension should be used.]

Answer 2

You have to preprocess the query populating two arrays:

$names = $grades = array();
for( $i=1; $i<=10; $i++ ) 
{
    $names[] = mysql_fetch_assoc( $result );
    $grades[] = mysql_fetch_assoc( $result2 );
}

Then, in our code:

<?php foreach( $names as $key => $name ) {?>
    <option name="subj"><?php echo $name['name']; ?></option>
<?php } ?>

and

<?php foreach( $grades as $key => $grade ) {?>
    <option name="grad"><?php echo $grade['grade']; ?></option>
<?php } ?>

mysql_fetch_assoc fetch the current row from results and then ‘go’ to next row, so when the first for loop is terminate, the cursor has reached the end of results and - in next for loops, there is nothing to fetch.

Please note:

mysql_ syntax was deprecated in PHP 5.5.0, and it was removed in PHP 7.0.0. Instead, the MySQLi or PDO_MySQL extension should be used.