i wrote a form with some php loop to insert array and numbers
a form inside a table with tr td to get fixed width
php array within form, then for loops with html tr td
at the end of the form, <input>
<table>
<form method='post'>
<?php
$row=array(
'databaseName',
'tableName',
'nColumns',
'columnName',
'dataType',
'dataSize',
'null',
'indexing');
for($k=0; $k<=7; $k++){
?>
<tr><td><?php echo $row[$k]; ?></td>
<td><input name='<?php echo $row[$k]; ?>' type='text'>
</input></td></tr>
<?php if($k==2){ ?>
<tr>
<td>Number of Columns</td>
<td><select onchange='nColumns()'>
<?php for($i=1; $i<=12; $i++){ ?>
<option id='<?php echo $i; ?>'><?php echo $i; ?></option>
<?php } ?>
</select></td></tr><?php } ?>
<?php } ?>
<input name='submit' type='submit'></input>
</form></table>
but the input is displayed first before php loops. why? and how do i make the submit button to be at the bottom?
2
Answers
You are facing this issue because your submit button is outside the table structure, specifically outside the
<tr>
and<td>
.Try the below code:
table need to in the form.
recommended to place the submit button outside the table