do while in do while in php - php

Question

That loop using do while, and it is working fine but when I add another do while in to this the second do while work but first do while only show one row not all 10 row. My code is below
<?php do { ?>
<tr>
<td><?php echo $row_Recordset1['date']; ?></td>
<?php do { ?>
<td><?php echo $row_Recordset1['nav']; ?></td>
<?php } while ($row_Recordset1= mysql_fetch_assoc($Recordset1)); ?>
</tr>
<?php } while ($row_Recordset1 = mysql_fetch_assoc($Recordset1)); ?>
My date should be displayed once in a row but all the other td will be took from nav;
This is what I want:
------------------------
|2014-02-26|5.5|3.2|3.5|
------------------------
|2014-02-25|3.1|1.2|1.5|
But I'm currently getting:
------------------------
|2014-02-26|5.5|3.2|3.5|
------------------------

Answer 1

You should look at http://php.net/mysql_fetch_assoc
It moves the internal pointer one step ahead, so each time you do mysql_fetch_assoc() you get the next value, hence only your enternal do while is executed. That is, only date from the first row is output, and all other values are output in second do while;
Try this for exercise:
$q = mysql_query("SOME MYSQL QUERY WITH MINIMUM THREE ROWS");
$f = mysql_fetch_assoc($q);
$f = mysql_fetch_assoc($q);
$f = mysql_fetch_assoc($q);
print_r($f);

Answer 2

First of all, please don't bounce in and out of PHP like that...
<?php
do {
echo '<tr>
<td>';
echo $row_Recordset1['date'];
echo '</td>';
do {
echo '<td>';
echo $row_Recordset1['nav'];
echo '</td>';
} while ($row_Recordset1 = mysql_fetch_assoc($Recordset1));
echo '</tr>';
} while ($row_Recordset1 = mysql_fetch_assoc($Recordset1));
?>
Where do you fetch your first $row_Recordset1? Somewhere earlier than this code? You're going to output table cells (<td>) until you run out of rows. I don't think you want that.

Answer 3

Like so:
<?php do { ?>
<tr>
<td><?php echo $row_Recordset1['date']; ?></td>
<td><?php echo $row_Recordset1['nav']; ?></td>
</tr>
<?php } while ($row_Recordset1 = mysql_fetch_assoc($Recordset1)); ?>
Because each time you use mysql_fetch_assoc it does the following
Returns an associative array that corresponds to the fetched row and moves the internal data pointer ahead.

Answer 4

Part of the reason this doesn't work is because you are re-assigning the value of the same variable. If you don't want to lose it then you cannot do it this way. At the very least it is a bad idea if you ever intend to do anything new with the code.
Next, I really recommend you populate an array with the data first before you ever do anything with outputting it. It is a very good idea to separate logic from output.
Your code should look more like this:
<?php
//Assumes you have already connected to a mysqli resource
$conditional_data_filtered = $mysqli->real_escape_string($conditional_data);
$sql = "SELECT date, nav FROM some_table WHERE some_column = '" . $conditional_data_filtered . "'";
if ($result = $mysqli->query($sql)) {
/* fetch associative array */
while ($row = $result->fetch_assoc()) {
$data = array(); //Make sure we start with a fresh array each time.
$data['date'] = $row['date'];
//Now I think you have a bunch of data in the nav you need to loop through
$nav_filtered = $mysqli->real_escape_string($row['nav']);
$subSql = "SELECT some_data FROM navigation WHERE some_identifier = '" . $nav_filtered . "'";
if ($subResult = $mysqli->query($subSql)) {
while ($subRow = $subResult->fetch_assoc()) {
$data['nav'][] = $subRow;
}
}
$rowList[] = $data;
}
/* free result set */
$result->free();
}
foreach ($rowList as $rowData) {
echo "<tr>";
echo "<td>" . $rowData['date'] . "</td>";
foreach ($rowData['nav'] as $navData) {
echo "<td>" . $navData['some_info'] . "</td>";
}
echo "</tr>";
}
?>
Just a note: If you really don't have sub-data and just want to output the next column then you don't need a sub-loop. You just need to echo the contents of the column like you did with the first one. Then you can get rid of the sub-loops I have shown above and just put it within the first loop.