Print specified rows mysql & PHP

Question

Well, very newbie question here, but I couldn't find the appropiate question to put it on google or here.

I have a very small table, with only 3 rows, every row with a defined id (1, 2, 3).

What I want to do is to print the row data of each one, but avoiding a loop function.

I.e.

The query would be:

$query = "SELECT field1, field2, field3 FROM table";
$var1 = mysql_query($query);
$var2 = mysql_fetch_assoc($var1);
$var3 = mysql_num_rows($var1);

The PHP:

<p><?php echo $var2['field1'];//how can i tell to the code "use the row with this id here -> id=1" ?></p>

<p><?php echo $var2['field1']; // "and use the row with the id "2" here" ?></p>

PS: Sorry about my poor english, argentine dude here.

Ok,

This is why I didn't want to do it as a loop:

I know I can do it this way:

<?php do { ?>

<p>bla bla <?php echo $var2[field1]; ?> </p> 

<?php } while ($var2 = mysql_fetch_assoc($var1)); ?>

But it can't be done like this, because the site design puts this data in different divs, with different positioning and CSS formatting.

I would like to do it the right way, avoiding the use o one query per row.

The rows will be forever 3, and the id of each one will always be the same (1, 2, 3).

@user1570159 See my answer, I posted an example of how to do it simply with a loop. — dririan
– dririan, Commented Aug 2, 2012 at 2:57

Hawili · Accepted Answer · 2012-08-02 03:03:08Z

1

you can use mysql_field_seek like this:

$query = "SELECT field1, field2, field3 FROM table";
$query_result = mysql_query($query);

$var1 = mysql_field_seek($query_result,0);
$var2 = mysql_field_seek($query_result,1);
$var3 = mysql_field_seek($query_result,2);

$resutl1 = mysql_fetch_field($var1);
$resutl2 = mysql_fetch_field($var2);
$resutl3 = mysql_fetch_field($var3);

answered Aug 2, 2012 at 3:03

Hawili

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Comments

dririan · Accepted Answer · 2012-08-02 02:54:06Z

0

There is no way to do it without a loop. Why do you want to avoid a loop, though? It should be fairly simple, for example:

$query = 'SELECT field1, field2, field3 FROM table';
$result = mysql_query($query);
while ($row = mysql_fetch_assoc($result))
    echo "<p>{$row['field1']}</p>";
mysql_free_result($result);

answered Aug 2, 2012 at 2:54

dririan

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Comments

Ben Rowe · Accepted Answer · 2012-08-02 02:57:25Z

0

Here's a simple function to get the fieldname and id. Then you can call it for any field/id combo you need.

function getField($fieldName, $id)
{ 
  $sql = sprintf("SELECT `%s` FROM `table` WHERE `id` = %d", $fieldName, $id);
  $result = mysql_query($sql);
  return mysql_num_rows($result) == 1 ? mysql_result($result, 0, 0) : null;
}

echo getField('field1', 1);
echo getField('field1', 2);

answered Aug 2, 2012 at 2:57

Ben Rowe

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Comments

Revent · Accepted Answer · 2012-08-02 03:07:54Z

I can't think of a reason not to use a loop... In general the mysql_fetch_assoc function returns a row of data from the result resource and then automatically increments the internal row pointer in the result resource. This would be the loop way to do it:

while ($row = mysql_fetch_assoc($var1)) {
    echo $row["field1"];
    echo $row["field2"];
    echo $row["field3"];
}

mysql_free_result($var1);

However, you could do this since the internal row pointer is automatically incremented.

$row1 = mysql_fetch_assoc($var1);
$row2 = mysql_fetch_assoc($var1);
$row3 = mysql_fetch_assoc($var1);

echo $row1["field1"];
echo $row1["field2"];
echo $row1["field3"];

echo $row2["field1"];
...

and that will essentially get you the same thing without the loop. Happy coding!

sephoy08 · Accepted Answer · 2012-08-02 03:08:45Z

0

How about this one? this might work for you.

$result = mysql_query("SELECT field1, field2, field3 FROM table");

$arr1 = (mysql_fetch_array($result));
$arr2 = (mysql_fetch_array($result));
$arr3 = (mysql_fetch_array($result));

switch ($x)
{
    case 1: echo $arr1[0] ." - ".$arr1[1] ." - ".$arr1[2]; break;
    case 2: echo $arr2[0] ." - ".$arr2[1] ." - ".$arr2[2]; break;
    case 3: echo $arr3[0] ." - ".$arr3[1] ." - ".$arr3[2]; break;
    default: echo "Unknown";
}

answered Aug 2, 2012 at 3:08

sephoy08

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