assign db fields to variables with dynamically built var names

[webandwe]

Hi,

I need to store database data in variables so I can work with it.
Please se below what I am trying to do in my script....

If you echo the stuff out is work fine. Meaning if I go echo $row['a5]; I get the data I want but I can't get the data to store in a vairable.

I need something that goes from $a1 to $a18 and in each of then store the data from one field in the database name a5.

Thanks

[PHP]
<?php
$con = mysql_connect(" localhost","*** **","******") ;
if (!$con)
{
die('Could not connect: ' . mysql_error());
}

mysql_select_db ("*******", $con);

$result = mysql_query("SE LECT * FROM hole ORDER BY a5");

while($row = mysql_fetch_arr ay($result))
{
$a1 = $row['a5'];
$a2 = $row['a5']; /////// needs to give second result
$a3 = $row['a5']; /////// needs to give third result

}

echo "<img name=\"Hole\" src=\"$a1.jpg\" width=\"32\" height=\"32\" alt=\"Hole\">";



mysql_close($co n);


?>
[/PHP]

[ronverdonk]

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This helps to ensure that other members, and also the general public, will have a better chance of finding answers to any similar questions.

MODERATOR

[Markus]

Hi,

I need to store database data in variables so I can work with it.
Please se below what I am trying to do in my script....

If you echo the stuff out is work fine. Meaning if I go echo $row['a5]; I get the data I want but I can't get the data to store in a vairable.

I need something that goes from $a1 to $a18 and in each of then store the data from one field in the database name a5.

Thanks

[PHP]
<?php
$con = mysql_connect(" localhost","*** **","******") ;
if (!$con)
{
die('Could not connect: ' . mysql_error());
}

mysql_select_db ("*******", $con);

$result = mysql_query("SE LECT * FROM hole ORDER BY a5");

while($row = mysql_fetch_arr ay($result))
{
$a1 = $row['a5'];
$a2 = $row['a5']; /////// needs to give second result
$a3 = $row['a5']; /////// needs to give third result

}

echo "<img name=\"Hole\" src=\"$a1.jpg\" width=\"32\" height=\"32\" alt=\"Hole\">";



mysql_close($co n);


?>
[/PHP]

This is untested (just some logic)
[php]
$_i = 0;
$_a = array();
while($_row = mysql_fetch_arr ay($_res))
{
$_a[$_i] = $_row['col'];
++$_i;
}
[/php]

[ronverdonk]

Your question is so not clear!

Do you want

1. to assign the content of column 'a5' in each subsequent row into a variable $a1

2. store the contents of column 'a5' of each row into 18 different variables starting with $a1-$a18?

3. store the content of column 'a5' in variables $a1-$a18, e.g. column 'a5' content from row 1 goes into $a1, from row 10 goes into $a10, from row 18 goes into $a18.

Please describe your problem a bit more clearer.

Ronald

[webandwe]

Your question is so not clear!

Do you want

1. to assign the content of column 'a5' in each subsequent row into a variable $a1

2. store the contents of column 'a5' of each row into 18 different variables starting with $a1-$a18?

3. store the content of column 'a5' in variables $a1-$a18, e.g. column 'a5' content from row 1 goes into $a1, from row 10 goes into $a10, from row 18 goes into $a18.

Please describe your problem a bit more clearer.

Ronald

Hi yes.

Number 3.

3. store the content of column 'a5' in variables $a1-$a18, e.g. column 'a5' content from row 1 goes into $a1, from row 10 goes into $a10, from row 18 goes into $a18.

Please describe your problem a bit more clearer.

[Markus]

Hi yes.

Number 3.

3. store the content of column 'a5' in variables $a1-$a18, e.g. column 'a5' content from row 1 goes into $a1, from row 10 goes into $a10, from row 18 goes into $a18.

Please describe your problem a bit more clearer.

Then my post should suffice.

[webandwe]

I am testing it but for some reason it does not want to go...

[Markus]

I am testing it but for some reason it does not want to go...

Post what youre using

[webandwe]

[PHP]
<?php
$con = mysql_connect(" l*****","*****" ,"*****");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}

mysql_select_db ("****datebase* ****", $con);

$result = mysql_query("SE LECT * FROM hole ORDER BY a5");

$i = 0;
$a = array();
while($row = mysql_fetch_arr ay($result))
{
$a[$i] = $row['a5'];
++$i;
}

mysql_close($co n);
////Testing
echo "$a1 $a2 $a3 ";

?>
[/PHP]

[Markus]

[PHP]
<?php
$con = mysql_connect(" l*****","*****" ,"*****");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}

mysql_select_db ("****datebase* ****", $con);

$result = mysql_query("SE LECT * FROM hole ORDER BY a5");

$i = 0;
$a = array();
while($row = mysql_fetch_arr ay($result))
{
$a[$i] = $row['a5'];
++$i;
}

mysql_close($co n);
////Testing
echo "$a1 $a2 $a3 ";

?>
[/PHP]

Remember, the variables are being assigned into an array - not seperate variables.
[php]
echo "{$_a[0]} {$_a[1]} {$_a[2]}";
[/php]

[webandwe]

Super, it is working.!!!!!!! !!!!!!!

Never worked with it before but now i'll know!

Thanks a mil again!

[ronverdonk]

Remember, the variables are being assigned into an array - not seperate variables.
[php]
echo "{$_a[0]} {$_a[1]} {$_a[2]}";
[/php]

That is cheating: the requirement was to assign them to separate variables $a1 - $a18. You can do that as follows:
[php]while($row = mysql_fetch_arr ay($result))
{
$var='a'.$i;
$$var = $row['a5'];
++$i;
}
mysql_close($co n);
////Testing
echo "$a1 $a2 $a3 ";[/php]Ronald

[Markus]

Super, it is working.!!!!!!! !!!!!!!

Never worked with it before but now i'll know!

Thanks a mil again!

Welcome!
Regards.
:)

[webandwe]

That is cheating: the requirement was to assign them to separate variables $a1 - $a18. You can do that as follows:
[php]while($row = mysql_fetch_arr ay($result))
{
$var='a'.$i;
$$var = $row['a5'];
++$i;
}
mysql_close($co n);
////Testing
echo "$a1 $a2 $a3 ";[/php]Ronald

This one work the best for me because I need to use the Vairables all over the place. Thanks!!!!!

[ronverdonk]

As Markus already said: you are welcome!. See ya again.

Ronald