Insert a PHP Variable in MySQL Database?

kumschick

Hi,

I would like to insert a PHP variable in a MySQL Database. I give you an example what I would like to insert in the db:

<table>
  <tr>
    <td>font face="$face" size="$size" color="$color">This is a test...</font></td>
  </tr>
</table>

But my problem is that when I make a "SELECT * FROM DBNAME" I get the value back in the same way I typed it in. But I want the the $phpvariable takes affect.

I don't know how I could do it and under what I could find it in the manual. Does someone know how to solve my problem?

leatherback

Hi,

Have a look at the eval() function, and see wether you could get it tow work for you

kumschick

I don't know how to use [ eval(); ] with this script:

                    $connection=mysql_connect($host, $user, $pwd)or die("<font color=\"#ff0000\">ERROR: Server Not Found</font>");
                    $database=mysql_select_db($db_name,$connection);
                    $list=mysql_query("select * from $tbl_doc where id='$pgn'",$connection);

                    while($row=mysql_fetch_array($list))
                    {
                    echo "<img width=\"1\" height=\"3\" src=\"/images/layout/blank.gif\"><h4>$row[title]</h4>$row[content]";
                    }

and in $row[content] is this in the mysql db:

<table>
<tr>
<td>font face="$face" size="$size" color="$color">This is a test...</font></td>
</tr>
</table>

leatherback

Hi,

Not sure. Think you'd have to put something like

echo "blablabla".$var."blablabla";

in your database. Then I think you can output using
eval($row[content]);

Just try this using variables which hold the entry you'd have in the database, and see wether it gives the right output.

kumschick

I tried that, but I always get an error msg:

<?
$connection=mysql_connect("localhost", "username", "password")or die("<font color=\"#ff0000\">ERROR: Server Not Found</font>");
$database=mysql_select_db("database",$connection);
$list=mysql_query("select * from tablename where id='$pgn'",$connection);

while($row=mysql_fetch_array($list))
{
eval($row[content]);
echo "$row[content]";
}
?>

Parse error: parse error, unexpected '<' in D:\www\kumschicknw.com\users\help\demos\eval.php(10) : eval()'d code on line 1

and if I put eval($row[content]); as an comment the page works without errors, but the variable doesn't work, but that's normal.

What I don't understand is the "eval()";

leatherback

Hi,

Tested this snippet, which works. SHould help you get going:

<?

$var = "NOW!";
$outputin = 'print this stupid thing $var';


eval("\$outputout = \"$outputin\";");

echo $outputout;

?>

for you it would become something like:

$outputin = $row[content];

eval("\$outputout = \"$outputin\";");

echo $outputout;

Let me know wether this works!?

kumschick

thx it works now, but i have changed a little line, cause with your script is a small problem.

If the DB has plain <HTML> code like this:
<font size="$font_size">Hello World</font>
it don't work cause of the ["]

So if you change it to:

$outputin=addslashes($row[content]);
eval("\$outputout = \"$outputin\";");
echo $outputout;

It takes the normal HTML code:
<font size=$font_size>Hello World</font>

😃

Thx