new to programming and have a problem.

Christopher_Peters · May 15, 2013

Hey guys new to programming here, learning because I want to (personal interest) been programming for 2-3 days now and started learning PHP today, below is the code i am using, I have successfully got it to access my database and display my results in a table in the order I want, however I want variable "$f6" to anchor the link it displays on the echo ($f6 is a website url).

First post here but probably won't be the last, if anyone has any suggestions please do tell

<html>

<body>

<?php

$username=" ";

$password=" ";

$database=" ";

mysql_connect(" ",$username,$password);

@mysql_select_db($database) or die("Database Error");

$query="SELECT * FROM legs";

$result=mysql_query($query);

$num=mysql_numrows($result);

mysql_close();

?>

<tr>

<td>Difficulty</td>

<td>Time Needed</td>

<td>Equipment Needed</td>

<td>Gender</td>

</tr>

<?php

$i=0;

while ($i < $num) {

$f1=mysql_result($result,$i,"Body Area");

$f2=mysql_result($result,$i,"Difficulty");

$f3=mysql_result($result,$i,"Time Needed");

$f4=mysql_result($result,$i,"Equipment Needed");

$f5=mysql_result($result,$i,"Gender");

$f6=mysql_result($result,$i,"Link");

?>

<tr>

</tr>

<?php

$i++;

}

?>

</body>

</html>

andrewerhardt · May 15, 2013

Hello!

If I understand your question correctly, you can do this.

<td><font face="Arial, Helvetica, sans-serif"><?php echo '<a href=". $f6 .">A Link</a>'; ?></font></td>

Hope that helps!

Christopher_Peters · May 15, 2013

mmm i have tried this and i don't think its what i mean no ill try to explain better (still getting used to terminology)

bascially variable $f6 is the "website url" column in my database so for example, it would currently display as www.google.com, but without an anchor to that website i want an anchor link to be there for whatever website $f6 brings up

thanks for the quick reply though

Q695 · May 15, 2013

Hello!

If I understand your question correctly, you can do this.
<td><?php echo '<a href=". $f6 .">A Link</a>'; ?></td>
Hope that helps!

He forgot the while loop:

<?php
while ($row){
$f1=$row['f1'];
$f6=$row['f6'];

echo"
<tr>
<td><font face='Arial, Helvetica, sans-serif'> $f1</font></td>
...
<td><font face='Arial, Helvetica, sans-serif'> $f6 </font></td>
</tr>";
}
?>

Edited May 15, 2013 by Q695

Christopher_Peters · May 15, 2013

hey guys, thanks for the replys again but i am still having issues with this, here is my code again with the alterations. keep getting "( ! ) Parse error: syntax error, unexpected T_STRING, expecting ',' or ';' in C:\wamp\www\newdatabase\index.php on line 43"

just so im fully clear with what im trying to achieve, $f6 will display a link like www.google.com (or whatever entry is in my db) but i want that link to be displayed with an anchor to the website too, instead of just plain text

thanks again for the help so far.

<?php

$username="";

$password="";

$database="";

mysql_connect("",$username,$password);

@mysql_select_db($database) or die("Unable to select database");

$query="SELECT * FROM legs";

$result=mysql_query($query);

$num=mysql_numrows($result);

mysql_close();

?>

<tr>

<td>Difficulty</td>

<td>Time Needed</td>

<td>Equipment Needed</td>

<td>Gender</td>

</tr>

<?php

$i=0;

while ($i < $num) {

$f1=mysql_result($result,$i,"Body Area");

$f2=mysql_result($result,$i,"Difficulty");

$f3=mysql_result($result,$i,"Time Needed");

$f4=mysql_result($result,$i,"Equipment Needed");

$f5=mysql_result($result,$i,"Gender");

$f6=mysql_result($result,$i,"Link");

?>

<?php

while ($row){

$f1=$row['f1'];

$f6=$row['f6'];

echo"

<tr>;

<td><?php echo $f1; ?></td>;

<td><?php echo $f2; ?></td>;

<td><?php echo $f3; ?></td>;

<td><?php echo $f4; ?></td>;

<td><?php echo $f5; ?></td>;

<td><?php echo '<a href=". $f6 .">A Link</a>'; ?></td>;

</tr>";

}

?>

<?php

$i++;

}

?>

</body>

</html>

Jessica · May 16, 2013

And the problem is?

Christopher_Peters · May 16, 2013

( ! ) SCREAM: Error suppression ignored for ( ! ) Parse error: syntax error, unexpected T_STRING, expecting ',' or ';' in C:\wamp\www\newdatabase\index.php on line 44

these are the errors i am currently getting (with the code above)

Jessica · May 16, 2013

Well, for one, remove the error surpressor.

Then solve the syntax error.

Post the code for lines 42-46 or so, in CODE TAGS.

Christopher_Peters · May 16, 2013

<?php
while ($row){
$f1=$row['f1'];
$f6=$row['f6'];
echo"
 
<tr>;
<td><font face="Arial, Helvetica, sans-serif"><?php echo $f1; ?></font></td>;
<td><font face="Arial, Helvetica, sans-serif"><?php echo $f2; ?></font></td>;
<td><font face="Arial, Helvetica, sans-serif"><?php echo $f3; ?></font></td>;
<td><font face="Arial, Helvetica, sans-serif"><?php echo $f4; ?></font></td>;
<td><font face="Arial, Helvetica, sans-serif"><?php echo $f5; ?></font></td>;
<td><font face="Arial, Helvetica, sans-serif"><?php echo '<a href=". $f6 .">A Link</a>'; ?></font></td>;
</tr>";
}
?>

Edited May 16, 2013 by Christopher_Peters

Jessica · May 16, 2013

Code tags means the little button with the <> so we can read the code. Not type the <>.

You have started a string with " - where you wrote echo"

Then you never properly end that string before trying to do more PHP. An editor with syntax highlighting would make this obvious

echo '<tr>
<td><font face="Arial, Helvetica, sans-serif">'.$f1.'</font></td>';

etc etc would work.

And for the love of god, get rid of those FONT tags.

Edited May 16, 2013 by Jessica

Christopher_Peters · May 16, 2013

Thanks again for the help, I have edited my code but I am now getting an undefined variable error on line 37 which is "while($row){"

<?php
while ($row){
$f1=$row['f1'];
$f6=$row['f6'];
echo '<tr>
<td><font face="Arial, Helvetica, sans-serif">' .$f1. '</td>';
'<td><font face="Arial, Helvetica, sans-serif">' .$f2. '</td>';
'<td><font face="Arial, Helvetica, sans-serif">' .$f3. '</td>';
'<td><font face="Arial, Helvetica, sans-serif">' .$f4. '</td>';
'<td><font face="Arial, Helvetica, sans-serif">' .$f5. '</td>';
'<td><font face="Arial, Helvetica, sans-serif">' .$f6. '</td>';
'</tr>'; 
 
}
?>

Jessica · May 16, 2013

Well, sounds like $row doesn't exist.

Not to mention, the code as shown would produce an infinate loop. Go read the php manual page for mysql_fetch_assoc or whatever function you want to use.

Christopher_Peters · May 16, 2013

ooh silly me, I have solved that error and my table is showing up fine, but my link row still isn't anchoring to the website.

I really appreciate all the help you've given me and for being patient with me.

Here's my code so far

<?php
$username=" ";
$password=" ";
$database=" ";
 
mysql_connect(" ",$username,$password);
@mysql_select_db($database) or die("Unable to select database");
$query="SELECT * FROM legs";
$result=mysql_query($query);
$num=mysql_numrows($result);
mysql_close();
?>
 
<table border="1" cellspacing="2" cellpadding="3">
 
<tr>
<td><font face="Arial, Helvetica, sans-serif"><b><u>Body Area</td>
<td><font face="Arial, Helvetica, sans-serif"><b><u>Difficulty</td>
<td><font face="Arial, Helvetica, sans-serif"><b><u>Time Needed</td>
<td><font face="Arial, Helvetica, sans-serif"><b><u>Equipment Needed</td>
<td><font face="Arial, Helvetica, sans-serif"><b><u>Gender</td>
<td><font face="Arial, Helvetica, sans-serif"><b><u>Link</td>
</tr>
 
<?php
$i=0;
while ($i < $num) {
 
$f1=mysql_result($result,$i,"Body Area");
$f2=mysql_result($result,$i,"Difficulty");
$f3=mysql_result($result,$i,"Time Needed");
$f4=mysql_result($result,$i,"Equipment Needed");
$f5=mysql_result($result,$i,"Gender");
$f6=mysql_result($result,$i,"Link");
?>
 
<?php
while ($row=mysql_fetch_assoc($result)){
$f1=$row['Body Area'];
$f2=$row['Difficulty'];
$f3=$row['Time Needed'];
$f4=$row['Equipment Needed'];
$f5=$row['Gender'];
$f6=$row['Link'];
 
echo '<tr>
<td><font face="Arial, Helvetica, sans-serif">' .$f1. '</td>' . 
'<td><font face="Arial, Helvetica, sans-serif">' .$f2. '</td>' .
'<td><font face="Arial, Helvetica, sans-serif">' .$f3. '</td>' .
'<td><font face="Arial, Helvetica, sans-serif">' .$f4. '</td>' .
'<td><font face="Arial, Helvetica, sans-serif">' .$f5. '</td>' . 
'<td><font face="Arial, Helvetica, sans-serif">' .$f6. '</td>';
'</tr>'; 
 
}
?>
 
<?php
$i++;
}
?>
 
</body>
</html>

Edited May 16, 2013 by Christopher_Peters

ignace · May 16, 2013

This is how your code should have looked. I, and so does the PHP manual, discourage the use of mysql_* functions and encourage the use of mysqli_* functions.

Learn CSS. Nobody use anymore nor do they use or .

<?php
$host   = "";
$user   = "";
$pass   = "";
$dbname = "";

$db = mysqli_connect($host, $user, $pass, $dbname);
if (mysqli_connect_errno()) {
    echo mysqli_connect_error();
    exit;
}

$query = "SELECT * FROM legs";
$result = mysqli_query($db, $query) or die(mysqli_error($db));
?>

<style>
/* this is CSS preferably this should be in a .css file and linked through a <link/> tag. */
body { font-family: Arial, Helvetica, sans-serif; }
table { border:1px solid gray; }
table thead tr th { font-weight: bold; text-decoration: underline; }
table tbody tr td, table thead tr th { padding: 3px; }
</style>

<table cellspacing="2">
<thead>
    <tr>
        <th>Body Area</th>
        <th>Difficulty</th>
        <th>Time Needed</th>
        <th>Equipment Needed</th>
        <th>Gender</th>
        <th>Link</th>
    </tr>
</thead>
<tbody>
    <?php while ($row = mysqli_fetch_assoc($result)) { ?>
    <tr>
        <td><?php echo $row['Body Area']; ?></td>
        <td><?php echo $row['Difficulty']; ?></td>
        <td><?php echo $row['Time Needed']; ?></td>
        <td><?php echo $row['Equipment Needed']; ?></td>
        <td><?php echo $row['Gender']; ?></td>
        <td><?php echo $row['Link']; ?></td>
    </tr>
    <?php } ?>
</tbody>
</table>

Edited May 16, 2013 by ignace

Christopher_Peters · May 16, 2013

Thanks for the reply, yet again , I have changed my code but my links still won't anchor to the website they belong to?

Here is a picture of the table (the data is sample data)

I want the www.whatevereverever.com to display as an actual link rather than just text

oh and yeah, I started learning CSS yesterday I will eventually be making style sheets for myself but getting this problem solved is my priority at the moment.

Thanks again.

Edited May 16, 2013 by Christopher_Peters

Jessica · May 16, 2013

So then wrap it in anchor tags like you did before.

ginerjm · May 16, 2013

Christopher_Peters · May 16, 2013

Ah you guys are amazing, iv finally managed to get the link to display and nearly everything is perfect, the one problem is the link is displaying as

http://localhost/newdatabase/www.google.com instead of just www.google.com

    <?php while ($row = mysqli_fetch_assoc($result)) { ?>
    <tr>
        <td><?php echo $row['Body Area']; ?></td>
        <td><?php echo $row['Difficulty']; ?></td>
        <td><?php echo $row['Time Needed']; ?></td>
        <td><?php echo $row['Equipment Needed']; ?></td>
        <td><?php echo $row['Gender']; ?></td>
        <td><a href="<?php echo $row['Link']?>"target=_blank">Link</a></td>
    </tr>
    <?php } ?>
</tbody>
</table>

Jessica · May 16, 2013

Try adding an "http://" in front of your link (either in the HTML or the DB.)

Christopher_Peters · May 16, 2013

Yeah that did it, again thanks a lot glad I can move on to the next task haha.

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