using links inside a php with

Question

I am unable to concatenate HERE it would be great if somebody can point out my mistake and let me know what I am doing wrong and how it should be done.

<?php
.
.
while ($query_row = mysql_fetch_assoc($query_run)){             

$id=$query_row['id'];
$firstname=$query_row['firstname'];

echo '<a class="profile" href="profile.php?id=<?php echo $id.' ">'ucfirst' ($firstname);?></a>'; // HERE
.
.
.       
?>

Random · Accepted Answer · 2015-11-09 13:36:03Z

2

You are in a <?php (line 1), so you can't use an other tag inside it (in your echo). It should look this way :

echo '<a class="profile" href="profile.php?id='.$id.'">'.ucfirst($firstname).'</a>';

edited Nov 9, 2015 at 13:36

answered Nov 9, 2015 at 13:29

Random

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8 Comments

kunz Over a year ago

/profile.php?id=$id thats what my url says but it should say a number not $id

Random Over a year ago

in a php string, $id will be interpreted. so it will be href="profile.php?id=12345"

kunz Over a year ago

echo '<a class="profile" href="profile.php?id='.$id.'">'.ucfirst($firstname).'</a>'; i have changed it to this now it works

kunz Over a year ago

i will mark this is correct since i only figured half of it and you guys told me how to do the rest and thanks for pointing out what i did wrong

Random Over a year ago

Yes it is, it it mostly the same. just change them, and see if you got errors when launching the code, you'll see if there is a missing/wrong parameter...

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Saty · Accepted Answer · 2015-11-09 13:33:44Z

2

You use php tag inside php tag without close it and your ucfirst treated as a string.

echo '<a class="profile" href="profile.php?id='.$id.'">'.ucfirst($firstname).'</a>'; // HERE

Note:- mysql is deprecated instead use mysqli or pdo

edited Nov 9, 2015 at 13:33

answered Nov 9, 2015 at 13:30

Saty

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