Hi there.
I'm very very new to php. :o so i hope i'm posting in the right forum.
and i really can't figure this out. probably really simple but i need to place a url into:
while ($row = mysql_fetch_array($result)) {
echo $row;
echo " " . $row . "<br>";
everytime i place <a href= etc i get errors. :sad:
I need the login to link to the user's profile page.
Please could anybody help??
Kind Regards
Ross
Jump to PostSuppose you have the code:
echo "$value";Everything within the quotes is printed out. However, now take the following code:
echo "<a href="$value">";PHP gets confused where you have quotes within quotes. Therefore, if you want to print out a real quote inside an …
Jump to PostPractice makes perfect.
echo "<a href=\"member.php?mid=$row[mid]\">$row[login]</a>";I'm not exactly sure what $m[id] is ? Where are you fetching that $m[] array from?
Jump to Postgenerally when u pull it from mysql you would reference the item you want by column number, so....
if your query was something to this effect:$result = mysql_query("SELECT login,age FROM some_table WHERE user_name='".$user.'"); $row = mysql_fetch_row($result); $login = $row[0]; $age = $row[1]; $link = "<a href=\"members.php?mid=".$login."\">".$age."</a>
Suppose you have the code:
echo "$value";Everything within the quotes is printed out. However, now take the following code:
echo "<a href="$value">";PHP gets confused where you have quotes within quotes. Therefore, if you want to print out a real quote inside an echo statement, you have to escape the " character with a backlash. Like this ...
echo "<a href=\"$value\">";:confused: so it would be something like:
echo "<a href=member.php?mid=$m[id]\"$row\">";
?
said i'm new at this, sorry :o
Thanks for the quick reply
Kind Regards
Ross
:sad: unfortunatly none of the ways i try seem to work.
i don't think i'm cut out for this php thing.
Ross
Practice makes perfect.
echo "<a href=\"member.php?mid=$row[mid]\">$row[login]</a>";I'm not exactly sure what $m[id] is ? Where are you fetching that $m[] array from?
generally when u pull it from mysql you would reference the item you want by column number, so....
if your query was something to this effect:
$result = mysql_query("SELECT login,age FROM some_table WHERE user_name='".$user.'");
$row = mysql_fetch_row($result);
$login = $row[0];
$age = $row[1];
$link = "<a href=\"members.php?mid=".$login."\">".$age."</a>Hello. thanks for all your help, both of you, :D
what i currently have is:
<?php
$result = mysql_query("select P.id
, M.login
, dayofmonth(P.birthdate) as birth_day
, month(P.birthdate) as birth_month
, year(current_date)
-year(P.birthdate) as age
from profiles as P
inner
join members as M
on P.id = M.id
where dayofmonth(current_date) = dayofmonth(P.birthdate)
and month(current_date) = month(P.birthdate)") or die("Query failed: " . mysql_error());
while ($row = mysql_fetch_array($result)) {
echo "<a href=\"member.php?mid=$row[mid]\">$row[login]</a>";
echo " " . $row['age'] . "<br>";
}
?>which does give me the linked login name that i was after ;)
but the link doesnt point to the users profile :sad:
I wanted it as a todays birthdays list with login name and age, but the tables for login and birthdate was in seperate locations so i ended up with the above.
can you see what i am doing wrong here? :cry:
Kind Regards
Ross
i don't know, maybe it's a problem with your sql query?
can you login via ssh/telnet and double check to see if that works?
if not here is a PHP interface:
<!-- Program Name: mysql_send.php
Description: PHP program that sends an SQL query to the
MySQL server and displays the results.
-->
<html>
<head>
<title>SQL Query Sender</title>
</head>
<body>
<?php
$user="root";
$host="localhost";
$password="";
/* Section that executes query */
if (@$form == "yes")
{
mysql_connect($host,$user,$password);
mysql_select_db($database);
$query = stripSlashes($query) ;
$result = mysql_query($query);
echo "Database Selected: <b>$database</b><br>
Query: <b>$query</b>
<h3>Results</h3>
<hr>";
if ($result == 0)
echo("<b>Error " . mysql_errno() . ": " . mysql_error() . "</b>");
elseif ( <email protected>($result) == 0)
echo("<b>Query completed. No results returned.</b><br>");
else
{
echo "<table border='1'>
<thead>
<tr>";
for ($i = 0; $i < mysql_num_fields($result); $i++)
{
echo("<th>" . mysql_field_name($result,$i) . "</th>");
}
echo " </tr>
</thead>
<tbody>";
for ($i = 0; $i < mysql_num_rows($result); $i++)
{
echo "<tr>";
$row = mysql_fetch_row($result);
for ($j = 0; $j < mysql_num_fields($result); $j++)
{
echo("<td>" . $row[$j] . "</td>");
}
echo "</tr>";
}
echo "</tbody>
</table>";
}
echo "<hr><br>
<form action=$PHP_SELF method=post>
<input type=hidden name=query value=\"$query\">
<input type=hidden name=database value=$database>
<input type=submit name=\"queryButton\" value=\"New Query\">
<input type=submit name=\"queryButton\" value=\"Edit Query\">
</form>";
unset($form);
exit();
}
/* Section that requests user input of query */
@$query = stripSlashes($query);
if (@$queryButton != "Edit Query")
{
$database = " ";
$query = " ";
}
?>
<form action=<?php echo $PHP_SELF ?>?form=yes method="post">
<table>
<tr>
<td align="right"><b>Type in database name</b></td>
<td>
<input type=text name="database" value=<?php echo $database ?> >
</td>
</tr>
<tr>
<td align="right" valign="top"><b>Type in SQL query</b></td>
<td><textarea name="query" cols="60" rows="10"><?php echo $query ?></textarea>
</td>
</tr>
<tr>
<td colspan="2" align="center"><input type="submit" value="Submit
Query"></td>
</tr>
</table>
</form>
</body>
</html>this has always proved a big help in my dire times of need with SQL debugging
you saved my life with this thread, , thanks !
you can use this tested ok
echo '<a href="'."/yourlink.php?id=".$row['id'].'">'.$row['type'].'</a>';We're a friendly, industry-focused community of developers, IT pros, digital marketers, and technology enthusiasts meeting, networking, learning, and sharing knowledge.