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Newbie Question. how to put url into echo $row
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Join Date: Jan 2005
Posts: 6
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Solved Threads: 0
Hi there.
I'm very very new to php. :o so i hope i'm posting in the right forum.
and i really can't figure this out. probably really simple but i need to place a url into:
while ($row = mysql_fetch_array($result)) {
echo $row['login'];
echo " " . $row['age'] . "<br>";
everytime i place <a href= etc i get errors.
I need the login to link to the user's profile page.
Please could anybody help??
Kind Regards
Ross
I'm very very new to php. :o so i hope i'm posting in the right forum.
and i really can't figure this out. probably really simple but i need to place a url into:
while ($row = mysql_fetch_array($result)) {
echo $row['login'];
echo " " . $row['age'] . "<br>";
everytime i place <a href= etc i get errors.
I need the login to link to the user's profile page.
Please could anybody help??
Kind Regards
Ross
Suppose you have the code:
Everything within the quotes is printed out. However, now take the following code:
PHP gets confused where you have quotes within quotes. Therefore, if you want to print out a real quote inside an echo statement, you have to escape the " character with a backlash. Like this ...
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echo "$value";
PHP Syntax (Toggle Plain Text)
echo "<a href="$value">";
PHP Syntax (Toggle Plain Text)
echo "<a href=\"$value\">";
Practice makes perfect.
I'm not exactly sure what $m[id] is ? Where are you fetching that $m[] array from?
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echo "<a href=\"member.php?mid=$row[mid]\">$row[login]</a>";
I'm not exactly sure what $m[id] is ? Where are you fetching that $m[] array from?
generally when u pull it from mysql you would reference the item you want by column number, so....
if your query was something to this effect:
if your query was something to this effect:
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$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>
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Join Date: Jan 2005
Posts: 6
Reputation:
Solved Threads: 0
Hello. thanks for all your help, both of you,
what i currently have is:
which does give me the linked login name that i was after
but the link doesnt point to the users profile
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
what i currently have is:
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<?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
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:
this has always proved a big help in my dire times of need with SQL debugging
can you login via ssh/telnet and double check to see if that works?
if not here is a PHP interface:
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<!-- 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
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so it would be something like: