doublea2k7
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Everything posted by doublea2k7
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that fix it. ty both of u =)
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thanks for the reply PaulRyan. I've tried your code but all the rows have the same class even the last one.
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thers a syntax error here is a snipet of my code. Its something wrong with the endwhilte but i dont know wat is it. <?php $row2 = mysql_fetch_array($result); while($row = mysql_fetch_array($result)){ ?> <tr id="<?php echo $row['statusid']?>" class="status"> <td><?php echo $row['date']?></td> <td id="<?php echo$row['statusid']?>" style="text-align:left;" class="editStatus"> <?php echo htmlspecialchars($row['statusTxt'], ENT_QUOTES)?></td> <td style="width:1%;"> <form action=""> <input type="image" src="img/cancel.png" onclick="makeVariable('<?php echo $row['statusid']?>')" class="deleteRowButton"/> </form> </td> </tr> <?php $row2 = $row;?> <?php } end while?> <tr id="<?php echo $row['statusid']?>" class="statusLast"> <td><?php echo $row['date']?></td> <td id="<?php echo$row['statusid']?>" style="text-align:left;" class="editStatus"> <?php echo htmlspecialchars($row['statusTxt'], ENT_QUOTES)?></td> <td style="width:1%;"> <form action=""> <input type="image" src="img/cancel.png" onclick="makeVariable('<?php echo $row['statusid']?>')" class="deleteRowButton"/> </form> </td> </tr> <?php } ?>
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Im not sure where to post this but since it includes php il post it here instead of in the mysql forum. ok so, i have a table and i get the values using while($row = mysql_fetch_array($result)){ and then echo them in rows. that works fine but i need to add a class to the last row of my table. I would need somehow to fetch the last row of the array and make it echo something different. Any help is appreciated Thank you
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Current Mysql Version I have a table that holds user account info. I have another table that holds status updates according to their account "id" in the clients table. Example |Table clients| id Name Last Name 1 Fred John 2 Frank Joe |Table Status| id status clientId 0 In Progress 1 1 Still In Progress 1 I want to be able to select all accounts with no entries in the status table. Here is what I have but its not working. SELECT id, name FROM clients as u INNER JOIN ( SELECT clientId, COUNT(*) AS statusCount FROM status HAVING statusCount > '0' ) AS statusSubQuery ON (statusSubQuery.clientId = u.id) ";
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I have a table that holds user account info. I have another table that holds status updates according to their account "id" in the account table. Example |Table account| id Name Last Name 1 Fred John 2 Frank Joe |Table Status| id status accountid 0 In Progress 1 How can would i select all the account ids that don't have any entries in the status table. So for the example above how would i select the account Frank Joe with id 2 as i has no status linked to it.
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i have a table called accounts inside there is ID NAMe password now users dont know their id so they would have to enter their name. So this code [code]if ($query=mysql_query("update accounts set loggedin='1' WHERE name='$name'")) { echo "Your account has been fix try to log in the game now!"; but this works because it finds the id which is the primayr index ($query=mysql_query("update accounts set loggedin='1' WHERE id='$id'")) { echo "Your account has been fix try to log in the game now!";[/code] but the user doesn know their id how could i get the name and then get the id so the update will work.
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echo not displaying and updating database not working
doublea2k7 replied to doublea2k7's topic in PHP Coding Help
they r on cause if a remove a bracket i get a syntax error ? what could be the problem -
echo not displaying and updating database not working
doublea2k7 replied to doublea2k7's topic in PHP Coding Help
i try $query = "update accounts set loggedin='2' where name=$name"; echo $query; mysql_query($query) or die(mysql_error()); i get the same thing just a blank page like i did before -
why isn't my echo showing and my database updating <? include "config2.php"; ?> <!doctype html public "-//w3c//dtd html 3.2//en"> <html> <head> </head> <body> <? $name=mysql_real_escape_string($name); if ($rec=mysql_fetch_array(mysql_query("SELECT name FROM accounts WHERE name='$name'"))) { if (($rec['name']==$name)) { if(mysql_query("update accounts set loggedin='2' WHERE name='$name'")){ echo "<font face='Verdana' size='12' ><center>OK <br> Your account has been fix try to log in the game now!</font></center>"; } } } ?> </body> </html>
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o ty but now i have another problem my message in echo does not display as well as the database field loggedin does not change to 1
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can anyone tell me whats wrong with this script i get this error Parse error: syntax error, unexpected $end in /home/urzone/public_html/smashlegacies.exofire.net/loginfix_check.php on line 28 <? include "config2.php"; ?> <!doctype html public "-//w3c//dtd html 3.2//en"> <html> <head> </head> <body bgcolor="#ffffff" text="#000000" link="#0000ff" vlink="#800080" alink="#ff0000"> <? $name=mysql_real_escape_string($name); if($rec=mysql_fetch_array(mysql_query("SELECT * FROM accounts WHERE name='$name'"))){ if(($rec['name']==$name)){ if(mysql_query("update accounts set loggedin='1' WHERE name='$name'")){ echo "<font face='Verdana' size='2' ><center>OK <br> Your account has been fix try to log in the game now!</font></center>";}} ?> </body> </html> im really new at php coding so please help
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hi im new to the forum and i need help how would i check if a user name doesn't exist i know how to check if user name exits if(mysql_num_rows(mysql_query("SELECT userid FROM plus_signup WHERE userid = '$userid'") but i don't know how to check if it doesn't exits. plz help