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iRoot121

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  1. Hi guys, I wanted to make a script login to the website Marktplaats.nl, but it doesn't seem to work.. At the moment I can't post the code as it's on my laptop. Does anyone know how to login and maybe do other stuff on Marktplaats.nl with cURL? Kind regards, Kevin Ruhl.
  2. Hi, I'm trying to make a program that filters data out of a website. Now I've got a problem. For example, I've this line: <td class="maintxt" width="200"><a href="profile.php?x=Stranger">Stranger</a></td><td class="maintxt" width="200">Godfather (96.76%)</td><td class="maintxt" width="200"><i><a href="clanprofile.php?x=Ettertjes">Ettertjes</a></i></td> Form this line I want to set the word "Stranger" in a variable, and do the same for the word "Godfather". What is the best method to do this? Thanks in advance, iRoot121
  3. Hi, I'm trying to let JavaScript check if a givin user exist in the database. It seems that the _check-user.php always returns 0, but if I fill in a name that doesn't exist, and echo out the result variable in JS, the echo will return 1. Is there someone who could help me? JavaScript part: function checkUser() { $.post("_check_user.php", { 'username': document.getElementById("username").value }, function(result) { if (result == 1) { document.getElementById("checkUser").className = "succes"; document.getElementById("checkUser").innerHTML = "Name is available"; }else{ document.getElementById("checkUser").className = "errormsg"; document.getElementById("checkUser").innerHTML = "Name is not available!"; } }); } _check-user.php: <?php include("config.php"); $result = mysql_query("SELECT login FROM users WHERE login='".clean_string($_POST["username"])."'"); if(mysql_num_rows($result)>0){ echo 0; }else{ echo 1; } ?>
  4. I've already got the solution: after the UPDATE query there was another query that said: update the cash of the user with the cash + an amount of cash. But the second query had the old cash value, not the updated one. Thanks for helping anyway guys!
  5. Yea, that could be possible. Here is the code if you want to take a look at it: http://pastebin.com/JDWZuf2b The update query is on line 431
  6. Do you mean the whole page? If so, do you want to have a pastebin, or should I just paste it here? EDIT: Here is the code for the error handling: $q = "UPDATE users SET cash='".clean_string($info["cash"]-$_POST["bet"])."' WHERE login='".$info["login"]."'"; echo "Query is <br>$q"; $qresults = MySQL_query($q); if (!$qresults) { echo "Error in query - msg is<br>" . MySQL_error(); exit(); } $q2 = "UPDATE objecten SET bank='".clean_string($object["bank"]+$_POST["bet"])."' WHERE type='4' AND land='".$info["land"]."'"; echo "Query is <br>$q2"; $qresults2 = MySQL_query($q2); if (!$qresults2) { echo "Error in query - msg is<br>" . MySQL_error(); exit(); }
  7. If I use that for both queries, I get: Query is UPDATE users SET cash='13000' WHERE login='Goed' Query is UPDATE objecten SET bank='9223372036854774807' WHERE type='4' AND land='1' as output, no errors or whatever.
  8. If I echo the var_dumps like this: echo '$info["login"]: '.var_dump($info["login"]); echo '$_POST["bet"]: '.var_dump($_POST["bet"]); echo '$info["cash"]: '.var_dump($info["cash"]); echo '$info["land"]: '.var_dump($info["land"]); I get string(4) "Goed" $info["login"]: string(4) "1000" $_POST["bet"]: string(5) "14000" $info["cash"]: string(1) "1" $info["land"]: as output. Do you need any more information?
  9. Both of the queries aren't working. And if I do: echo "UPDATE users SET cash='".clean_string($info["cash"]-$_POST["bet"])."' WHERE login='".$info["login"]."'"; echo "UPDATE objecten SET bank='".clean_string($object["bank"]+$_POST["bet"])."' WHERE type='4' AND land='".$info["land"]."'"; I get UPDATE users SET cash='12000' WHERE login='Goed' AND UPDATE objecten SET bank='775807' WHERE type='4' AND land='1' as output, so the variables are set. By the way, if I throw these outputs in PHPMyAdmin, the query works perfectly.
  10. Hi, When I run an UPDATE query in my PHP code, nothing happends, no errors or anything. The weardest thing is, is that if I echo the query, it's the correct output. Any other queries that come after them are still executed. This is my query code: mysql_query("UPDATE users SET cash='".clean_string($info["cash"]-$_POST["bet"])."' WHERE login='".$info["login"]."'") or die(mysql_error()); mysql_query("UPDATE objecten SET bank='".clean_string($object["bank"]+$_POST["bet"])."' WHERE type='4' AND land='".$info["land"]."'") or die(mysql_error()); Thanks in advance, iRoot121.
  11. Hi guys, I was wondering, is it possible to split an array into pages with 10 values on each page? And how can I make a navigation bar with Page 1 2 3 in it? And if it's possible, how would I do that? Thanks in advance, iRoot121
  12. Hi guys, I had a question about how to do a safe query. For example, I've a query like SELECT * FROM users WHERE login='".$_GET["x"]."' Do I need to do more then just adding mysql_real_escape_string(), or is that one just enough? And how goes it for the INSERT, UPDATE, DELETE statement? And do you need to parse the output from a database before displaying it? Thanks in advance,
  13. Thanks for the info, I'll change the MySQL values .
  14. I asume you want to have some sort of userlist, with a link to each user's profile. So, if this is the case, you can do something like this: Members.php: <?php $sql = mysql_query("SELECT * FROM `users` ORDER BY `id`"); if ($sql) { while ($info = mysql_fetch_array($sql)) { //Some userinfo here. //The link to the user his profile. echo '<a href="profile.php?user='.$info["id"].'">More info about this user.</a>'; } }else{ die(mysql_error()); } ?> Profile.php: if (!isset($_GET['user']) || $_GET['user'] == "") { echo 'Please insert a user to lookup..'; exit; }else{ $sql = mysql_query("SELECT * FROM users WHERE id='".$_GET['user']."'"); if (mysql_num_rows($sql)>0) { while ($info= mysql_fetch_array($sql)) { //User information to display. } }else{ echo 'This user isn\'t in our database!'; exit; } }
  15. I don't think there is one. Except that the table needs to be sorted again after a cache clean.
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