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PhpxZ

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Everything posted by PhpxZ

  1. I will give you example there is a textdraw with 4 rows. after you type 8 characters at the first row it's automatic go to the next row (row 2) after you type 8 characters at the second row it's automatic go to the next row (row 3) and.. more..
  2. and how can i do it with javascript?
  3. How to Limit number of character in each line in textarea Let say script should allow maximum 4 rows and should allow 8 character in each line. Also If character reached the limit then it should move to next line. its will look like this: 12345678 90123456 78901234 this is my code what i need to add <textarea name="text" cols="76" rows="15" lang="lt"></textarea>
  4. How to Limit number of character in each line in textarea Let say script should allow maximum 4 rows and should allow 8 character in each line. Also If character reached the limit then it should move to next line. its will look like this: 12345678 90123456 78901234 this is my code what i need to add <textarea name="text" cols="76" rows="15" lang="lt"></textarea>
  5. So what is yes? and yes its connected to the databate
  6. No nothing.. at my phpmyadmin the "Skin" row is smallint(3) UNSIGNED Null - No. Default 0
  7. mysql_query("INSERT INTO players (Skin) VALUES('233' ) ") ; It's doesn't change the value... to 233
  8. I did something else.. Thanks you all
  9. Parse error: syntax error, unexpected '<' in The line after the if
  10. Here is the full code while ($row = mysql_fetch_array($result)) { if ($row['reason'] == NULL) { $count = true; echo '<div style="margin:5px;padding:15px;'; if ($bright) { echo 'background-color:#f4f3f0;border:1px solid #a0a0a0'; $bright = false; } else { echo 'background-color:#e5ecf9;border:1px solid #9da9bf'; $bright = true; } $accid = $row['accid']; $result2 = mysql_query('SELECT * FROM accounts WHERE id='. $accid .''); $row2 = mysql_fetch_array($result2); $accNAME = $row2['name']; if($row['submited'] != NULL){ $submited = $row['submited']; } else { $submited = "informacijos n?ra"; } $metai = date('Y') - $row['age']; echo '"> <p align="right">Date: '.$submited.'</p> <b>Character Name:</b> '. $row['charname'] .'<br /> if($row['sex'] == 1) { <b>Sex:</b> Male<br /> } else{ <b>Sex:</b> Female<br /> } <b>Age:</b> '. $row['age'] .' (' .$metai. ')<br /><br /> I get the error for the line if($row['sex'] == 1) {
  11. if($row['sex'] == 1){ but i get this error: Parse error: syntax error, unexpected T_STRING, expecting ',' or ';' in what is whe problem?
  12. this is my line if (mysql_query('INSERT INTO `pendingchars` (`accid`,`charname`,`sex`,`age`,`pgmg`,`dmrktt`,`history`,`carcrashrp`,`carscrash`) VALUES ('. $_SESSION['Logged'] .',\''. $_POST['charname'] .'\','. $_POST['sex'] .'\','. $_POST['age'] .',\''. $_POST['pgmg'] .'\',\''. $_POST['dmrktt']. '\',\''. $_POST['history'] .'\',\''. $_POST['carcrashrp'] .'\',\''. $_POST['carscrash'] .'\')', $con)) { i get a mysql error I dont know why.. i just added the "sex" in it.. it's something with that. I have all the rows and lines in the Mysql
  13. this is my line if (mysql_query('INSERT INTO `pendingchars` (`accid`,`charname`,`sex`,`age`,`pgmg`,`dmrktt`,`history`,`carcrashrp`,`carscrash`) VALUES ('. $_SESSION['Logged'] .',\''. $_POST['charname'] .'\','. $_POST['sex'] .'\','. $_POST['age'] .',\''. $_POST['pgmg'] .'\',\''. $_POST['dmrktt']. '\',\''. $_POST['history'] .'\',\''. $_POST['carcrashrp'] .'\',\''. $_POST['carscrash'] .'\')', $con)) { i get a mysql error I dont know why.. i just added the "sex" in it.. it's something with that. I have all the rows and lines in the Mysql
  14. But this need to be in if.. couse i need to use it after
  15. The line: if (mysql_query('INSERT INTO `chars` (`accid`,`charname`,`sex`,`age`,`pgmg`,`dmrktt`,`history`,`carcrashrp`,`carscrash`) VALUES ('. $_SESSION['Logged'] .',\''. $_POST['charname'] .',\''.$_POST['sex'] .',\''. $_POST['age'] .',\''. $_POST['pgmg'] .'\',\''. $_POST['dmrktt']. '\',\''. $_POST['history'] .'\',\''. $_POST['carcrashrp'] .'\',\''. $_POST['carscrash'] .'\')', $con)) I have the rows in the mysql.. but i still get error.. the error is: Error! You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near
  16. I have a problem with the if in my function the if in switch function IsValidSkin($skinid) { switch ($skinid) { case 3; case 4; case 5; case 6; case 8; case 17; case 42; case 50; case 61; case 65; case 71; case 74; case 86; case 92; case 268; case 99; case 119; case 149; if($row['Sex'] == "1") echo' case14'; return false; break; } return true; } The if isn't work. what is the problem?
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