FilipKrstic

Ohhhhhhhhhh............... "Unknown column 'msg' in 'field list'" .... To tired, I'm going to sleep. Thanks of lot...

Hello everybody, I have a problem selecting rows from db. Copied error message: I have searched for this problem, and I found tons of posts but I didn't do anything. Here is a code: <?php include("include/functions.php"); db_connect(); lang_en(); lang_smiles(); lang_links(); $sql= mysql_query("SELECT msg_id,msg from messages order by msg_id desc"); while($row=mysql_fetch_array($sql)) { $msg=$row['msg']; $msg_id=$row['msg_id']; ?> <li class="bar<?php echo $msg_id; ?>"> <div align="left"> <span ><?php echo $msg; ?> </span> <span class="delete_button"><a href="#" id="<?php echo $msg_id; ?>" class="delete_update">X</a></span> </div> </li> <?php } db_close(); ?> This line 223 is this: $sql= mysql_query("SELECT msg_id,msg from messages order by msg_id desc"); ... I have tried to replace msg_id,msg with * but on this way I don't have dump from table...

Same... Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/w0500004/public_html/razvoj/vtsa/index.php on line 521

Hello everybody, I`m coding something, and after hours and hours of coding, I`m goona in trapping. Here is code... <?php $query1 = mysql_query("select * from v2_ispitnirok order by id desc"); while($data1 = mysql_fetch_array($query1)) { echo "<table width='86%' border='1' align='center' cellspacing='0' bordercolor='#E9E5C0' style='border-collapse: collapse' colspan='2'> <tbody> <tr class='head'> <td bgcolor='#E9E5C0' style='text-align: left;'><strong>$data1[naslov]</strong></td> </tr> <tr class='head'><td width='170%' style='text-align: left;'>"; $ispitnirok = "$data1[id]"; $query2 = mysql_query("select * from v2_rezispita where $ispitnirok = `ispitnirok` and order by id desc"); Line 521 if($data2 = mysql_fetch_array($query2)) { echo" $data2[naslov], $data2[datum], $data2[rezultat]"; } echo"</td></tr> </tbody> </table><br />"; } ?> and line 521 is problem... Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/w0500004/public_html/razvoj/vtsa/index.php on line 521 What I try to do. I tried to list data from table v2_ispitnirok, use their id, and compare him in second query with "select * from v2_rezispita where $ispitnirok = `ispitnirok` and order by id desc", and list data from second table. but after that I have a problem with if($data2 = mysql_fetch_array($query2)) ??? Anyone to help?

Solved... tnx

I do it, but nothing... Now function likes on this: function zameni($rec) { $rec = str_replace(")","<img src='../images/smiles/icon_e_biggrin.gif'>",$rec); $rec = str_replace(":-D)","<img src='../images/smiles/icon_e_biggrin.gif'>",$rec); $rec = str_replace(":grin:)","<img src='../images/smiles/icon_e_biggrin.gif'>",$rec); $rec = str_replace(": )","<img src='../images/smiles/icon_e_smile.gif'>",$rec); $rec = str_replace(":-\)","<img src='../images/smiles/icon_e_smile.gif'>",$rec); $rec = str_replace(":smile:)","<img src='../images/smiles/icon_e_smile.gif'>",$rec); $rec = str_replace(":?","<img src='../images/smiles/icon_e_confused.gif'>",$rec); $rec = str_replace(" :-?","<img src='../images/smiles/icon_e_confused.gif'>",$rec); $rec = str_replace(":\(","<img src='../images/smiles/icon_e_sad.gif'>",$rec); $rec = str_replace(":-\(","<img src='../images/smiles/icon_e_sad.gif'>",$rec); $rec = str_replace(":sad:","<img src='../images/smiles/icon_e_sad.gif'>",$rec); $rec = str_replace("\","<img src='../images/smiles/icon_e_wink.gif'>",$rec); $rec = str_replace("\;-\)","<img src='../images/smiles/icon_e_wink.gif'>",$rec); $rec = str_replace(":wink:","<img src='../images/smiles/icon_e_wink.gif'>",$rec); $rec = str_replace("","<img src='../images/smiles/icon_lol.gif'>",$rec); $rec = str_replace(":mrgreen:","<img src='../images/smiles/icon_mrgreen.gif'>",$rec); $rec = str_replace("","<img src='../images/smiles/icon_razz.gif'>",$rec); $rec = str_replace(":-P","<img src='../images/smiles/icon_razz.gif'>",$rec); $rec = str_replace(":razz:","<img src='../images/smiles/icon_razz.gif'>",$rec); return $rec; }

Fix url? Structure is: functions.php is in public_html smiles are in public_html/images/smiles/ and wall.php file is in public_html/mcp/wall.php ...

I`m writing simple message wall... I have two php files. 1. functions.php where I put functions like connect_db(), check_login() etc... and i write this function for smiles and badwords (here I copy ereg_replace for smiles): function zameni($rec) { $rec = ereg_replace(")","<img src='../images/smiles/icon_e_biggrin.gif'>",$rec); $rec = ereg_replace(":-D)","<img src='../images/smiles/icon_e_biggrin.gif'>",$rec); $rec = ereg_replace(":grin:)","<img src='../images/smiles/icon_e_biggrin.gif'>",$rec); $rec = str_replace("","<img src='../images/smiles/icon_e_smile.gif'>",$rec); $rec = ereg_replace(":-)","<img src='../images/smiles/icon_e_smile.gif'>",$rec); $rec = ereg_replace(":smile:)","<img src='../images/smiles/icon_e_smile.gif'>",$rec); $rec = ereg_replace(":?","<img src='../images/smiles/icon_e_confused.gif'>",$rec); $rec = ereg_replace(" :-?","<img src='../images/smiles/icon_e_confused.gif'>",$rec); $rec = ereg_replace("","<img src='../images/smiles/icon_e_sad.gif'>",$rec); $rec = ereg_replace(":-(","<img src='../images/smiles/icon_e_sad.gif'>",$rec); $rec = ereg_replace(":sad:","<img src='../images/smiles/icon_e_sad.gif'>",$rec); $rec = ereg_replace("","<img src='../images/smiles/icon_e_wink.gif'>",$rec); $rec = ereg_replace(";-)","<img src='../images/smiles/icon_e_wink.gif'>",$rec); $rec = ereg_replace(":wink:","<img src='../images/smiles/icon_e_wink.gif'>",$rec); $rec = ereg_replace("","<img src='../images/smiles/icon_lol.gif'>",$rec); $rec = ereg_replace(":mrgreen:","<img src='../images/smiles/icon_mrgreen.gif'>",$rec); $rec = ereg_replace("","<img src='../images/smiles/icon_razz.gif'>",$rec); $rec = ereg_replace(":-P","<img src='../images/smiles/icon_razz.gif'>",$rec); $rec = ereg_replace(":razz:","<img src='../images/smiles/icon_razz.gif'>",$rec); return $rec; } and I have second php file named wall.php file starts <?php session_start(); include("../functions.php"); connect_db(); zameni($rec); but smiles codes don`t wants to replace with smiles... Can somebody help? What`s problem?

Never mind, i fixed it up!

Hi.. This is what's bodering me.. I have some data in mysql database, and when i'm reading from database and, while printing, use nl2br function it works fine.. And I created inline ajax editor to edit that data. When i click "Edit" and 'throw' data through xmlhttp req, my php script get data, update database, and echo some html so the js script could update html.. And there is my problem.. My php script echo nl2br($text); but it seems like nl2br doesnt have any effect to $text. It outputs with '\n' inside text.. How to fix this?

heh... Solved with classic if, again... <?php if(isset($_SESSION['mcp'])) { print"Wlc Mod"; } if(isset($_SESSION['username'])) { print"Wlc Adm"; } ?>

Here we are... Session username is session for admin and mcp is session for moderator_control_panel. And I want make that the admin can come in mcp with his login (with his session). That is the reason why I put OR. Informations about $_SESSIONS['username'] comes from mysql, table admins, and $_SESSIONS['mcp'] from mysql too, table mods... Code for mcp page... <?php session_start(); require_once("../config.php"); // Check his status. if (!empty($_SESSION[mcp]) or ($_SESSION[username])) { // MOD AREA } else // bad info { echo "Need to login first"; } ?> Login page... <?php session_start(); // Check if he wants to login: if (!empty($_POST[mcp])) { require_once("../config.php"); // Check if he has the right info. $query = mysql_query("SELECT * FROM mods WHERE username = '$_POST[mcp]' AND password = '$_POST[password]'") or die ("<Wrong username or password"); $row = mysql_fetch_array($query) or die ("Wrong username or password"); if (!empty($row[username])) // he got it. { $_SESSION[mcp] = $row[username]; echo "Success"; exit(); } else // bad info. { echo "Please login"; exit(); } } ?>

When I try this peace of code, and when I login as $_SESSION[username], server shows me: "Need to login first", missing _$SESSION[bla] OR _$SESSION_[blabla]... Maybe: <?php session_start(); require_once("../config.php"); // Check his status. if (isset($_SESSION[mcp]) or isset($_SESSION[username])) { ?> ? What is ||?

Huh, guys many helpful informations.. Amazing... I have one more question, and I can start with writing... I have two sessions... Session $_SESSION[username] and $_SESSION[mcp]. At the start I have classic if... session_start(); require_once("../config.php"); // Check his status. if (!empty($_SESSION[mcp]) or ($_SESSION[username])) { // There is place where I need to put loggin function... } else // bad info { echo "Need to login first"; } And I don`t how to make fuction logged? "if user logged as $_SESSION[username] print Welcome $_SESSION[username] or if logged as $_SESSION[mcp] print Welcome $_SESSION[mcp]" ??? P.S I`m php starter...

Thanks... That`s enough...