Alicia Posted November 20, 2009 Share Posted November 20, 2009 Hi guys, May I know whats wrong with this php mysql query : $checknewinvt = mysql_query("SELECT * FROM `f_chat_mask_users` WHERE `chat_mask_user_user_id`='{$_COOKIE['memberID']}' AND `mobilestat` != '1'") or die(mysql_error()); Error displayed : 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 '' at line 1 When I change the $_COOKIE['memberID'] to any numbers directly, it loads fine back. Please advise, thanks. Link to comment https://forums.phpfreaks.com/topic/182331-whats-prob/ Share on other sites More sharing options...
Maq Posted November 20, 2009 Share Posted November 20, 2009 The string syntax looks valid. Echo out the cookie value to see what it actually is. Also if the member_id is an integer than you don't need single quotes around it, but that shouldn't cause an error. Link to comment https://forums.phpfreaks.com/topic/182331-whats-prob/#findComment-962155 Share on other sites More sharing options...
Alicia Posted November 20, 2009 Author Share Posted November 20, 2009 I did echo out the cookie, it is only normal integer ID. When I replace the cookie with the ID, it works fine. I am wondering what is the prob Link to comment https://forums.phpfreaks.com/topic/182331-whats-prob/#findComment-962159 Share on other sites More sharing options...
Mchl Posted November 20, 2009 Share Posted November 20, 2009 Change your code to this $memberID = mysql_real_escape_string($_COOKIE['memberID']); $query = "SELECT * FROM `f_chat_mask_users` WHERE `chat_mask_user_user_id`='$memberID' AND `mobilestat` != '1'"; $checknewinvt = mysql_query($query) or trigger_error(mysql_error().": $query",E_USER_WARNING); Link to comment https://forums.phpfreaks.com/topic/182331-whats-prob/#findComment-962163 Share on other sites More sharing options...
PFMaBiSmAd Posted November 20, 2009 Share Posted November 20, 2009 I did echo out the cookie, it is only normal integer ID. I'll bet that the actual value in the cookie also includes single-quotes as part of the value. Link to comment https://forums.phpfreaks.com/topic/182331-whats-prob/#findComment-962165 Share on other sites More sharing options...
Mchl Posted November 20, 2009 Share Posted November 20, 2009 Quote possible, but I'd expect slightly different MySQL error message Link to comment https://forums.phpfreaks.com/topic/182331-whats-prob/#findComment-962167 Share on other sites More sharing options...
Alicia Posted November 20, 2009 Author Share Posted November 20, 2009 I changed the code to $memberID = mysql_real_escape_string($_COOKIE['memberID']); $query = "SELECT * FROM `f_chat_mask_users` WHERE `chat_mask_user_user_id`='$memberID' AND `mobilestat` != '1'"; $checknewinvt = mysql_query($query) or trigger_error(mysql_error().": $query",E_USER_WARNING); Still the same error. I echoed the cookie and there is no single code displayed. please advise.. Link to comment https://forums.phpfreaks.com/topic/182331-whats-prob/#findComment-962169 Share on other sites More sharing options...
Maq Posted November 20, 2009 Share Posted November 20, 2009 Do this and paste the exact output: $query = "SELECT * FROM `f_chat_mask_users` WHERE `chat_mask_user_user_id`='$memberID' AND `mobilestat` != '1'"; echo $query; Link to comment https://forums.phpfreaks.com/topic/182331-whats-prob/#findComment-962171 Share on other sites More sharing options...
Mchl Posted November 20, 2009 Share Posted November 20, 2009 Still the same error. That's impossible. You should have full query text after the error message, and it would be really helpful if you pasted it in here. Link to comment https://forums.phpfreaks.com/topic/182331-whats-prob/#findComment-962173 Share on other sites More sharing options...
PFMaBiSmAd Posted November 20, 2009 Share Posted November 20, 2009 The query that is failing is NOT the one you have been posting or modifying. Link to comment https://forums.phpfreaks.com/topic/182331-whats-prob/#findComment-962175 Share on other sites More sharing options...
Alicia Posted November 20, 2009 Author Share Posted November 20, 2009 This is the output : SELECT * FROM `f_chat_mask_users` WHERE `chat_mask_user_user_id`='1749' AND `mobilestat` != '1' 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 '' at line 1 The query that is failing is NOT the one you have been posting or modifying. you mean I used the wrong code ? please advise. Thanks Link to comment https://forums.phpfreaks.com/topic/182331-whats-prob/#findComment-962176 Share on other sites More sharing options...
PFMaBiSmAd Posted November 20, 2009 Share Posted November 20, 2009 Perhaps if you posted all the code in the file. xxxxx out any sensitive information. Link to comment https://forums.phpfreaks.com/topic/182331-whats-prob/#findComment-962181 Share on other sites More sharing options...
Mchl Posted November 20, 2009 Share Posted November 20, 2009 We have kind of twilight zone situation here Ok... what happens when you run this query? SELECT * FROM `f_chat_mask_users` WHERE `chat_mask_user_user_id`='1749' Link to comment https://forums.phpfreaks.com/topic/182331-whats-prob/#findComment-962182 Share on other sites More sharing options...
Alicia Posted November 20, 2009 Author Share Posted November 20, 2009 Still the same error when I direct put the number. Here is the full script : <? session_start(); include"includes/mydb.php"; ?> <? $srrt1 = mysql_query("SELECT * FROM `f_chat_users` WHERE `chat_user_id`= '{$_COOKIE['memberID']}'") or die(mysql_error()); $srrt2 = mysql_fetch_array($srrt1); if($srrt2['chat_user_status'] == '1') { $stat = "Online"; } else { $stat = "Offline"; } ?> <!DOCTYPE html PUBLIC "-//WAPFORUM//DTD XHTML Mobile 1.2//EN" "http://www.openmobilealliance.org/tech/DTD/xhtml-mobile12.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="application/xhtml+xml; charset=utf-8" /> <meta http-equiv="refresh" content="3;url=im_user.php"> <title>Test</title> </head> <style type="text/css"> <!-- .style2 {font-size: 12px} --> </style> <style type="text/css"> <!-- div.pagination { padding: 3px; margin: 3px; } div.pagination a { padding: 2px 5px 2px 5px; margin: 2px; border: 1px solid #AAAADD; text-decoration: none; /* no underline */ color: #000099; } div.pagination a:hover, div.pagination a:active { border: 1px solid #000099; color: #000; } div.pagination span.current { padding: 2px 5px 2px 5px; margin: 2px; border: 1px solid #000099; font-weight: bold; background-color: #000099; color: #FFF; } div.pagination span.disabled { padding: 2px 5px 2px 5px; margin: 2px; border: 1px solid #EEE; color: #DDD; } refre { font-size: 0.5em; } --> </style> <body> <div class="mainwrapper"> <div id="content"> <div class="min-width"> <table width="260" border="0" cellspacing="5" cellpadding="0"> <tr> <td> <span class="text2"><span class="style2">This page will refresh every : 3seconds<br /> ReadInvitation :<br /> <? $memberID = mysql_real_escape_string($_COOKIE['memberID']); //$memberID = '1749'; //$query = "SELECT * FROM `f_chat_mask_users` WHERE `chat_mask_user_user_id`='1749' AND `mobilestat` != '1'"; $query = "SELECT * FROM `f_chat_mask_users` WHERE `chat_mask_user_user_id`='$memberID' AND `mobilestat` != '1'"; echo $query; $checknewinvt = mysql_query($query) or trigger_error(mysql_error().": $query",E_USER_WARNING); //$checknewinvt = mysql_query("SELECT * FROM `f_chat_mask_users` WHERE `chat_mask_user_user_id`={$_COOKIE['memberID']} AND `mobilestat` != '1'") or die(mysql_error()); $checknewinvt2 = mysql_num_rows($checknewinvt); if($checknewinvt2 > 0) { ?> <table border="0" cellspacing="0" cellpadding="7"> <tr> <td width="37" valign="top"><img src="images/new.gif" /></td> <td width="227" valign="top"> <? while($fetchcheck = mysql_fetch_array($checknewinvt)) { $userreq = mysql_query("SELECT DISTINCT `chat_mask_user_user_id` FROM f_chat_mask_users WHERE chat_mask_user_mask_id='{$fetchcheck['chat_mask_user_mask_id']}' AND chat_mask_user_user_id != {$row['ID']}") or die(mysql_error()); $userreq2 = mysql_fetch_array($userreq); $sendName = mysql_query("SELECT * FROM Profiles WHERE ID='{$userreq2['chat_mask_user_user_id']}'"); $sendName2 = mysql_fetch_array($sendName); echo "{$sendName2['NickName']} -<a href=\"imR.php?pid={$fetchcheck['chat_mask_user_mask_id']}\" target=\"_blank\">ReadNow</a><br>"; } // close while loop ?> </td> </tr> </table> <? } else { echo 'XXX Message'; } ?> </body> </html> Link to comment https://forums.phpfreaks.com/topic/182331-whats-prob/#findComment-962188 Share on other sites More sharing options...
PFMaBiSmAd Posted November 20, 2009 Share Posted November 20, 2009 The query that is failing is the one later on the page containing the following term because there is no code setting $row - chat_mask_user_user_id != {$row['ID']} The reason for the mismatch in information being outout is because you are echoing $query then getting the mysql_error() output from a following die() instruction. As previously written (long long ago...), what trigger_error() does is dependent on error_reporting/display_errors/log_errors settings and should not be used in a debugging environment unless accompanied by instructions that would cause its output to become visible. Since we don't really know what you intended by the chat_mask_user_user_id != {$row['ID']} term, it will be a little hard to help with what it should be. Link to comment https://forums.phpfreaks.com/topic/182331-whats-prob/#findComment-962213 Share on other sites More sharing options...
Mchl Posted November 20, 2009 Share Posted November 20, 2009 As previously written (long long ago...), what trigger_error() does is dependent on error_reporting/display_errors/log_errors settings and should not be used in a debugging environment unless accompanied by instructions that would cause its output to become visible. Link to comment https://forums.phpfreaks.com/topic/182331-whats-prob/#findComment-962215 Share on other sites More sharing options...
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