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PHP - Please Help! I Go Live Tomorrow! (mysql/php Unknown 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 '["rid"]., SELECT refid1 FROM oto_members WHERE id='13', SELECT refid2 FR' at line 5
With this script area: $affiliate1=('.$_REQUEST["rid"].'); $affiliate2=("SELECT refid1 FROM oto_members WHERE id='".$_REQUEST["rid"]."'"); $affiliate3=("SELECT refid2 FROM oto_members WHERE id='".$_REQUEST["rid"]."'"); $affiliate4=("SELECT refid3 FROM oto_members WHERE id='".$_REQUEST["rid"]."'"); $affiliate5=("SELECT refid4 FROM oto_members WHERE id='".$_REQUEST["rid"]."'"); $affiliate6=("SELECT refid5 FROM oto_members WHERE id='".$_REQUEST["rid"]."'"); $affiliate7=("SELECT refid6 FROM oto_members WHERE id='".$_REQUEST["rid"]."'"); $affiliat8e=("SELECT refid7 FROM oto_members WHERE id='".$_REQUEST["rid"]."'"); $affiliate9=("SELECT refid8 FROM oto_members WHERE id='".$_REQUEST["rid"]."'"); $affiliate10=("SELECT refid9 FROM oto_members WHERE id='".$_REQUEST["rid"]."'"); $qry="INSERT INTO ".$prefix."members(firstname,lastname,email,address,city,state,postcode,country,telephone,username,password,refid1,refid2,refid3,refid4,refid5,refid6,refid7,refid8,refid9,refid10,geo,paypal_email,joindate,mtype,groupid,cb_id,status,signupip) VALUES('".$_REQUEST["firstname"]."','".$_REQUEST["lastname"]."','".$_REQUEST["email"]."','".$_REQUEST["address"]."','".$_REQUEST["city"]."','".$_REQUEST["state"]."','".$_REQUEST["postcode"]."','".$_REQUEST["country"]."','".$_REQUEST["telephone"]."','".$_REQUEST["username"]."','".md5($_REQUEST["password"])."', $affiliate1, $affiliate2, $affiliate3, $affiliate4, $affiliate5, $affiliate6, $affiliate7, $affiliate8, $affiliate9, $affiliate10 ,'".$_REQUEST["geo"]."','".$_REQUEST["paypal_email"]."',NOW(),".$_REQUEST["mtid"].",$eogroup,'".$_REQUEST["cb_id"]."','$memberstatus','$signupip')"; Similar TutorialsHey all, I keep getting this 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 '' at line 2 When I use the script below. I'm finding it a bit confusing because everything about it continues to work, it's just it gives me an error. When the script runs, the outcome is "Success! Your dog now looks far more energetic! Looks like this food is all used up." Followed by the error, which cuts the remainder of the page off. Does anybody know why it's doing this? $dogyay = $_POST['dogid']; $checkenergy = "SELECT energy FROM dogs WHERE id=$dogyay"; $energylevel = mysql_query($checkenergy) or die(mysql_error()); $row = mysql_fetch_array($energylevel) or die(mysql_error()); if($row['energy'] >= 100) { echo "<b>Oops!</b> Looks like your dog is full right now...";} else{ echo "<b>Success! Your dog now looks far more energetic!</b><br><br>"; $sql11="UPDATE dogs SET energy=energy + 50 WHERE id=$dogyay"; $result11=mysql_query($sql11); $sql12="UPDATE items SET uses = uses - 1 WHERE itemid=$id"; $result12=mysql_query($sql12); $checkuses = "SELECT uses FROM items WHERE itemid=$id"; $useslevel = mysql_query($checkuses) or die(mysql_error()); $row = mysql_fetch_array($useslevel) or die(mysql_error()); if($row['uses'] == 0) { echo "Looks like this food is all used up.<bR><br>"; mysql_query("DELETE FROM items WHERE itemid='$id'") or die(mysql_error());} Thanks ! I'm a newbie, can you please help? I'm getting a mysql error "Unknown column 'E0000001' in 'where clause'" The id is in the URL: ...user-profile.php?id=E0000001 My Query $query = mysqli_query($con, "SELECT * FROM UserList WHERE UserID=".$id) or die (mysqli_error($con)); Thank you Edited September 18, 2019 by ACBMSEHey, So what im trying to do is put my database variables into a session array. So this is what im trying to accomplish... $_SESSION['Name_of_Row'] = $value This is the script I wrote: Code: [Select] Function setupSession(){ session_start(); $query = "SELECT * FROM users WHERE u_id ='{$this->u_id}'"; $result = mysql_query($query); $row = mysql_fetch_array($result); foreach($row as $key => $value){ if(!empty($value)){ $_SESSION[$key] = $value; } } } When that runs I get the following warning. Can anyone tell me what this means and how to fix it? Error: Notice: Unknown: Skipping numeric key 0 in Unknown on line 0 Hello! I have a strange error on my PHP script and i dont how to fix it. If someone can help me, please help me then! Here is my error: Code: [Select] logout(); } else { $iq = mysql_query("SELECT * FROM users WHERE username='{$signin_username}' AND password='{$signin_password}' AND suspended='0' LIMIT 1;"); $ir = mysql_fetch_array($iq); $_SESSION['me'] = $ir; } } } } else { die("The configuration did not recieve appropriate variables to accept your request."); } if ($set['next_clearup'] < time ()) { $next_clearup = time () + 60 * 60 * 24; mysql_query ('' . 'UPDATE settings SET set_value=\'' . $next_clearup . '\' WHERE set_name=\'next_clearup\' LIMIT 1;'); mysql_query ('UPDATE users SET ads_clicked=\'\' WHERE ads_clicked!=\'\''); } } ?> Warning: include(THDIRindex.php) [function.include]: failed to open stream: No such file or directory in C:\xampp\htdocs\Upload\index.php on line 16 Warning: include() [function.include]: Failed opening 'THDIRindex.php' for inclusion (include_path='.;\xampp\php\PEAR') in C:\xampp\htdocs\Upload\index.php on line 16 And here is the PHP file the error is in: Code: [Select] <?php session_start(); include_once('lib/lib.php'); include_once('lib/configuration.php'); $ddir = THDIR.$do->get_file_url(); include($ddir); if(file_exists(HEADER)) { include_once(HEADER); } if($contents) { print $contents; } if(file_exists(FOOTER)) { include_once(FOOTER); } ?> Help ASAP if you can! I Am getting along with php better than I was previously. But this 68 year old brain still refuses to learn very fast! Here is the error I'm receiving when I'm trying to open the db: Parse error: syntax error, unexpected T_VARIABLE in /home/taft65/public_html/memProtest.php on line 197 <?php error_reporting(E_ERROR | E_PARSE | E_CORE_ERROR); $host = "localhost"; $dbname="database;" Failing ------>$username = "user"; $password="drDedf#hj"; I understand you do not need to declare varibles in PHP, Correct? I checked the db to ensure that I'm calling the correct value. NuSpherePhpEd to validate the code. I also check it with DSV PHP Editor. Both come up with the same error. I'm also using MyPhpAdmin to create the database and tables. I know also to place this calling info in another folder and include it by calling it with a php include statement. I just have it within the code to quickly test it. Thank you for any assistance. Bob... im using 000webhost as a test site and when i run this code it redirects me to there err page but with no error message. the sql query works fine in phpmyadmin and i added the rest of the code to try the php side. i "think" the problem is the echo $rows"value's" as im unsure of what the $vars should be <?php include("config.php"); // Connect to server and select database. mysql_connect("$host", "$username", "$password")or die("cannot connect"); mysql_select_db("$db_name")or die("cannot select DB"); $sql = "SELECT make, COUNT(*) AS total, SUM(IF(comments = \'pass\', 1, 0)) AS withComments FROM dsgi_serval GROUP BY make ORDER BY COUNT(*) DESC"; $result=mysql_query($sql); echo "$sql"; echo "$result"; ?> <table><tr> <td colspan="4"><strong>List data from mysql </strong> </td> </tr> <tr> <td align="center"><strong>make</strong></td> <td align="center"><strong>Total</strong></td> <td align="center"><strong>Validated</strong></td> </tr> <?php while($rows=mysql_fetch_array($result)){ ?> <tr> <td><?php echo $rows['make']; ?></td> <td><?php echo $rows['total']; ?></td> <td><?php echo $rows['withcomments']; ?></td> </tr> <?php } ?> </table> <?php mysql_close(); ?> i am trying to upload a .sql file on godaddy database and i am getting this error . Help me.
I have searched this forum as well as over 200 other forums and have not found the answer that is specific to my question. I have shortened my code drastically to assist in resolving this quickly -
I have a search form that has criteria for the search criteria with "virtual" "columns" in an array but it's not working. If I search one column at a time it works just fine but when I try to search 8 columns with one select I get the following error: SELECT Error: Unknown column 'achievements' in 'where clause'.
When a user selects search in Achievements, I need it to look at all 8 columns that are associated with achievements and bring back the results that match - the same as if the user selects search in Associations, I need it to look at all 5 columns and bring back the results that match.
My shortened code is as follows:
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Search</title>
</head>
<body>
<form name="search" action="" method="POST">
<p>Search:</p>
<p>
Achievements/Associations: <input type="text" name="find1" /> in
<Select NAME="field1">
<Option VALUE="achievements">Achievements</option>
<Option VALUE="associations">Associations</option>
</Select>
<br><br>
Secondary Education: <input type="text" name="find2" /> in
<Select NAME="field2">
<Option VALUE="edu1sectype">Highest Certificate Attained</option>
<Option VALUE="edu1secname">Highest Grade Passed</option>
<Option VALUE="edu1secinst">Name of High School</option>
<Option VALUE="edu1secdate">Date Completed</option>
<Option VALUE="edu1secinsttyp">Type of Institution</option>
<Option VALUE="subjects">Subjects</option>
</Select>
<br><br>
<input type="hidden" name="searching" value="yes" />
<input type="submit" name="search" value="Search" />
</p>
</form>
<?php
$searching = $_POST['searching'];
$find1 = $_POST['find1'];
$field1 = $_POST['field1'];
$find2 = $_POST['find2'];
$field2 = $_POST['field2'];
if ($searching =="yes")
{
echo "<br><b>Searched For:</b> $find1 $find2<br>";
echo "<br><h2>Results</h2><p>";
//If they did not enter a search term we give them an error
// Otherwise we connect to our Database
include_once "connect_to_mysql.php";
mysql_select_db("table_name") or die(mysql_error());
// We preform a bit of filtering
$find = strtoupper($find);
$find = strip_tags($find);
$find = trim($find);
$find = mysql_real_escape_string($find);
$field = mysql_real_escape_string($field);
$data = mysql_query("SELECT * FROM table_name
WHERE upper(".$field1.") LIKE '%$find1%'
AND upper(".$field2.") LIKE '%$find2%'
") or die("SELECT Error: ".mysql_error());
$result = mysql_query("SELECT * FROM table_name
WHERE upper($field1) LIKE '%$find1%'
AND upper($field2) LIKE '%$find2%'
") or die("SELECT Error: ".mysql_error());
$num_rows = mysql_num_rows($result);
echo "There are $num_rows records:<br>";
echo '<center>';
echo "<table border='1' cellpadding='5' width='990'>";
// set table headers
echo "<tr><th>Reference</th>
<th>First Name</th>
<th>Last Name</th>
</tr>";
//get images and names in two arrays
$name= $row["name"];
$surname= $row["surname"];
$achieve1 = $row["achieve1"];
$achieve2 = $row["achieve2"];
$achieve3 = $row["achieve3"];
$achieve4 = $row["achieve4"];
$achieve5 = $row["achieve5"];
$achieve6 = $row["achieve6"];
$achieve7 = $row["achieve7"];
$achieve8 = $row["achieve8"];
$assoc1 = $row["assoc1"];
$assoc2 = $row["assoc2"];
$assoc3 = $row["assoc3"];
$assoc4 = $row["assoc4"];
$assoc5 = $row["assoc5"];
$edu1sectype = $row["edu1sectype"];
$edu1secinst = $row["edu1secinst"];
$edu1secname = $row["edu1secname"];
$edu1secdate = $row["edu1secdate"];
$edu1secinsttyp = $row["edu1secinsttyp"];
$subject1 = $row["subject1"];
$subject2 = $row["subject2"];
$subject3 = $row["subject3"];
$subject4 = $row["subject4"];
$subject5 = $row["subject5"];
$subject6 = $row["subject6"];
$subject7 = $row["subject7"];
$subject8 = $row["subject8"];
$compsoft1name = $row["compsoft1name"];
$compsoft2name = $row["compsoft2name"];
$compsoft3name = $row["compsoft3name"];
$compsoft4name = $row["compsoft4name"];
$compsoft5name = $row["compsoft5name"];
$compsoft6name = $row["compsoft6name"];
$achievements = array('achieve1', 'achieve2', 'achieve3', 'achieve4', 'achieve5', 'achieve6', 'achieve7', 'achieve8');
$associations = array('assoc1', 'assoc2', 'assoc3', 'assoc4', 'assoc5');
$subjects = array('subject1', 'subject2', 'subject3', 'subject4', 'subject5', 'subject6', 'subject7', 'subject8' );
$compsoft = array('compsoft1name', 'compsoft2name', 'compsoft3name', 'compsoft4name', 'compsoft5name', 'compsoft6name');
while ($row = mysql_fetch_array($result))
{
echo "<tr>";
echo "<td ALIGN=LEFT>" . $row['id'] . "</td>";
echo "<td ALIGN=LEFT>" . $row['name'] . "</td>";
echo "<td ALIGN=LEFT>" . $row['surname'] . "</td>";
echo "</tr>";
}
echo "</table>";
//This counts the number or results - and if there wasn't any it gives them a little message explaining that
$anymatches=mysql_num_rows($data);
if ($anymatches == 0) {
echo "Sorry, but we can not find an entry to match your query";
}
}
?>
</body>
</html>
Any assistance will be greatly appreciated as I have been working on this website for the past 4 months which has totalled over 150 pages and this is one of the last pages left to program and it's taken 6 days to get to this search page to this point.
Hi there I'm trying to insert audi files' records into mysql database this is how the records insertion looks like: Code: [Select] $fileName = $_FILES['uploaded']['name'];//name $tmpName = $_FILES['uploaded']['tmp_name'];//temp location $fileSize = $_FILES['uploaded']['size'];//size of the file $fileType = $_FILES['uploaded']['type'];//type of file $error = $_FILES['uploaded']['error'];//verifys errprs $ext = substr($fileName, strrpos($fileName, '.') +1); //this will get the extention out of the file name e.g. .mp3 //check that a file is passed by and no errors if(isset($fileName) && $error == 0 && $fileSize != 0){ //condition to accept only certain file types/extentions if($ext == "mp3" || $ext == "wma" || $ext == "wav"){ //get file content $fp = fopen($tmpName, 'r'); $content = fread($fp, filesize($tmpName)); $content = addslashes($content); fclose($fp); if(!get_magic_quotes_gpc()) { $fileName = addslashes($fileName); } //query to retrieve user's id $userid = mysql_query("select userID from user where username = '$username'"); //to get id by username $row = mysql_fetch_assoc($userid); $userid = $row['userID']; $query = "INSERT INTO tracks (trackName, userID, tag, price, file, fileName, fileSize, fileType) VALUES ('$tname','$userid','$tag','$price','$content','$fileName','$fileSize' , '$fileType)"; mysql_query($query) or die (mysql_error()); when i send it i get this error: Unknown column 'application' in 'field list' i did set data type for the fileType field to varchar. when i remove the fileType all together the record is inserted successfully into database though. ?!! can anyone help please Just a heads up, my OOP skills are in development so sorry for the roughness of things...(any tips always appreciated!) I'm basically making my own little repository site that will have snippets (for quick copy paste anywhere) (I know there are other services, but mine is a little different and needs to be private). I'll try to just post the necessary code: HTML trying to pass: Code: [Select] <p><br /><span class="something">Hey buddy</span></p> Javascript/JQuery: Code: [Select] //send info $.ajaxFileUpload({ url:'./upload_zip.php', secureuri:false, fileElementId:'form_zip', dataType: 'json', data: {... 'description': htmlEntities($('#modularContent form textarea[name="description"]').val())...}, success:function(data, status){ var html = ... html+= ...'<div class="description"><p>'+data.description+'<br /><br />';... $('#list').append(html); ... $('#modularOverlay,#modularContent').remove(); }, error:function(data, status, e){ alert(e); } ... function htmlEntities(str) { return String(str).replace(/&/g, '&').replace(/</g, '<').replace(/>/g, '>').replace(/"/g, '"'); } //found via: http://css-tricks.com/snippets/javascript/htmlentities-for-javascript/ PHP: Code: [Select] <?php //Start object and modify initial values public function __construct($title,$description,$url,$snippet){ ... if($this->snippet == 'yes'){ $this->handle_snippet(); } } public function handle_snippet(){ $code = htmlentities($this->description, ENT_QUOTES | ENT_IGNORE, 'UTF-8',false); $this->description = $code; } //returning value public function getJSON($action){ //temp fix for js breaking if remove_tag_ids is empty/null if(empty($this->remove_tag_ids)){ $this->remove_tag_ids = 'null'; } if($this->snippet == 'yes'){ $this->description = stripslashes($this->description); } switch($action){ case 'delete': //not important default: $json = array(...'description'=>$this->description...); break; return $json; } ?> POST Value: (via firebug) Code: [Select] description: <p><br /><span class="something">Hey buddy</span></p> JSON Return value: (via firebug) Code: [Select] description: <p><br /><span class="something">Hey buddy</span></p> DB value: Code: [Select] <p><br /><span class=\"something\">Hey buddy</span></p> As far as error, I have no idea what the error is haha. It does not get through to the error code cause the plugin tosses an error b4 that (using an ajaxUpload plugin http://www.phpletter.com/Our-Projects/AjaxFileUpload/, (this does not affect the upload, it uploads just fine, but it tosses an error for some reason)). Anyone have any insights/ideas? Thanks for any and all help, Justin I have this function I use to simplify things.
function search_string( $needle, $haystack ) {
if ( preg_match_all( "/$needle/im", $haystack ) || strpos( $haystack, $needle ) ) {
return TRUE;
}
return FALSE;
}
I keep getting this error in my PHP logs, and it comes in a sequence: [07-Nov-2020 05:34:14 America/Los_Angeles] PHP Warning: preg_match_all(): Unknown modifier 'G' in /home/baser-b/public_html/include/functions.php on line 791 [07-Nov-2020 05:34:14 America/Los_Angeles] PHP Warning: preg_match_all(): Unknown modifier 'g' in /home/baser-b/public_html/include/functions.php on line 791 Meaning, it will come with one with the small g, then three with the big G, then one with the small g, then five with the big G, and so on.... My question is, how can I stop getting this error. It won't show me the functions being called to arrive at this answer, as this is likely an error generated by another function calling this one. I was wondering if anyone knew what to change in the search_string function to stop getting this error, why this error is happening, or why the strange repetitive sequence. Is it someone trying to do a hack? The only variable that would be changeable by a visitor would be the $needle variable, so what could they type that has something to do with 'g' to get this? Anyway, thanks. This topic has been moved to Ajax Help. http://www.phpfreaks.com/forums/index.php?topic=327780.0 In drive.php
public function insert($postBody, $optParams = array()) I am not very advanced in web programming I need help..
I am using an API and my sql database. I would to implement live/ dynamic updates in the text field where user will input their text, and the web should first check the database then the api live. I would also like to retrieve the user input without refreshing the page, so the retrieved information regarding the inputted text should be automatically loaded without refreshing the page.
please help .
I have been pulling my hair out for the lasy 3 hours i am trying to update a MySql table but i cant get it too work, i just keep getting MySql error #1064 - You have an error in your SQL syntax; if i just update 1 field it works fine but if i try to update more than 1 field it dosent work, Help Please! <?php $root = $_SERVER['DOCUMENT_ROOT']; require("$root/include/mysqldb.php"); require("$root/include/incpost.php"); $con = mysql_connect("$dbhost","$dbuser","$dbpass"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("$dbame", $con); mysql_query("UPDATE Reg_Profile_p SET build='$build' col='$col' size='$size' WHERE uin = '$uinco'"); ?> does anyone know who to resolve this issue of importing a CSV file from excel into sql? I get this error when I do. LOAD DATA LOCAL INFILE '/tmp/phpq2aAbU' INTO TABLE `Events` FIELDS TERMINATED BY ',' ENCLOSED BY '\\"' ESCAPED BY '\\\\' LINES TERMINATED BY '\r\n' Hey, I've got a query in which a variable is interpreted as a column and I don't why this is caused. $upgrade_time_sql = "SELECT * FROM todo_upgrades WHERE profile_id = ".$profile_id." AND level = ".$profile_data['level2_'.$show.'']." AND type = ".$show.""; $upgrade_time_res = mysql_query($upgrade_time_sql) or die (mysql_error()); $show is filled with the content "storage" The mysql_error is "Unknown column 'storage' in 'where clause'" Thanks for helping. Hey im kind of new to PHP me and my mate was working on a script and we can not seem to get this working: Its connected through mysql. And it shows Quote Notice: Undefined index: username in C:\xampp\htdocs\stick_arena.php on line 21 Notice: Undefined index: userpass in C:\xampp\htdocs\stick_arena.php on line 22 Notice: Undefined index: action in C:\xampp\htdocs\stick_arena.php on line 23 Notice: Undefined index: usercol in C:\xampp\htdocs\stick_arena.php on line 24 Notice: Undefined index: stats in C:\xampp\htdocs\stick_arena.php on line 25 Here are the lines Quote // declare variables $username = sanitize($_POST['username']); $password = sanitize(md5($_POST['userpass'])); $action = sanitize($_POST['action']); $usercol = sanitize($_POST['usercol']); $stats = sanitize($_POST['stats']); Here is the full script for that page: Click here please help as soon as possible. Hey I have a script that builds an array but for some reason there is a loose integer in the variable which is confusing me and i think its the cause of my syntax errors in javascript. But i got no idea where it is coming from =/ Here is the function i use: Code: [Select] <?php function img($id) { $img = array(); $img['tiles'] = $this->db->getAll("SELECT DISTINCT t.id, CONCAT('data/tiles/',filename) AS f FROM `map_tiles` AS mt INNER JOIN `tiles` AS t ON (mt.tile=t.id) WHERE mt.map_id={$this->db->qstr($id)}"); echo print_r($img); die; //exit(json_encode($img)); // turned off for testing ?> The echo returns this: Code: [Select] Array ( [tiles] => Array ( [0] => Array ( [id] => 10 [f] => data/tiles/floor.png ) ) ) 1 <----- why does this appear? And could this cause a syntax error if sent back for JS processing? As you can see there is 1 showing at the end but i don't see why =/ Any ideas if thats suppose to happen ? |
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