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PHP - Mysql Error: Unknown Column … In 'where Clause'
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 ACBMSESimilar TutorialsI 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 I've a simple Inset List (jQuery Mobile) of several items with a Flip switch, Drop down, check box and Slider elements. I have wrapped this Inset List inside a HTML Form. <code> <div data-role="fieldcontain"> <ul data-role="listview" data-inset="true"> <li data-role="list-divider"> Inhouse capabilities </li> <li> <label for="flip-b">Project Manager</label> <select name="slider" id="flip-b" data-role="slider"> <option value="no">No</option> <option value="yes">Yes</option> </select> </li> <li> <label for="flip-b">Site Supervisor</label> <select name="slider" id="flip-b" data-role="slider"> <option value="no">No</option> <option value="yes">Yes</option> </select> </li> <li><label for="flip-b">Architect</label> <select name="slider" id="flip-b" data-role="slider"> <option value="no">No</option> <option value="yes">Yes</option> </select> </li> <li><label for="flip-b">Builderworks</label> <select name="slider" id="flip-b" data-role="slider"> <option value="no">No</option> <option value="yes">Yes</option> </select> </li> <li><label for="flip-b">Electrical</label> <select name="slider" id="flip-b" data-role="slider"> <option value="no">No</option> <option value="yes">Yes</option> </select> </li> <li><label for="flip-b">Mechanical</label> <select name="slider" id="flip-b" data-role="slider"> <option value="no">No</option> <option value="yes">Yes</option> </select> </li> <li><label for="flip-b">Hydraulics</label> <select name="slider" id="flip-b" data-role="slider"> <option value="no">No</option> <option value="yes">Yes</option> </select> </li> <li><label for="flip-b">Joinery</label> <select name="slider" id="flip-b" data-role="slider"> <option value="no">No</option> <option value="yes">Yes</option> </select> </li> <li> <label for="select-choice-1" class="select">Main Location:</label> <select name="select-choice-1" id="select-choice-1"> <option value="NSW">NSW</option> <option value="ACT">ACT</option> <option value="VIC">VIC</option> <option value="SA">SA</option> <option value="QLD">QLD</option> <option value="NT">NT</option> <option value="WA">WA</option> <option value="TAS">TAS</option> </select> </li> <!-- Select Menus: Main Location--> <li> <div data-role="fieldcontain"> <fieldset data-role="controlgroup"> <legend>CBA Reference</legend> <input type="checkbox" name="checkbox-1a" id="checkbox-1a" class="custom" /> <label for="checkbox-1a">Commercial</label> <input type="checkbox" name="checkbox-2a" id="checkbox-2a" class="custom" /> <label for="checkbox-2a">Retail</label> </fieldset> </div> </li> </ul> </div> <!-- End of Field Contain div tag --> <a href="#additionalinfo" data-role="button" data-inline="true" data-iconpos="left" data-theme="b">Additional Info</a> <a href="./search2.html" rel="external" data-ajax="false" data-role="button" data-inline="true" data-icon="search" data-iconpos="left" data-theme="e">Search</a> </code> I want to know, how I can include only the elements the user selected in my SQL query. Lets say, User wants to find the Electrical workers in QLD, then he will select Yes for Electrical and QLD as Main Location and the other items he would not touch. Now, how can I include only those two values in the select clause. Say, Select Electrical, Main Location from table where Main location='QLD'. I used POST method to gather users input. I use PHP, MySQL for this project. Any ideas on how to achieve this. Regards Sandeep Hey 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 ! 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')"; Hey, 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 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()); $show is filled with the content "storage" The mysql_error is "Unknown column 'storage' in 'where clause'" Thanks for helping. Hi,
Im having a problem and I can't seem to figure it out or find anything on the net.
If I use the following code the script successfully updates every row in the table:
mysqli_query($con,"UPDATE Ads SET Ads_LocalArea='Stroud'");However if I try updating the table using the WHERE clause in any of the combinations below nothing happens. mysqli_query($con,"UPDATE Ads SET Ads_LocalArea='Stroud' WHERE Ads_ID=$DBROWID");---------------------------------------------------------------------- My Script:
mysqli_query($con,"INSERT INTO Ads (Ads_ID, Ads_AID, Ads_Title)
VALUES ('', '$Polished_AdRef', '$Polished_AdTitle')");
$DBROWID = mysqli_insert_id($con);
mysqli_query($con,"UPDATE Ads SET Ads_AID='Stroud' WHERE Ads_ID=$DBROWID");
// TRIED THESE TOO
// mysqli_query($con,"UPDATE Ads SET Ads_AID='Stroud' WHERE Ads_ID='$DBROWID'");
// mysqli_query($con,"UPDATE Ads SET Ads_AID='Stroud' WHERE Ads_ID='5'");
Does any one know where I am going wrong?
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 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... 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! This topic has been moved to MySQL Help. http://www.phpfreaks.com/forums/index.php?topic=342695.0 Hi guys, I need your help. I am trying to insert the rows in the mysql database as I input the values in the url bar which it would be like this: Code: [Select] www.mysite.com/testupdate.php?user=tester&pass=test&user1=tester&email=me@shitmail.com&ip=myisp However i have got a error which i don't know how to fix it. Error: Column count doesn't match value count at row 1 <?php session_start(); define('DB_HOST', 'localhost'); define('DB_USER', 'mydbusername'); define('DB_PASSWORD', 'mydbpassword'); define('DB_DATABASE', 'mydbname'); $errmsg_arr = array(); $errflag = false; $link = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD); if(!$link) { die('Failed to connect to server: ' . mysql_error()); } $db = mysql_select_db(DB_DATABASE); if(!$db) { die("Unable to select database"); } function clean($var){ return mysql_real_escape_string(strip_tags($var)); } $username = clean($_GET['user']); $password = clean($_GET['pass']); $adduser = clean($_GET['user1']); $email = clean($_GET['email']); $IP = clean($_GET['ip']); if($username == '') { $errmsg_arr[] = 'username is missing'; $errflag = true; } if($password == '') { $errmsg_arr[] = 'PASSWORD is missing'; $errflag = true; } if($errflag) { $_SESSION['ERRMSG_ARR'] = $errmsg_arr; echo implode('<br />',$errmsg_arr); } else { $sql = "INSERT INTO `members` (`username`,`email`,`IP`) VALUES ('$adduser','$email','$IP')"; if (!mysql_query($sql,$link)) { die('Error: ' . mysql_error()); } echo "The information have been updated."; } ?> Here's the name of the columns i have got in my database: Code: [Select] username IP I have input the correct columns names, so I can't correct the problem I am getting. Please can you help? i am trying to upload a .sql file on godaddy database and i am getting this error . Help me.
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 $needle variable, so what could they type that has something to do with 'g' to get this? Anyway, thanks. In drive.php
public function insert($postBody, $optParams = array()) My Php Buddies, I have mysql tbl columns these:
id: Now, I want to display their row data by excluded a few columns. Want to exclude these columns: date_&_time account_activation_code account_activation_status id_verification_video_file_url password
So, the User's (eg. your's) homepage inside his account should display labels like these where labels match the column names but the underscores are removed and each words' first chars CAPITALISED:
Id: 1
For your convenience only PART 1 works. Need help on Part 2 My attempted code:
PART 1
<?php
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
// Query to get columns from table
$query = $conn->query("SELECT COLUMN_NAME FROM INFORMATION_SCHEMA.COLUMNS WHERE TABLE_SCHEMA = 'members' AND TABLE_NAME = 'users'");
while($row = $query->fetch_assoc()){
$result[] = $row;
}
// Array of all column names
$columnArr = array_column($result, 'COLUMN_NAME');
foreach ($columnArr as $value) {
echo "<b>$value</b>: ";?><br><?php
}
?>
PART 2
<?php
//Display User Account Details
echo "<h3>User: <a href=\"user.php?user=$user\">$user</a> Details</h3>";?><br>
<?php
$excluded_columns = array("date_&_time","account_activation_code","account_activation_status","id_verification_video_file_url","password");
foreach ($excluded_columns as $value2)
{
echo "Excluded Column: <b>$value2</b><br>";
}
foreach ($columnArr as $value)
{
if($value != "$value2")
{
$label = str_replace("_"," ","$value");
$label = ucwords("$label");
//echo "<b>$label</b>: "; echo "$_SESSION[$value]";?><br><?php
echo "<b>$label</b>: "; echo "${$value}";?><br><?php
}
}
?>
PROBLEM: Columns from the excluded list still get displayed. Edited November 19, 2018 by phpsaneHi all, Just curious why this works: Code: [Select] while (($data = fgetcsv($handle, 1000, ",")) !== FALSE){ $import="INSERT into $prodtblname ($csvheaders1) values('$data[0]','$data[1]','$data[2]','$data[3]','$data[4]','$data[5]','$data[6]')"; } And this does not: $headdata_1 = "'$data[0]','$data[1]','$data[2]','$data[3]','$data[4]','$data[5]','$data[6]'"; while (($data = fgetcsv($handle, 1000, ",")) !== FALSE){ $import="INSERT into $prodtblname ($csvheaders1) values($headdata_1)"; }it puts $data[#'s] in the database fields instead of the actual data that '$data[0]','$data[1]'... relates to. I wrote a script to create the values in $headdata_1 based on the number of headers in $csvheaders1 but can't seem to get it working in the sql statement. Thanks hi guys how to fix this error. i have this query $sql = " SELECT take.StudentID ,student.StudentName ,take.CourseID ,course.CourseName FROM take ,student ,course WHERE take.StudentID = student.StudentID AND take.CourseID = course.CourseID AND StudentID LIKE '$_POST[sid]%' ORDER BY StudentID ASC "; $result = mysql_query($sql) or trigger_error('MySQL Error: ' . mysql_error(), E_USER_ERROR); error msg: Fatal error: MySQL Error: Column 'StudentID' in where clause is ambiguous in C:\wamp\www\dis_take.php on line 98 |
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