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PHP - Insert Id Manually
i was working on modifying a code written by a previous user.
i have field called adid in the ads table which should have been a primary key and autoincremented. But since we already have lot of data across different sites we cannot modify the table structure. The problem is i want to check if the following code is adding adid manually.is there some problem with this code? Code: [Select] $get_ad_id = mysql_query("SELECT adid FROM ads ORDER BY adid DESC LIMIT 1", $con) or die("Error Getting AD ID: ".mysql_error()); while($get_ad_id_results = mysql_fetch_assoc($get_ad_id)) { $adid=$get_ad_id_results['adid']; } if($adid!="") { $adid=$adid+1; } else { $adid="1"; } if($bannertype=="120x60" || $bannertype=="468x60") { $New_URL = $_POST['URL']; $active='Y'; $insert_image_banner_query="INSERT INTO ads (adid, URL, image_path, date, userid, active, size) VALUES ('".$adid."','$New_URL', '$newname', '$Date', '$loggedinuserid','".$active."','".$bannertype."')"; mysql_query($insert_image_banner_query, $con) or die("Inserting Image Ad Failed: ".mysql_error()); $update_banner_cache="YES"; } Similar TutorialsIs it possible to manually edit a SESSION (not cookie) from client side? Sort of how a user can easily edit a cookie's value. I've got a script that's been running away happily as a cron job for the last few days, but it's suddenly decided to stop working. When I run the script manually it works perfectly, but not when I run it as a cron job. I've done some debugging and I've found that the cause of the problem appears to be an ftp_get function that I'm using. I'm guessing I need to either change a file path, or a file permission, but I'm not sure to what. Here's the info: Code: [Select] $local_dir = '../../col_protected/'; $c = ftp_connect('www.mydomain.net') or die("Can't connect"); ftp_login($c,'ftp_username','ftp_password') or die("Can't login"); ftp_get($c, $local_dir.$filename, FTP_ASCII) or die("Can't transfer"); The FTP connect and login are still working fine, it's the ftp_get that's causing the problem. The '../../col_protected/' directory is on the same level as the public_html (ie. outside the public_html) with file permissions set to 777. The cron is running at /home/username/public_html/col/proc1.php Can anyone advise please? Many thanks, Chris Server version: 5.1.53-log I have the following queries Code: [Select] SELECT user_id FROM phpbb_profile_fields_data WHERE pf_rsname = "Atroxide" LIMIT 1 SELECT user_id FROM phpbb_profile_fields_data WHERE pf_rsname = "Delia Smith" LIMIT 1 SELECT user_id FROM phpbb_profile_fields_data WHERE pf_rsname = "espinozagabe" LIMIT 1 SELECT user_id FROM phpbb_profile_fields_data WHERE pf_rsname = "Jaunty1" LIMIT 1 SELECT user_id FROM phpbb_profile_fields_data WHERE pf_rsname = "lvoos" LIMIT 1 All 5 of these queries are executed at a different time (in a foreach loop). All 5 except for the one below returned a result. Code: [Select] SELECT user_id FROM phpbb_profile_fields_data WHERE pf_rsname = "Delia Smith" LIMIT 1 I couldn't figure out why it wasn't working so I copy pasted it into PHPMyAdmin and it returned the result I was looking for. What could cause for PHPMyAdmin to work but not the exact same query in a php script to not? It didn't return an error using mysql_error() either. Pretty sure its irreverent but here is the php script. Code: [Select] foreach ($online as $username => $activity) { $query = " SELECT user_id FROM phpbb_profile_fields_data WHERE pf_rsname = \"" . $username . "\" LIMIT 1 "; $result = $db->query($query); } The table is Code: [Select] user_id mediumint(8) UNSIGNED No 0 pf_rsname varchar(255) utf8_bin Yes NULL I have a strange problem. When a guest visits my contact-user.php page, they get a message telling them the must login before viewing the page. After the guest logs in, they view the same page and it tells them they have to login again (keeps on looping). But if they manually refresh that page with the "you must be logged in" message, it recognizes the login and lets them in. How can I get this page to immediately recognize that the user is logged in and not require them to refresh the page manually? Here is my code for contact-user.php <?php session_start(); header("Cache-Control: private, max-age=10800, pre-check=10800"); header("Pragma: private"); header("Expires: " . date(DATE_RFC822,strtotime("+2 day"))); include("connection.php"); mysql_select_db("database"); if (isset($_SESSION['username'])) { ******** MY HTML PAGE CONTENT ******** } else { echo "<meta http-equiv='REFRESH' content='2;url=http://www.mysite.com/login.php'> <center><font color='#EE0000'><p>You must be logged in before negotiating. You will now be redirect to the login page.</p></font></center>"; } ?> Here is my code for login.php script: <?php include("connection.php"); mysql_select_db("database"); session_start(); if(isset($_POST['login'])){ $username = mysql_real_escape_string($_POST['username']); $password = mysql_real_escape_string($_POST['password']); $tUnixTime = time(); $sGMTMySqlString = gmdate("Y-m-d H:i:s", $tUnixTime); if (!$username || !$password) { print "Please fill out all fields."; exit; } $logres = mysql_num_rows(mysql_query("SELECT * FROM members WHERE username = '$username' and password = '$password'")); if ($logres <= 0) { print "Login failed. If you have not already, please signup. Otherwise, check your spelling and login again."; exit; } else { $_SESSION['username'] = $username; if (isset($_SESSION)) { echo'You are now logging in'; mysql_query("UPDATE members SET activity = '$sGMTMySqlString' WHERE username = '$username'"); } else { echo "You are not logged in!"; } echo'<html><head><meta http-equiv="REFRESH" content="1;url=http://www.mysite.com/members/' . $_SESSION['username'] . '/"></head><body></body></html>'; exit; } } ?> Hi, I am trying to make some adjustments to uploadify.php which comes with the latest version of uploadify (3.0 beta), so that it works with a session variable that stores the login username and adds it to the path for uploads. Here is uploadify.php as it currently looks: Code: [Select] <?php session_name("MyLogin"); session_start(); $targetFolder = '/songs/' . $_SESSION['name']; // Relative to the root if (!empty($_FILES)) { $tempFile = $_FILES['Filedata']['tmp_name']; $targetPath = $_SERVER['DOCUMENT_ROOT'] . $targetFolder; $targetFile = rtrim($targetPath,'/') .'/'. $_FILES['Filedata']['name']; // Validate the file type $fileTypes = array('m4a','mp3','flac','ogg'); // File extensions $fileParts = pathinfo($_FILES['Filedata']['name']); if (in_array($fileParts['extension'],$fileTypes)) { move_uploaded_file($tempFile,$targetFile); echo '1'; } else { echo 'Invalid file type.'; } } echo $targetFolder; ?> I added Code: [Select] echo $targetFolder; at the bottom so that I could make sure that the string returned was correct, and it is, i.e. '/songs/nick'. For some reason though, uploads are not going to the correct folder, i.e. the username folder, but instead are going to the parent folder 'songs'. The folder for username exists, with correct permissions, and when I manually enter Code: [Select] $targetFolder = '/songs/nick';all works fine. Which strikes me as rather strange. I have limited experience of using php, but wonder how if the correct string is returned by the session variable, the upload works differently than with the manually entered string. Any help would be much appreciated. It's the last issue with a website that was due to go live 2 days ago! Thanks, Nick I'm missing something here. I have a form, and when the submit is pressed, the relevant post data inserts into table one, then I want the last insert id to insert along with other form data into a second table. The first table's still inserting fine, but I can't get that second one to do anything. It leapfrogs over the query and doesn't give an error. EDIT: I forgot to add an error: I get: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'usage, why VALUES ('14', '', '123', '','1234', '', '')' at line 1 query:INSERT INTO tbl_donar (donar_fname, donar_name, donar_address, donar_address2, donar_city, donar_state, donar_zip, donar_email, donar_phone, donar_fax, donar_company) VALUES ('test 14', 'asdfa', 'asdf', 'adf','asdf', '', '', '', '123', '', '') Code: [Select] if (empty($errors)) { require_once ('dbconnectionfile.php'); $query = "INSERT INTO tbl_donar (donar_fname, donar_name, donar_address, donar_address2, donar_city, donar_state, donar_zip, donar_email, donar_phone, donar_fax, donar_company) VALUES ('$description12', '$sn', '$description4', '$cne','$description5', '$description6', '$description7', '$description8', '$description9', '$description10', '$description11')"; $result = @mysql_query ($query); if ($result) { $who_donated=mysql_insert_id(); $query2 = "INSERT INTO tbl_donation (donor_id, donor_expyear, donor_cvv, donor_cardtype, donor_authorization, amount, usage, why) VALUES ('$who_donated', '$donate2', '$donate3', '$donate4','$donate5', '$donate6', '$donate7')"; $result2 = @mysql_query ($query2); if ($result2) {echo "Info was added to both tables! yay!";} echo "table one filled. Table two was not."; echo $who_donated; //header ("Location: http://www.twigzy.com/add_plant.php?var1=$plant_id"); exit(); } else { echo 'system error. No donation added'; Can anyone tell me why this is not INSERTing? My array data is coming out just fine.. I've tried everything I can think of and cannot get anything to insert.. Ahhhh! <?php $query = "SELECT RegionID, City FROM geo_cities WHERE RegionID='135'"; $results = mysqli_query($cxn, $query); $row_cnt = mysqli_num_rows($results); echo $row_cnt . " Total Records in Query.<br /><br />"; if (mysqli_num_rows($results)) { while ($row = mysqli_fetch_array($results)) { $insert_city_query = "INSERT INTO all_illinois SET state_id=$row[RegionID], city_name=$row[City] WHERE id = null" or mysqli_error(); $insert = mysqli_query($cxn, $insert_city_query); if (!$insert) { echo "INSERT is NOT working!"; exit(); } echo $row['City'] . "<br />"; echo "<pre>"; echo print_r($row); echo "</pre>"; } //while ($rows = mysqli_fetch_array($results)) } //if (mysqli_num_rows($results)) else { echo "No results to get!"; } ?> Here is my all_illinois INSERT table structu CREATE TABLE IF NOT EXISTS `all_illinois` ( `state_id` varchar(255) NOT NULL, `city_name` varchar(255) NOT NULL ) ENGINE=MyISAM DEFAULT CHARSET=latin1; Here is my source table geo_cities structu CREATE TABLE IF NOT EXISTS `1` ( `CityId` varchar(255) NOT NULL, `CountryID` varchar(255) NOT NULL, `RegionID` varchar(255) NOT NULL, `City` varchar(255) NOT NULL, `Latitude` varchar(255) NOT NULL, `Longitude` varchar(255) NOT NULL, `TimeZone` varchar(255) NOT NULL, `DmaId` varchar(255) NOT NULL, `Code` varchar(255) NOT NULL ) ENGINE=MyISAM DEFAULT CHARSET=latin1; Hello, I'm having a bit of a problem here, all help to this issues would be much appreciated I am trying to use text boxes to insert numbers into the database based on what is inputed. If I have a string, like this for example: $variable = 09385493; And I want to insert it into the database like this: mysql_query("INSERT INTO integers(number) VALUES ('$variable')"); When checking the integers table in my database, looking at the number field, the $variable that was inserted is outputted as 9385493 Notice the number zero was taken out of the front of the number. If the number is double 0's (009385493), both of those zero's would disappear, too. Thanks i need the insert code Would someone be able to tell me why I can't run a successful mysql_query($sql) with $sql="INSERT INTO myDB.Titles (Title, Year, cID) values('". $title . "', '" . $year . "', '" . $cID . "')"; $sql echoes out to "INSERT INTO myDB.Titles (Title, Year, cID) values('Test', '2000', '2')" It looks right but I'm probably off on the quotes somewhere. Thanks in advance. Dear Sir/Madame I am trying to insert comments into a table called comments using echo ID in the input form but so far i can insert the comments with the post id but unable re-comment again kindly help me what do i do?
This is the input form:
<div>
<form action="ehscomment.php" method="post">
<input type="hidden" name="id" value="<?php echo $id ?>">
<div>
<label>Add comment</label>
<div>
<textarea rows="6" cols="110" name="comment" placeholder="comment"></textarea>
</div>
</div>
<input type="submit" name="postcomment" value="comment"></form>
</div>
And this is ehscomment.php
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "registration";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
if(isset($_POST['postcomment'])){
$id = $_POST['id'];
$comment = $_POST['comment'];
$sql = "INSERT INTO comments (id,comment)
VALUES ('$id','$comment')";
if ($conn->query($sql) === TRUE) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
$conn->close();}
?>
Error message: Error: INSERT INTO comments (id,comment) VALUES ('242','asddd') I'm trying to use PDO and get used to doing things this way. I've been away from php/mysql for a few years, so, I'm crusty. I'm not getting any error messages back on this code, but the insert just doesn't happen. My first guess is that I'm doing something wrong with the datetime now() function. But, I may not have the PDO code right. I tried the script the old fashion way with mysql_query() and that worked. So, it has to be something in this code. I believe my server is set up to do PDO as it shows: PDO PDO support enabled PDO drivers mysql, sqlite pdo_mysql PDO Driver for MySQL, client library version 5.0.45 My php version is 5.2.14 and Mysql is 5.0.45. Any help would be appreciated. Code: [Select] $DBH = new PDO("mysql:host=$host;dbname=$dbname", $user, $pass); $DBH->setAttribute( PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION ); $sql=$DBH->prepare("INSERT INTO assets(asset_name,date_added,short_desc) VALUES (:asset_name,NOW(),:short_desc)"); $sql->bindParam(':asset_name',$asset_name); $sql->bindParam(':short_desc',$short_desc); $name=$_POST["input1"]; $short_desc=$_POST["input2"]; $DBH->exec(); echo $name; echo "\nPDO::errorInfo():\n"; print_r($DBH->errorInfo()); } catch(PDOException $e) { echo "Syntax Error: ".$e->getMessage(); } I am working with a insert query... Code: [Select] $query="INSERT INTO user_info_more (website) VALUES ('".$website."') WHERE id='$_SESSION[id]'"; mysql_query($query) or die ('Unable to register an account with following details 1'); but the output is " Unable to register an account with following details 1" Basically i have a row with 6 cloumns in mysql database in which only the id column and name column is filled... Now i want to insert $website in the "website" column of the database... Neither i can update as the website column initially is null... I do not know what to do.... A lot of cnfusion I am wondering about the following problem: I have two sessions, one for the user ($_SESSION['user_id']) and second one for the products inside cart ($_SESSION['cart']). I need to insert data into table racun. I did it this way and it works, but i want to know if this can be done better? Code: [Select] while ($row=mysql_fetch_array($query)){ $id = $_SESSION['korisnik_id']; $idp = $row['product_id']; $kol = $_SESSION['cart'][$row['product_id']]['quantity'] ."<br>"; $insert = "INSERT INTO racun (product_id, quantity, korisnik_id) VALUES ('$idp', '$kol', '$id') "; $result = mysql_query($insert); } hi; Code: [Select] $giden = $_GET['sehir_part']+1; $url = '31.php?sehir_part='.$giden.''; //$erkek_top = 7 for($q=1;$q<=7;$q++) { //i want insert into to mysql in here ; but its doing only one inserting so i want be 7 more insert } header( 'refresh:1;url=31.php?sehir_part='.$giden)and in every page loading i want inserting to mysql againg pls helpme thanks Hey, who can help? Example: I want to insert a row with a input and a select option. I am inserting now with select option, but i want to insert with a input too, because if i need to insert a link instead of choosing from select option. --------- (puts a link)Link: ...link... (chooses nothing)Select: Choose one... Submit --------- I need something like if LINK is filled insert it. Nothing happens to SELECT, but they have to have the same row names. Hope its clear enough. Cheers hey developers,
i have problem to get last insert id. btw.. i try to put that code on my mvc creation.
on register_model page i write like this:
public function numbering(){ $sth = $this->db->prepare('SELECT * FROM member_data'); $result = $this->db->lastInsertId(); return $result; } and for the controllers page, i write the code like this: $c = $this->model->numbering(); foreach ($c as $key => $value) { $data = $result('id'); } $abc['member_id'] = $data; but i didn't get the results,.. can you help me? Not being able to upload images in the directory with a path name in the database? 1) This is my code for insert into the database and directory:
<?php
$host="localhost";
$username="root";
$pass="";
$db="registration";
$conn=mysqli_connect($host,$username,$pass,$db);
if(!$conn){
die("Database connection error");
}
// insert query for register page
if(isset($_POST['ronel'])){
$images = $_FILES['file']['name'];
$target_dir = "uploads/";
$target_file = $target_dir . basename($_FILES["file"]["name"]);
// Select file type
$imageFileType = strtolower(pathinfo($target_file,PATHINFO_EXTENSION));
// Valid file extensions
$extensions_arr = array("jpg","jpeg","png","gif","pdf");
// Check extension
if( in_array($imageFileType,$extensions_arr) )
{
$details=$_POST['details'];
$location=$_POST['location'];
$checkbox=$_POST['checkbox'];
$injured=$_POST['injured'];
$agegender=$_POST['agegender'];
$contact=$_POST['contact'];
$empid=$_POST['empid'];
$dept=$_POST['dept'];
$organization=$_POST['organization'];
$summary=$_POST['summary'];
$name=$_POST['name'];
$outcome=$_POST['outcome'];
$cause=$_POST['cause'];
$action=$_POST['action'];
$reportedname=$_POST['reportedname'];
$position=$_POST['position'];
$organisation=$_POST['organisation'];
$reportedcontact=$_POST['reportedcontact'];
$reporteddept=$_POST['reporteddept'];
$status="Pending";
$comment=$_POST['comment'];
$query="INSERT INTO `proposals` (`details`,`location`,`date`,`time`,`checkbox`,`injured`,`agegender`,`contact`,`empid`,`dept` ,`organization`,`summary`,`image`,`outcome`,`cause`,`action`,`reportedname`,`position`,`organisation`,`reportedcontact`,`reporteddept`,`status`,`comment`) VALUES ('$details','$location', current_timestamp(),current_timestamp(),'$checkbox','$injured','$agegender','$contact','$empid','$dept' ,'$organization','$summary','$name','$outcome','$cause','$action','$reportedname','$position','$organisation','$reportedcontact','$reporteddept','$status','$comment')";
$res=mysqli_query($conn,$query);
if($res){
$_SESSION['success']="Not Inserted successfully!";
header('Location:');
}else{
echo "<script>alert('Proposal not applied!');</script>";
}
// Upload file
move_uploaded_file($_FILES['file']['tmp_name'],$target_dir.$image);
}
}
date_default_timezone_set("Asia/Kolkata");
?>
2) Here is the input file:
<form class="form-horizontal" method="post" action="" enctype="multipart/form-data">
<input type="hidden" name="ronel" value="">
<div class="form-group">
<label style="position:absolute; left:63%; top:425px;" for="inputEmail" class="col-lg-3"><b>Upload Images Here :</b></label><br><br>
<div class="col-lg-9">
<input style="position:absolute; left:78%; top:420px;" type="file" name="file" enctype="multipart/form-data" class="form-control" name="incident_reference" onchange="document.getElementById('inc_ref').src = window.URL.createObjectURL(this.files[0]); document.getElementById('inc_ref').className +='_active'; document.getElementById('inc_ref_span').className += '_hidden'">
</div><iframe id="inc_ref" class="form-group" width="220px" height="130px" style="position:absolute; left:78%; top:32%;"></iframe></div>
</form>
3) error code: Notice: Undefined index: name in /opt/lampp/htdocs/create-nearmiss.php on line 57 ONGC TRIPUR Edited March 28, 2020 by Ronel change of var 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 'like (comment_id, name, like) VALUES ('102','Steven','1')' at line 1 $sql="INSERT INTO like (comment_id, name, like) VALUES ('$_POST[comment_id]','$_POST[name]','$_POST[like]')"; if (!mysql_query($sql,$con)) { die('Error:' . mysql_error()); } echo "You have succesfully liked this comment.<br /><INPUT TYPE=\"button\" VALUE=\"Back\" onClick=\"history.go(-1);\">"; mysql_close($con) Table: comment_id name like For some reason, I cannot spot this error. Can you find it? My $con = mysql_connect is a part I left out but it connects fine! The data form is echo'ing fine but it's not inserting into the MySQL Table. Thanks in Advanced! I get the following message: Quote name4Query failed: Unknown column 'late' in 'field list' with this code. What does it mean? Code: [Select] <?php $apt=$_POST['search_term']; $stat = mysql_connect("localhost","root",""); $stat = mysql_select_db("prerentdb"); $query = "SELECT name FROM payments WHERE late = 'L'"; $stat = @mysql_fetch_assoc(mysql_query($query)); echo $stat["name"]; $name=$_POST['name']; $apt=$_POST['apt']; $amtpaid=$_POST['amtpaid']; $rentdue=$_POST['rentdue']; $prevbal=$_POST['prevbal']; $hudpay=$_POST['hudpay']; $tentpay=$_POST['tentpay']; $datepaid=$_POST['datepaid']; $late=$_POST['late']; $comments=$_POST['comments']; $paidsum=$_POST['paidsum']; $query = " INSERT INTO payhist (name,apt,amtpaid,rentdue,prevbal, hudpay,tentpay,datepaid,late,comments,paidsum) VALUES('$name','$apt','$amtpaid','$rentdue','$prevbal', '$hudpay','$tentpay','$datepaid','$late','$comments','$paidsum')"; $stat = mysql_query($query) or die('Query failed: ' . mysql_error()); mysql_close(); echo "data inserted<br /><br />"; ?> |
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