following trigger was working fine, but nows after change in table and schema gives error:
DROP TRIGGER IF EXISTS `invite`;
DELIMITER //
CREATE TRIGGER `invite` AFTER INSERT ON `Invite_page`
FOR EACH ROW BEGIN
Insert into userpost(userid,url,title,preview,sentiment,time) values(NEW.userid,NEW.url,NEW.title,NEW.preview,NEW.sentiment,NEW.time);
Insert into urlcontent(url,title,preview,sentiment,time) values(NEW.url,NEW.title,NEW.preview,NEW.sentiment,NEW.time);
END
//
DELIMITER ;
Error is:
Error: Unknown column 'userid' in 'NEW'
Table contains columns correctly:
PHP code:
<?php
$url=$_POST['url'];
$id=$_POST['id'];
$title=$_POST['title'];
$preview=$_POST['preview'];
$sentiment=$_POST['sent'];
$con = mysqli_connect('127.0.0.1', 'root', '', 'karim');
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
return;
}
$date = strtotime(date("Y-m-d H:i:s"));
//echo $datel
$insertQuery1 = "INSERT INTO invite_page(`userid`, `url`, `title`, `preview` ,`sentiment`,`time`) VALUES ('".$id."','".$url."','".$title."','".$preview."','".$sentiment."','".$date."')";
//$insertQuery2 = "INSERT INTO userpost(`url`, `title`, `preview` ,`sentiment`,`time`) VALUES ('".$url."','".$title."','".$preview."','".$sentiment."')";
if (!mysqli_query($con,$insertQuery1))
{
die('Error: ' . mysqli_error($con));
}
echo "Record added successful ..";
?>
where is the issue I could not find
Trigger and everything is correct.
Error was due to some another trigger existing in the same schema. I was not aware about.
Apologize for inconvenience.
Related
The code below is for a project I'm working on. I'm having some issues with my PHP code and am in need of help.
I have a button on my data table named "Export". When the button is clicked, I wish to copy the data on that row and move it to an archive.
<?php
function val($data) {
$data = trim($data);
$data = stripslashes($data);
$data = htmlspecialchars($data);
return $data;
}
$servername = "localhost";
$username = "username";
$password = "password";
$dbname = "databasename";
$ticket_id = $_GET["ticket_id"];
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT * FROM activeticket WHERE ticket_id='$ticket_id' INSERT INTO `ticketarchive`(`name`, `account_num`, `department`, `ticket_desc`, `email`, `assigned`, `status`, `fibre_site`) VALUES ([name],[account_num],[department],[ticket_desc],[email],[assigned],[status],[fibre_site])";
if ($conn->query($sql) === TRUE) {
echo "New record created successfully. Record ID is: ";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
$conn->close();
?>
Below is the error that this produces:
Error: SELECT * FROM activeticket WHERE ticket_id='1' INSERT INTO
archiveticket(name, account_num, department, ticket_desc,
email, assigned, status, fibre_site) VALUES
([name],[account_num],[department],[ticket_desc],[email],[assigned],[status],[fibre_site])
You have an error in your SQL syntax; check the manual that
corresponds to your MariaDB server version for the right syntax to use
near 'INSERT INTO archiveticket(name, account_num, department,
ticket_desc, `' at line 1
First execute the select query and after that execute the insert one. Right now you are trying to run them both and this is the problem.
This isn't really a PHP issue you're having, you just seem to be unfamiliar with SQL.
What you're trying to do is insert the result of a SELECT query into a table. This isn't the way to do it at all.
What you're looking for is :
$sql = "INSERT INTO `ticketarchive`(
`name`,
`account_num`,
`department`,
`ticket_desc`,
`email`,
`assigned`,
`status`,
`fibre_site`
)
SELECT
`name`,
`account_num`,
`department`,
`ticket_desc`,
`email`,
`assigned`,
`status`,
`fibre_site`
FROM
`activeticket`
WHERE
`ticket_id` = $ticket_id"
For more information, read here.
I'd also advice you look into parametrized queries to avoid SQL injections.
I'm stuck with a syntax error for a SQL query to insert data into a table.
Ah, the syntax error, most useless of all errors!!
Using the code modified from PHP Insert Data Into MySQL using both mysqli and PDO methods.
e.g.:
<?php
$servername = "localhost";
$username = "4w_write";
$password = "GjByhJzrQueHgTzw";
$dbname = "4w_test";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "INSERT INTO 4w (email) VALUES ($email)";
if ($conn->query($sql) === TRUE) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
$conn->close();
?>
Having stripped it down to a single variable which isn't using a keyword, I'm pretty sure the problem is with my table.
SQL Query:
INSERT INTO 4w (email) VALUES (myemail#gmail.com)
Error:
Error: INSERT INTO 4w (email) VALUES (myemail#gmail.com) You have an
error in your SQL syntax; check the manual that corresponds to your
MariaDB server version for the right syntax to use near '#gmail.com)'
at line 1
SQL table (4w):
# Name Type Default
1 id [Primary,Index] int(11)
2 email varchar(255)
3 whatIs tinytext
4 whereIs text
5 whattodo text
6 imageURL text
7 whenRep timestamp CURRENT_TIMESTAMP
The email value is a string, so you need to surround it with quotes:
$sql = "INSERT INTO 4w (email) VALUES ('$email')";
Or, better yet, use a prepared statement and bind it's value.
Couldn't see the problem, missing quotes....
Original code:
$sql = "INSERT INTO 4w (email) VALUES ($email)";
Fixed code:
$sql = "INSERT INTO 4w (email) VALUES ('$email')";
I get the message that the new record was created but when I reload phpmyadmin the table is the same. Also I have retrieved information from the same DB,
from the same table, with SELECT command, so the connection works..(plainly said). I have no clue why is not updating. Please help. Thank you in advance.
<html>
<head>
</head>
<body>
<?php
define('DB_NAME', 'appointments');
define('DB_USER', 'root');
define('DB_PASSWORD', '');
define('DB_HOST', 'localhost');
$link = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME);
if (!$link) {
die('Could not connect: ' . mysql_error());
}
$hos=$_POST['hos'];
echo $hos;
echo "<br/>";
$doc=$_POST['doc'];
echo $doc;
$date=$_POST['fdate'];
echo $date;
$time=$_POST['time'];
echo $time;
$pat=5;
echo $pat;
$sql = "INSERT INTO rantevou ('app_id','patient_id','date','time','hos','doc') VALUES ('4','$pat','$date','$time','$hos','$doc');";
if ($sql) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
mysqli_close($link);
?>
</body>
</html>
There are many mistake in your code
1. use of mysql_error()
you can't use mysql_error because you use mysqli for data base connection.second thing mysql is no more supported
Solution use mysqli_error($link);
2. use of $conn->error
You can't us of $conn->error beacuse you connect with mysqli procedure way not like object oriented way and you also not define a $conn instead you used $link
Solution use mysqli_error($link);
Correct Code
if(!mysqli_query($link, $sql)){
printf("Errormessage: %s\n", mysqli_error($link));
die;
}else{
echo "New record created successfully";
}
Why Data Not Inserted
because you declare variable $sql but you didn't executed that
the new record was created
You get this message all ways because your if condition check that variable have a value (not 0) and yes $sql have value
1.You must use prepare statement,if you don't wan't any sql injection in insert statement SQL INJECTION
2.'' single quote or "" apply only on a string not on id if your app_id is a int don't use ('' or "") quote instead of that convert '4' to int
3.handle error log https://stackoverflow.com/a/3531852/3234646
4.Please clear Concept use of Database Extension
http://php.net/manual/en/class.mysqli.php
You forgot to execute the query, if ($sql) { merely evaluates the variable.
if (mysqli_query($link, $sql)) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
Also, you need to use backticks for SQL-related variables, not single quotes:
$sql = "INSERT INTO rantevou (`app_id`,`patient_id`,`date`,`time`,`hos`,`doc`) VALUES ('4','$pat','$date','$time','$hos','$doc');";
You're not actually executing your query. If you add the line $result = mysqli_query($link, $sql); after declaring $sql you will execute the query.
You can then assess whether it worked using the same if, but change that line to be
if ($result) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . mysqli_error($link);
}
In the above example, I have also changed your error reporting as it was referencing $conn, a variable you had not declared before. It now uses the same $link variable as the rest of your code.
Also, I would highly recommend escaping your data since you're inserting the contents of posted data. Escaping your data will help protect against SQL Injection. It's not comprehensively safe, but it's a good start.
To add in escaping, change each $var = $_POST['var'] line to read $var = mysqli_real_escape_string($link, $_POST['var']);
For example, $hos=$_POST['hos']; becomes $hos = mysqli_real_escape_string($link, $_POST['hos']);
This helps prevent moments like this wonderful example by XKCD
1) Remove single quotes (') from column name to backtick (`)
2) Execute your query. You didn't executed.
3) If app_id column is auto incremented and primary key. Then, no need to pass value. Leave it blank.
<?php
$sql = "INSERT INTO rantevou (`app_id`,`patient_id`,`date`,`time`,`hos`,`doc`) VALUES ('','$pat','$date','$time','$hos','$doc');";
$query = mysqli_query($link,$sql) ;
if ($query) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
Instead of
"INSERT INTO rantevou ('app_id','patient_id','date','time','hos','doc') VALUES ('4','$pat','$date','$time','$hos','$doc');"
unquote the columns
"INSERT INTO rantevou (app_id, patient_id, date, time, hos, doc) VALUES ('4','$pat','$date','$time','$hos','$doc');"
or use backticks
"INSERT INTO rantevou (`app_id`, `patient_id`, `date`, `time`, `hos`, `doc`) VALUES ('4','$pat','$date','$time','$hos','$doc');"
you've forgot to execute your query
mysqli_execute($con, "INSERT INTO rantevou (`app_id`, `patient_id`, `date`, `time`, `hos`, `doc`) VALUES ('4','$pat','$date','$time','$hos','$doc')");
EDIT: What luweiqi said: the statement has yet to be executed!
It seems like you know what you are doing. Are you sure that the paramaters here:
$sql = "INSERT INTO rantevou (**'app_id','patient_id','date','time','hos','doc'**) VALUES ('4','$pat','$date','$time','$hos','$doc');";
if ($sql) {
exactly match your column titles in your database?
Another good way to check your statements, is to go to phpmyadmin and go to the SQL notepad and enter the query with the same structure and see what is being returned.
Your query may be returning a message, but a message saying that it has failed... which would still trigger your echo "New record created successfully";
This is how i've structured my most recent insert to DB:
<?php
// to get data from android app
$gardenID=$_POST["gardenID"];
$vID=$_POST["vID"];
$quantity = $_POST["quantity"];
$timePlanted = date("Y/m/d");
// establishes connection to database
require "init.php";
echo "here";
echo $timePlanted;
echo $quantity;
$query = "insert into garden_veg (gardenID, vID, quantity, timePlanted) values ('".$gardenID."','".$vID."',
'".$quantity."', '".$timePlanted."' );";
$result = mysqli_query($con,$query);
$response = array();
$code = "addItem_success"; //changed code
$message = "Item(s) added!";
array_push($response,array("code" => $code, "message"=>$message));
echo json_encode(array("server_response"=>$response));
mysqli_close($con);
?>
First of all, don't use single quotes for column names, either use nothing or use backticks.
Secondly, you forgot to execute the query.
Also, using OOP is better.
Please try:
$mysqli = new mysqli(DB_HOST,DB_USER,DB_PASSWORD,DB_NAME);
and
$query = "INSERT INTO rantevou (app_id,patient_id,date,time,hos,doc) VALUES ('4','$pat','$date','$time','$hos','$doc');";
if ($mysqli->query($query)) echo "New record created";
else echo "Error: ".$mysqli->error;
Hello i have a problem with my query ill keep getting errors from my query
this is my error;
Error: BEGIN; INSERT INTO our_work (id) VALUES ('6'); INSERT INTO
our_work_portf_img (portf_id, img_id) VALUES ('6', '7'); INSERT
INTO our_work_images (img_id, image) VALUES ('7', 'adawd.jpg');
COMMIT; 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 'INSERT INTO our_work (id) VALUES ('6'); INSERT INTO `our_wo'
at line 3
i've tried many things but i noticed one thing if i copy the $query string and i posted the query directly in mysql the problem will not accorded and it works just how i hoped it would.
Does anyone noticed the problem in my query cause im literal out of ideas.
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
if(isset($_POST['submit_new_img'])){
$pjt_dtls = $_POST['project_details'];
$categories = $_POST['categories'];
$link = $_POST['link'];
$image_path = "adawd.jpg";//$_POST['file']; //$_POST['image'];
$row_id ='6';//++$num_rows['i'];
$image_id ='7'; //++$num_rows['ii'];
$sql = "
BEGIN;
INSERT INTO `our_work`
(`id`)
VALUES
('{$row_id}');
INSERT INTO `our_work_portf_img`
(`portf_id`, `img_id`)
VALUES
('{$row_id}', '{$image_id}');
INSERT INTO `our_work_images`
(`img_id`, `image`)
VALUES
('{$image_id}', '{$image_path}');
COMMIT;
";
if ($conn->query($sql) === TRUE) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
$conn->close();
}
$conn->query($sql) does not work with multi-query like yours
you need to use multi_query instead
also here is nice comment:
Please note that there is no need for the semicolon after the last
query. That wasted more than hour of my time...
I'm getting a non-descriptive syntax error on a MYSQL query from PHP. If I "echo" the text of the query and paste it into a MySQL query window, the code works. Here is the SQL for the query, the error code, and the error message...
INSERT INTO ADVERTISEMENTS (`user_id`, `ad_name`, `click_url`, `img_url`, `bg_color`, `start_date`, `end_date`, `timer_delay`, `add_date`) VALUES (2, 'Test New Ad', 'http://www.google.com', 'red_arrow.png', '#000000', '1980-05-11 00:00:00', '2020-05-01 00:00:00', 5, '2013-07-14 22:21:59');
Error Code: 1064
Error Msg: 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
Here is the PHP code I am using...
$link = mysqli_connect($UM_Settings["database_options"]["server_name"], $UM_Settings["database_options"]["username"], $UM_Settings["database_options"]["password"], $UM_Settings["database_options"]["database_name"]);
$advertisementNameNew = mysqli_real_escape_string($link, $_POST['advertisementNameNew']);
$destinationURLNew = mysqli_real_escape_string($link, $_POST['destinationURLNew']);
$dropboxUploadFile = mysqli_real_escape_string($link, $_POST['dropboxUploadFile']);
$backgroundColorNew = mysqli_real_escape_string($link, $_POST['backgroundColorNew']);
$bannerStartDateNew = DateStringToMySQL($_POST['bannerStartDateNew']);
$bannerEndDateNew = DateStringToMySQL($_POST['bannerEndDateNew']);
$bannerSetTimerNew = intval($_POST['bannerSetTimerNew']);
$tmpUserID = UM_GetCookie("UM_UserID");
$tmpAddDate = DateStringToMySQL('now');
echo "INSERT INTO ADVERTISEMENTS(`user_id`, `ad_name`, `click_url`, `img_url`, `bg_color`, `start_date`, `end_date`, `timer_delay`, `add_date`) VALUES ($tmpUserID, '$advertisementNameNew', '$destinationURLNew', '$dropboxUploadFile', '$backgroundColorNew', '$bannerStartDateNew', '$bannerEndDateNew', $bannerSetTimerNew, '$tmpAddDate');<br />";
if (!mysqli_query($link, "INSERT INTO ADVERTISEMENTS(`user_id`, `ad_name`, `click_url`, `img_url`, `bg_color`, `start_date`, `end_date`, `timer_delay`, `add_date`) VALUES ($tmpUserID, '$advertisementNameNew', '$destinationURLNew', '$dropboxUploadFile', '$backgroundColorNew', '$bannerStartDateNew', '$bannerEndDateNew', $bannerSetTimerNew, '$tmpAddDate');")) {
printf("Error Code: %s\n", mysqli_errno($link));
echo "<br />";
printf("Error Msg: %s\n", mysqli_error($link));
}
I know that the database connection is working. I am able to select and update tables. I can also insert into other tables with different queries.
I am open to any suggestions.
Thank you in advance for your help!
I see a few errors in your query strings.
First, all your variables are passed as literal strings: "... VALUES ($tmpUserID, '$advertisementNameNew', ..." should be "... VALUES (".$tmpUserID.", '".$advertisementNameNew."', ...".
Second, I see missing quotes around $bannerSetTimerNew.
Third, there is an extra ;.
here's how I would write the query:
if (!mysqli_query($link, "INSERT INTO ADVERTISEMENTS (user_id, ad_name, click_url, img_url, bg_color, start_date, end_date, timer_delay, add_date) VALUES (".$tmpUserID.", '".$advertisementNameNew."', '".$destinationURLNew."', '".$dropboxUploadFile."', '".$backgroundColorNew."', '".$bannerStartDateNew."', '".$bannerEndDateNew."', '".$bannerSetTimerNew."', '".$tmpAddDate."')")) { ...
I didnt test it though.
hope this helps.
I see a ; at the end of the query. Are you sure that should be there?
There are two things
1. Remove the ; from at the end of the query.
2. I hope timer_delay field has datatype "Int" if its a VARCHAR then you will have to include quotes for that field value.
I hope this will help.
Passerby, thank you for your comment. This was my first experience with using mysqli, I changed my query to use the "bind_param" method, and everything works now. For anyone else with a similar problem, here is the corrected code...
$mysqli = new mysqli($UM_Settings["database_options"]["server_name"], $UM_Settings["database_options"]["username"], $UM_Settings["database_options"]["password"], $UM_Settings["database_options"]["database_name"]);
if ($mysqli->connect_errno) {
echo "Failed to connect to MySQL: (" . $mysqli->connect_errno . ") " . $mysqli->connect_error;
}
$advertisementNameNew = $_POST['advertisementNameNew'];
$destinationURLNew = $_POST['destinationURLNew'];
$dropboxUploadFile = $_POST['dropboxUploadFile'];
$backgroundColorNew = $_POST['backgroundColorNew'];
$bannerStartDateNew = DateStringToMySQL($_POST['bannerStartDateNew']);
$bannerEndDateNew = DateStringToMySQL($_POST['bannerEndDateNew']);
$bannerSetTimerNew = intval($_POST['bannerSetTimerNew']);
$tmpUserID = UM_GetCookie("UM_UserID");
$tmpAddDate = DateStringToMySQL('now');
/* Prepared statement, stage 1: prepare */
if (!($stmt = $mysqli->prepare("INSERT INTO `ADVERTISEMENTS` (`user_id`, `ad_name`, `click_url`, `img_url`, `bg_color`, `start_date`, `end_date`, `timer_delay`, `add_date`) VALUES (?,?,?,?,?,?,?,?,?)"))) {
echo "Prepare failed: (" . $mysqli->errno . ") " . $mysqli->error;
}
if (!$stmt->bind_param("issssssis",$tmpUserID, $advertisementNameNew, $destinationURLNew, $dropboxUploadFile, $backgroundColorNew, $bannerStartDateNew, $bannerEndDateNew, $bannerSetTimerNew, $tmpAddDate)) {
echo "Binding parameters failed: (" . $stmt->errno . ") " . $stmt->error;
}
if (!$stmt->execute()) {
echo "Execute failed: (" . $stmt->errno . ") " . $stmt->error;
}
$_GET['ad_id'] = $stmt->insert_id;
$stmt->close();