I'm trying to insert received values into postgresql table using php. I can't figure out why this statement doesn't work
$query = "INSERT INTO user_info (name, emailAddress, phoneNumber, jobDesc) VALUES ('" . $name . "," . $emailAddr . "," . $phoneNumber . "," . $jobDesc ."')";
I get this error:
Query failed: ERROR: column "emailaddress" of relation "user_info" does not exist
However, I tried this one:
$query = "INSERT INTO user_info VALUES ('" . $name . "," . $emailAddr . "," . $phoneNumber . "," . $jobDesc ."')";
It works, but it inserts all values into first column!
I'm not sure what I'm missing here!
I think you are missing a whole host of single quotes in your VALUES list...
$query = "INSERT INTO user_info (name, emailAddress, phoneNumber, jobDesc) VALUES ('" . $name . "','" . $emailAddr . "','" . $phoneNumber . "','" . $jobDesc ."')";
Related
Problem
With a php website, I have a form to collect information which will then be inserted into the MySQL database, but there are these three columns that have the wrong values inserted into them. The rest are all in the correct order.
Values inserted as php variables via MySQL transaction.
Thank you for your time.
phpmyadmin display (first row is manually corrected)
Code:
<?php
function registerPatient($ptUsername, $ptPassword, $ptFirstName, $ptLastName, $ptSalutation, $ptEmail, $ptDOB, $ptPostCode, $ptHouseNo, $ptTelNo, $link)
{
$accType = "Patient";
$dtID = $_COOKIE["ID"];
$errors = "";
$SQL_patientInsert =
"START TRANSACTION;
INSERT INTO accDetails (`username`, `hashPassword`, `accType`)
VALUES ('" . $ptUsername . "',
'" . $ptPassword . "',
'" . $accType . "');
INSERT INTO ptProfile (`firstName`, `lastName`, `salutation`, `email`, `DOB`, `postCode`, `houseNo`, `telephoneNo`, `dtID`, `ptID`)
VALUES ('" . $ptFirstName . "',
'" . $ptLastName . "',
'" . $ptSalutation . "',
'" . $ptEmail . "',
'" . $ptDOB . "',
'" . $ptPostCode . "',
'" . $ptHouseNo . "',
'" . $ptTelNo . "',
'" . $dtID . "',
LAST_INSERT_ID());
COMMIT;";
if (mysqli_multi_query($link, $SQL_patientInsert)) {
$errors .= "";
} else {
$errors .= "MYSQL Error: ". mysqli_error($link);
}
return $errors;
}
?>
Var_Dump of $SQL_patientInsert
string(495) "START TRANSACTION; INSERT INTO accDetails (`username`, `hashPassword`, `accType`) VALUES ('bingbong', '$2y$10$WDvSHSxzIxaYB8dPGLRIWOFyIdPXxSw5JDXagOxeYuJUtnvFhI.lO', 'Patient'); INSERT INTO ptProfile (`firstName`, `lastName`, `salutation`, `email`, `DOB`, `postCode`, `houseNo`, `telephoneNo`, `dtID`, `ptID`) VALUES ('Dr', 'Bing', 'Bong', 'EMAIL REMOVED FOR SO', '1996-08-02', 'POSTCODE REMOVED FOR SO', '7', '83824', '1256', LAST_INSERT_ID()); COMMIT;"
Table Structure
Table Structure in PHPMyAdmin, no autoincrements, all values allowed to be null
Your are calling your function with wrong parameters order.
Change this line ($ptFirstName <-> $ptSalutation);
function registerPatient($ptUsername, $ptPassword, $ptFirstName, $ptLastName, $ptSalutation, $ptEmail, $ptDOB, $ptPostCode, $ptHouseNo, $ptTelNo, $link)
with
function registerPatient($ptUsername, $ptPassword, $ptSalutation, $ptFirstName, $ptLastName, $ptEmail, $ptDOB, $ptPostCode, $ptHouseNo, $ptTelNo, $link)
I think you just mixed up your variables somewhere. Have you checked the form? Try printing out all the variables right before you build the query and check if they correspond correctly.
I am trying to run an mySQL insert statement like so:
function insertAppointment($connection, $id, $firstname, $lastname, $email, $phone, $date, $time){
$sql = "INSERT INTO `appointments` (firstname, lastname, email, phone, app_date, app_time) VALUES ('" . $id . "', '" . $firstname . "', '" . $lastname . "', '" . $email . "', " . $date . ", " . $time . ")";
$connection->query($sql);
}
$connection is my connection string, which is not the problem. I am able to use it for select statement like so:
function getTakenDates($connection){
$query = mysqli_query($connection, "SELECT app_date, app_time FROM `appointments`");
$results = array();
while($row = mysqli_fetch_assoc($query)){
$results[] = $row;
}
return $results;
}
You are vulnerable to SQL injection attacks, and are creating an incorrect query with your $date/$time values:
INSERT .... VALUES (..., 2014-11-10, 14:58:00)
since your date value is unquoted, you'll actually be trying to insert the result of that math operation (remember - is SUBTRACTION if it's not in a string), and 14:58:00 is a totally invalid number - mysql has no idea what those : chars are.
You want
$sql = "[..snip..] "', '" . $date . "', '" . $time . "')";
^-------------^--^-------------^----
instead. note the extra quotes. That'll produce
INSERT .... VALUES (..., '2014-11-10', '14:58:00')
I am trying to get POST'ed form variables and mySQL is throwing an error when trying to insert them. I can't figure for the life of me why. Hopefully someone can help out.
function submitFound(){
global $dbc;
if ($_SERVER['REQUEST_METHOD'] == 'POST'){
$query =
"INSERT INTO found1
(fname, lname, email, phone, name, color, make, model, sizes, info, location)
VALUES
(" .
mysql_real_escape_string($_POST['fname']) . "," .
mysql_real_escape_string($_POST['lname']) . "," .
mysql_real_escape_string($_POST['email']) . "," .
mysql_real_escape_string($_POST['phone']) . "," .
mysql_real_escape_string($_POST['name']) . "," .
mysql_real_escape_string($_POST['color']) . "," .
mysql_real_escape_string($_POST['make']) . "," .
mysql_real_escape_string($_POST['model']) . "," .
mysql_real_escape_string($_POST['size']) . "," .
mysql_real_escape_string($_POST['info']) . "," .
mysql_real_escape_string($_POST['location']). ")";
$results = mysqli_query($dbc, $query);
check_results($results);
//return $mysqli_insert_id($dbc);
mysqli_free_result($results);
}
}
That is the function that is submitting the information. (Generic information about an item. This is the mySQL error getting thrown.
MySQL 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 '#gmail.com,444-444-4444,Book,#000040,NA,NA,Large,Nothing special,Byrne House)' at line 4
This is what I input in the form. http://puu.sh/8l35Z.png
So yeah, any help would be great. Not sure if it is just something stupid. My eyes are starting to cross :P
Thanks in advance.
EDIT*
Fixed the Strings but am still getting an error:
New Code:
$query =
"INSERT INTO found1
(fname, lname, email, phone, name, color, make, model, sizes, info, location)
VALUES
('" .
mysql_real_escape_string($_POST['fname']) . "','" .
mysql_real_escape_string($_POST['lname']) . "','" .
mysql_real_escape_string($_POST['email']) . "','" .
mysql_real_escape_string($_POST['phone']) . "','" .
mysql_real_escape_string($_POST['name']) . "','" .
mysql_real_escape_string($_POST['color']) . "','" .
mysql_real_escape_string($_POST['make']) . "','" .
mysql_real_escape_string($_POST['model']) . "','" .
mysql_real_escape_string($_POST['size']) . "','" .
mysql_real_escape_string($_POST['info']) . "','" .
mysql_real_escape_string($_POST['location']). "')'";
MySQL error: MySQL 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 4
As John Conde mentioned, you need to add quotes around your values, your error message shows this.
#gmail.com,444-444-4444,Book,#000040,NA,NA
Notice the missing quotes, it should look like this
'#gmail.com','444-444-4444','Book','#000040','NA','NA'
Try this
VALUES
('" .
mysql_real_escape_string($_POST['fname']) . "','" .
mysql_real_escape_string($_POST['lname']) . "','" .
mysql_real_escape_string($_POST['email']) . "','" .
For example:
array('u_ad'=>'example name','u_mail'=>'example#mail.com','u_sifre'=>'exapmlepass')
Required query:
$sql = "INSERT INTO uyeler
(u_ad,u_mail,u_sifre)
VALUES
('example name','example#mail.com','examplepass')";
How I do that?
$sql = "INSERT INTO uyeler (". implode(",", array_keys($array)) .") VALUES ('". implode("','", $array) ."')";
Quick/dirty/unsafe:
$sql = "INSERT INTO uyeler (u_ad,u_mail,u_sifre) VALUES ('" . $theArray['u_ad'] . "','" . $theArray['u_mail'] . "','" . $theArray['u_sifre'] . "')";
Better:
$ad = mysql_real_escape_string($theArray['u_ad']);
$mail = mysql_real_escape_string($theArray['u_mail']);
$sifre = mysql_real_escape_string($theArray['u_sifre']);
$sql = "INSERT INTO uyeler (u_ad,u_mail,u_sifre) VALUES ('" . $ad . "','" . $mail . "','" . $sifre . "')";
Don't mess around with escaping! You should be using prepared statements where possible, and using PDO is a good way to do it.
See:
Why you Should be using PHP’s PDO for Database Access
ext/mysqli: Part I - Overview and Prepared Statements
I'm looking to use SELECT LAST_INSERT_ID()
Am using a form to have a user input values. With the first insert I need to get the last inserted id for the next insert... I have not figured out how to get the last selected id and then pass it into my 2nd insert statement
I have updated my code though I still can not get the id to post into the table
include("config.inc.php");
$link = mysql_connect($db_host,$db_user,$db_pass);
if(!$link) die ('Could not connect to database: '.mysql_error());
mysql_select_db($db_name,$link);
$query = "INSERT into `".$db_table."` (producer_id,series_id,lang_id,title_name,title_public_access) VALUES ('" . $_POST['producer_id'] . "','" . $_POST['series_id'] . "','" . $_POST['lang_id'] . "','" . $_POST['title_name'] . "','" . $_POST['title_public_access'] . "')";
$last_id = mysql_insert_id();
$query = "INSERT into `".$db_table2."` (seg_id, file_video_UNC,file_video_URL) VALUES ('" . '$last_id' . "','" . $_POST['file_video_UNC'] . "','" . $_POST['file_video_URL'] . "')";
mysql_query($query);
mysql_close($link);
There's a function for that, called mysql_insert_id().
... first query here ...
$last_id = mysql_insert_id();
$sql = "INSERT INTO $db_table SET
file_video = " . $_POST['file_video_UNC'].",
file_video_URL = " . $_POST['file_video_URL'] . ",
insert_id_of_first_query = $last_id";
...
Your updated code doesn't send the query to database - as a result no INSERT, so no LAST_INSERT_ID
$query = "INSERT into ".$db_table."
(producer_id,series_id,lang_id,title_name,title_public_access) VALUES
('" . $_POST['producer_id'] . "','"
. $_POST['series_id'] . "','"
. $_POST['lang_id'] . "','" . $_POST['title_name'] . "','"
. $_POST['title_public_access'] . "')";
mysql_query($query); /* YOU FORGOT THIS PART */
$last_id = mysql_insert_id();
You can't just dump a query into a string on its own in a line of PHP. You should have used LAST_INSERT_ID() inside your second query or, better, use PHP's mysql_insert_id() function which wraps this for you in the API.
In the line:
$query = "INSERT into `".$db_table2."` (seg_id, file_video_UNC,file_video_URL) VALUES ('" . '$last_id' . "','" . $_POST['file_video_UNC'] . "','" . $_POST['file_video_URL'] . "')";
I think VALUES ('" . '$last_id' . "', should just be VALUES ('" . $last_id . "', without the single quotes around the variable.