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Closed 9 years ago.
In the code below I am erring out with a T_CONSTANT_ENCAPSED_STRING error
$query = "INSERT INTO Accounts (FirstName,LastName,Email,Phone,salt,passwd,City,State)
VALUES ('"
. $_POST['FirstName'] . "','"
. $_POST['LastName'] . "','"
. $_POST['Email'] . "','"
. $_POST['Phone'] . "','"
. $salt . "','"
. hashPasswd($_POST['passwd'], $salt "','"
. ($_POST['City']
. "','"
. ($_POST['State'] . "'
);
Edit: Thanks for all the help on this question everyone. Sorry about the poor question, I was new to S/O at the time and didn't read up on how to ask a good question on S/O, which I know am much more familiar with. Again, thanks for the patience.
Try this . You were missing the " at the end to close the Insert ". I have broke the lines so that you can identify the missing quotes easily
$query = "INSERT INTO Accounts (FirstName,LastName,Email,Phone,salt,passwd,City,State)
VALUES ('". $_POST['FirstName'] . "',
'". $_POST['LastName'] . "',
'". $_POST['Email'] . "',
'". $_POST['Phone'] . "',
'". $salt . "',
'". hashPasswd($_POST['passwd'], $salt "',
'". ($_POST['City']. "',
'". ($_POST['State'] . "'
)";
You are missing a " at the end of line:
. ($_POST['State'] . "'"
and three closing )'s and a .:
. hashPasswd($_POST['passwd'], $salt "','")
. ($_POST['City'])
. "','"
. ($_POST['State'] . "'")
)
Cleand up the last three lines should look like:
. hashPasswd($_POST['passwd'], $salt) ."','"
. $_POST['City'] . "','"
. $_POST['State'] . "'"
)
Related
i'm using php to insert URLs into oracle.
i'm having a problem inserting urls in this format in the column ENDPOINT : http://lx2939:37080/Cougar/prod/api
it's saved in this format in the database: http://lx2939?/Cougar/prod/api
This is the PHP code i'm using
$query = "INSERT INTO RECENTTESTS
(EMAIL, FILENAME, SOR, RECIPIENTS, LAUNCHALL,
ENDPOINT, ORGOID, ASSOCIATEOID, ROLECODE, REALM, CONSUMERAPPOID)
VALUES ('" . $email . "', '" . $fileNameInput . "', '" . $sor . "',
'" . $recipients . "', '" . $launchAll . "','" . $endpoint . "',
'" . $orgOID . "','" . $associateOID . "','" . $rolecode . "',
'" . $realm . "','" . $consumeOID . "')";
$stid = oci_parse($conn, $query);
oci_execute($stid);
Could you please help me with this issue?
PS : When i execute the query directly into oracle it works as expected
You could fix 2 issues in one here.
Using a prepared parameterised query would almost definitely fix the issue you are having with the conversion of :37080 to a ? and also remove the likely SQL Injection Issues in your code
$query = "INSERT INTO RECENTTESTS
(EMAIL, FILENAME, SOR, RECIPIENTS, LAUNCHALL,
ENDPOINT, ORGOID, ASSOCIATEOID, ROLECODE, REALM, CONSUMERAPPOID)
VALUES (:EMAIL, :FILENAME, :SOR, :RECIPIENTS, :LAUNCHALL,
:ENDPOINT, :ORGOID, :ASSOCIATEOID, :ROLECODE, :REALM,
:CONSUMERAPPOID)";
$stid = oci_parse($conn, $query);
oci_bind_by_name($stid, ":EMAIL", $email);
oci_bind_by_name($stid, ":FILENAME", $fileNameInput);
oci_bind_by_name($stid, ":SOR", $sor);
oci_bind_by_name($stid, ":RECIPIENTS", $recipients);
oci_bind_by_name($stid, ":LAUNCHALL", $launchAll);
oci_bind_by_name($stid, ":ENDPOINT", $endpoint);
oci_bind_by_name($stid, ":ORGOID", $orgOID);
oci_bind_by_name($stid, ":ASSOCIATEOID", $associateOID);
oci_bind_by_name($stid, ":ROLECODE", $rolecode);
oci_bind_by_name($stid, ":REALM", $realm);
oci_bind_by_name($stid, ":CONSUMERAPPOID", $consumeOID);
oci_execute($stid);
You may also find it useful to read a little about catching and processing errors https://docs.oracle.com/cd/E17781_01/appdev.112/e18555/ch_seven_error.htm#TDPPH165
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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 ')'
$sql = "INSERT INTO
topics(topic_subject,
topic_date,
topic_cat,
topic_by)
VALUES('" . mysql_real_escape_string($_POST['topic_subject']) . "',
NOW(),
" . mysql_real_escape_string($_POST['topic_cat']) . ",
" . $_SESSION['fullname'] . "
)";
You are missing quotes around string values here is your new sql
ql = "INSERT INTO
topics(topic_subject,
topic_date,
topic_cat,
topic_by)
VALUES('" . mysql_real_escape_string($_POST['topic_subject']) . "',
NOW(),
'" . mysql_real_escape_string($_POST['topic_cat']) . "',
'" . $_SESSION['fullname'] . "'
)";
You need to put quotes around the value that are stored as VARCHAR, I guess $_SESSION['fullname'] in this case. Like this:
$sql = "INSERT INTO
topics(topic_subject,
topic_date,
topic_cat,
topic_by)
VALUES('" . mysql_real_escape_string($_POST['topic_subject']) . "',
NOW(),
" . mysql_real_escape_string($_POST['topic_cat']) . ",
'" . $_SESSION['fullname'] . "'
)";
Try the following:
You were missing single quotations.
$sql = "INSERT INTO
topics(topic_subject,
topic_date,
topic_cat,
topic_by)
VALUES('" . mysql_real_escape_string($_POST['topic_subject']) . "',
NOW(),
'" . mysql_real_escape_string($_POST['topic_cat']) . "',
'" . $_SESSION['fullname'] . "'
)";
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 ."')";
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']) . "','" .
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.