[My query wont insert into a single coloum in my phpmyadmin?]

You were told in the other topic to remove the WHERE from an an insert new record query

 

 

 

 

 

And you were given the syntax for an UPDATE query

 

Yet you still persist and complain about syntax errors

 

It isn't that you did not get the answer you wanted, it is more that you chose to ignore the answers.

 

 

okay that did help me mix my code up but now im getting this error??

 

error inserting new recordUnknown column 'TEST' in 'field list'

 

this happens when i enter information and press enter?

[My query wont insert into a single coloum in my phpmyadmin?]

This question is covered in your other topic

 

http://forums.phpfreaks.com/topic/277493-using-query-does-not-insert-into-the-phpmyadmin-table/?do=findComment&comment=1427501

 

so why have you started another topic with the same problem?

it not covered as i didnt get the answer i wanted?

[My query wont insert into a single coloum in my phpmyadmin?]

 

How is your table defined?

SHOW CREATE TABLE forumposts

 

oh sorry i posted in wrong code that code works perfectly:

 

have a look at this one please

 

 

if (isset($_POST['submitted'])) {

    

    

    $name = $_POST['entername'];

    $usernameorguest =$_POST['usernameorguest'];

    $reply =$_POST['reply'];

    $sqlinsert = "UPDATE `forumposts`(`REPLY`) VALUES('$reply') WHERE `POSTID`=1";

 

$query = mysql_query($sqlinsert);

    

    if (!$query) {

        echo "error inserting new record" . mysql_error(); 

    }

    

} // end of mani if statment 

 

 

//$newrecord ="1 record added to the database";

 

 

my table is defined all correctly not posted the connection code as i didnt think that was relevant 

[My query wont insert into a single coloum in my phpmyadmin?]

try echo $sqlinsert;

 

you forgot the e-mail variable on the insert.

oh sorry i posted in wrong code that code works perfectly:

 

have a look at this one please

 

 

if (isset($_POST['submitted'])) {

    

    

    $name = $_POST['entername'];

    $usernameorguest =$_POST['usernameorguest'];

    $reply =$_POST['reply'];

    $sqlinsert = "UPDATE `forumposts`(`REPLY`) VALUES('$reply') WHERE `POSTID`=1";

 

$query = mysql_query($sqlinsert);

    

    if (!$query) {

        echo "error inserting new record" . mysql_error(); 

    }

    

} // end of mani if statment 

 

 

//$newrecord ="1 record added to the database";

 

 

 

 

 

?>

 

i know i have missed a couple of vaiables out but i am just testing it to make sure it works first? 

 

any ideas

[My query wont insert into a single coloum in my phpmyadmin?]

hey guys rite i have created a website and i am trying to insert a indivual piece of data into 1 column of my phpmyadmin page table. But i am doing a TV show website and just creating a easy and basic forumn.. i am trying to insert a reply code which when you enter the details in the reply box and sumit it insert it into ID number 1 of column reply.. at the moment all it does it insert it into the end of another ID which it is creating automatically can anyone help me .. i have attached my table on to this message so you can see what the table is like!!

 

here is my code?

 

<?php

if (isset($_POST['submitted'])) {

    

    

    $posttopic = $_POST['POSTTOPIC'];

    $postdeatils =$_POST['POSTDEATILS'];

    $postauthor=$_POST['POSTAUTHOR'];

    $sqlinsert = "INSERT INTO forumposts  (`POSTTOPIC`, `POSTDEATILS`, `POSTAUTHOR`) VALUES('$posttopic', '$postdeatils', '$postauthor')";

 

$query = mysql_query($sqlinsert);

    

    if (!$query) {

        echo "error inserting new record" . mysql_error(); 

    }

    

} // end of mani if statment 

 

 

//$newrecord ="1 record added to the database";

 

 

 

 

 

?>

[using query does not insert into the phpmyadmin table?]

Very true.

Josh, read over the manual about sql syntax for a better understanding of how it works. http://dev.mysql.com/doc/refman/5.5/en/handler.html

okay thanks will do

[using query does not insert into the phpmyadmin table?]

 

ok theres progress, simply change your SQL syntax to UPDATE: 

$sqlinsert = "UPDATE `forumposts` SET `REPLY` = $reply WHERE `POSTID` = 1";

i have done and it brings up this error ? 

 

error inserting new recordYou have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '(`REPLY`) VALUES('pdijspdjf') WHERE `POSTID`=1' at line 1

[using query does not insert into the phpmyadmin table?]

Are you trying to INSERT into a row that already exists?

If is the case, you need to switch to an UPDATE query or, if it is a new row, remove the WHERE

 

it is a column that already exsists in my phpmyadmin table.. it is empty and this code should post it in to that column.. 

 

i change it to INSERT and it comes up with this error message

 

error inserting new recordYou have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'INTO `forumposts`(`REPLY`) VALUES('test') WHERE `POSTID`=1' at line 1

 

saying the code is incorrect.. when i have ran it without the insert of POSTID it insert but not on the one i want it to it creates a new ID as that is on ai and posts it in that section??

 

any ideas??

 

thanks

josh

[using query does not insert into the phpmyadmin table?]

<?php

if (isset($_POST['submitted'])) {

    

    

    $name = $_POST['entername'];

    $usernameorguest =$_POST['usernameorguest'];

    $reply =$_POST['reply'];

    $sqlinsert = "INSERT INTO `forumposts`(`REPLY`) VALUES('$reply') WHERE `POSTID`=1";

 

$query = mysql_query($sqlinsert);

    

    if (!$query) {

        echo "error inserting new record" . mysql_error(); 

    }

    

} // end of mani if statment 

 

 

//$newrecord ="1 record added to the database";

 

 

 

 

 

?>

 

 

this is my code now 

 

 

 

your problem is your query string:

$sqlinsert = "INSERT INTO `forumposts` WHERE (`POSTID`=1) INTO (`REPLY`) VALUES('$reply')";

shoudl be:

$sqlinsert = "INSERT INTO `forumposts`(`REPLY`) VALUES('$reply')";

also, look into prepared statements http://php.net/manual/en/mysqli.prepare.php

[using query does not insert into the phpmyadmin table?]

your problem is your query string:

$sqlinsert = "INSERT INTO `forumposts` WHERE (`POSTID`=1) INTO (`REPLY`) VALUES('$reply')";

shoudl be:

$sqlinsert = "INSERT INTO `forumposts`(`REPLY`) VALUES('$reply')";

also, look into prepared statements http://php.net/manual/en/mysqli.prepare.php

 

 

I've never seen SQL like that before. What DB are you using?

MySQL (and every other SQL engine I've seen) would be: INSERT INTO `table` (column, column) VALUES (value, value)

 

 

 

sorry guys i knew i wrote it wrong.. okay i have replaced the code and now it comes up with this error?

 

error inserting new recordYou have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'WHERE `POSTID`=1' at line 1

 

what does this mean?? just want to insert the data in my valables into the result column in ID1 ? i dont understand why it will not work?

 

thanks a lot

josh

[using query does not insert into the phpmyadmin table?]

sorry guys i knew i wrote it wrong.. okay i have replaced the code and now it comes up with this error?

 

error inserting new recordYou have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'WHERE `POSTID`=1' at line 1

 

what does this mean?? just want to insert the data in my valables into the result column in ID1 ? i dont understand why it will not work?

 

thanks a lot

josh

[using query does not insert into the phpmyadmin table?]

i know i have undefined variables at the moment just trying to get it to insert into the correct box 

[using query does not insert into the phpmyadmin table?]

hey guys rite i have created a website and i am trying to insert a indivual piece of data into 1 column of my phpmyadmin page table. But i am doing a TV show website and just creating a easy and basic forumn.. i am trying to insert a reply code which when you press reply it insert it into ID number 1 of column reply.. at the moment all it does it insert it into the end of another ID which it is creating can anyone help me 

 

here is my code?

 

 

<?php

if (isset($_POST['submitted'])) {

    

    

    $name = $_POST['entername'];

    $usernameorguest =$_POST['usernameorguest'];

    $reply =$_POST['reply'];

    $sqlinsert = "INSERT INTO `forumposts` WHERE (`POSTID`=1) INTO (`REPLY`) VALUES('$reply')";

 

$query = mysql_query($sqlinsert);

    

    if (!$query) {

        echo "error inserting new record" . mysql_error(); 

    }

    

} // end of mani if statment 

 

 

//$newrecord ="1 record added to the database";

 

 

 

 

 

?>

 

[PHP coding help .. ? query and if statements playing up]

 

You added a second mysql_query in there. Do this...

 $sqlinsert = "INSERT INTO reviews (`SHOWTITLE`, `SHOWREVIEW`) VALUES('$showtitle', '$showreview')";

$query = mysql_query($sqlinsert);
    
    if (!query) {
        echo "error inserting new record" . mysql_error(); 
    }

BOOM fixed thanks man!!!! you superstar

[PHP coding help .. ? query and if statements playing up]

at least one of the versions of the code is running mysql_query() on the actual query, then running mysql_query() again on the result resource from the first mysql_query(). your original code didn't have the first mysql_query() statement and you randomly added it at some point for no reason.

 

 

Well while I look at it again, did you close this if statement like I told you to before??

if (!mysql_query($sqlinsert)) {
        die('error inserting new record'); 

}

You need that closing bracket on this one

 

 

yes the bracket is done on it.. 

 

okay i understand a little bit mac but i still dont have a clue how to fix it as soon as i can fix this im laughing! lolll sorry i sound so dumb

[PHP coding help .. ? query and if statements playing up]

Are you still getting the index error? The index error means that there is nothing in the query to insert.

 

no errors at all.. only thing i get is 

 

'error inserting new record'

 

which as you can see is my If statments output

 

if (!mysql_query($sqlinsert)) {

        die('error inserting new record'); 

 

so i may be rite by saying the code is doing it but it will only ever die because thats the only way it can go.. there is no else clauise or anything the if statment is saying even if successful Die?? am i correct?

[PHP coding help .. ? query and if statements playing up]

You are missing the point. It doesnt have to do with the code right now. The form that you are submitting the information from either doesn't have any information in it or it is the wrong form. Put information in the form and hit submit.

 

the database form has information it .. it is a review page.. all i want it to do is when someone rites in the field it inserts into the database.. simple as that.. i dont know why it is doing it when there is information in the database and all im asking of it is to just place it in the slots i have i issued within the code i.e SHOWTITLE ETC.. ???

[PHP coding help .. ? query and if statements playing up]

Ooo lol! It means that those two values don't have anything in them, so it can't insert anything. Put something in the input.

 

 

the errors are referring to the two $_POST variables. the form you posted apparently isn't the one that is submitting to that code and those two fields don't exist or are invalid html in the form.

 

 

okay so here is my code:

 

 

 

<?php

if (isset($_POST['submitted'])) {

    

    include('connect database.php');

    

    

    $showtitle = $_POST['SHOWTITLE'];

    $showreview =$_POST['SHOWREVIEW'];

    $sqlinsert =mysql_query ("INSERT INTO `reviews` (SHOWTITLE,SHOWREVIEW) VALUES('$showtitle', '$showreview'");

    

    if (!mysql_query($sqlinsert)) {

        die('error inserting new record'); 

        

        

        

    }

 

what changes would you make to it ?? im very terrible at this lol !!.. 

 

the names on the database on SHOWTITLE and SHOWREVIEW 

[error message help? :((]

Too many dollar signs. You should only have one dollar sign on $sqlinsert.

 what the dollar signs are making it not work :S

 

and in the if statement do i need a result factor as at the moment it only says die if you look at my code!

[PHP coding help .. ? query and if statements playing up]

Are you column names SHOWTITLE and SHOWREVIEW?? In the database.

 do you think it could be this??

 

 

if (isset($_POST['submitted'])) {

    

    include('connect database.php');

    

    

    $showtitle = $_POST['SHOWTITLE'];

    $showreview =$_POST['SHOWREVIEW'];

    $sqlinsert =mysql_query ("INSERT INTO `reviews` (SHOWTITLE,SHOWREVIEW) VALUES(['$showtitle'], ['$showreview']");

    

    if (!mysql_query($sqlinsert)) {

        die('error inserting new record'); 

 

 

at the bottom it says the if statment but it only says die not if it works etc??? 

 

do i need a result key or anything to finilise it 

[PHP coding help .. ? query and if statements playing up]

Are you column names SHOWTITLE and SHOWREVIEW?? In the database.

yes they are.. thats what they are called .. not sounding cheeky but could you write a code for inserting into a database with my fields or something to help me out and i try that? .. i just dont get what i am doing wrong ((((((((((((((

 

thanks a lot for your help 

[PHP coding help .. ? query and if statements playing up]

Ok, I can help explain this to you. So you have a query variable $sqlinsert, which is good. You if not statement is wrong though. It should something like this. 

 $sqlinsert = "INSERT INTO reviews (`SHOWTITLE`, `SHOWREVIEW`) VALUES('$showtitle', '$showreview')";

$query = mysql_query($sqlinsert);
    
    if (!query) {
        echo "error inserting new record" . mysql_error(); 
    }

I also added some backticks `` to your query for your column names. If there are any errors in your query now it will tell you.

 

 

 

i changed my code to this

 

 

<?php

 

    

   

    

    

    $showtitle = $_POST['SHOWTITLE'];

    $showreview =$_POST['SHOWREVIEW'];

    $sqlinsert = ("INSERT INTO reviews (`SHOWTITLE`, `SHOWREVIEW`) VALUES('$showtitle', '$showreview'");

    

    $query =mysql_query($sqlinsert);

    

    if (!$query) {

        echo "error inserting new record" . mysql_error();

    }

 

 

now when i run it comes up with this error ??

 

Notice: Undefined index: SHOWTITLE in C:\xampp\htdocs\PhpProject1\New Review.php on line 23

 

Notice: Undefined index: SHOWREVIEW in C:\xampp\htdocs\PhpProject1\New Review.php on line 24

error inserting new recordYou 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

 

 

please help me to get this sorted?

[PHP coding help .. ? query and if statements playing up]

mysql_error() not mysql_error. Add the parentheses 

 

 

im new to php can you expand... i have changed to mysql_error()); 

 

all i want to do is get it to insert and it is not working can you look at my code and suggest a way round it? 

[PHP coding help .. ? query and if statements playing up]

You actually have to query the database...

$query = mysql_query($sqlinsert, $dbConn);

if (!$query) {
        die('error inserting new record'); 

}

ALSO, close your if bracket. You close your if isset, but not your if not query.

 

 

change this line     die('error inserting new record');

to be     die('error inserting new record<br />' . mysql_error());

 

 

 

 

<?php

if (isset($_POST['submitted'])) {

    

    include('connect database.php');

    

    

    $showtitle = $_POST['SHOWTITLE'];

    $showreview =$_POST['SHOWREVIEW'];

    $sqlinsert =mysql_query ("INSERT INTO reviews (SHOWTITLE,SHOWREVIEW) VALUES('$showtitle', '$showreview'");

    

    if (!mysql_query($sqlinsert, $dbConn)) {

        die('error inserting new record</b>'. mysql_error); 

        

        

        

    }

 

 

thats is my code for the insert statment... now the query is in that.. 

 

here what it brings up

 

Notice: Use of undefined constant mysql_error - assumed 'mysql_error' in C:\xampp\htdocs\PhpProject1\New Review.php on line 28

error inserting new recordmysql_error

 

 

line 27 is die('error inserting new record</b>'. mysql_error);

 

now what am i doing wrong.. my SQL code maybe that is wrong but their is something in this code which wont allow it to work... pleasee help me guys haaa i have been dying with this code for ages

 

if you know can you put it in my code for me so i dont get confused like a example etc? 

 

thanks for your help 

[PHP coding help .. ? query and if statements playing up]

The mysql_query() function goes mysql_query($string, $connection). It seems you have it backwards.

 

 

That error typically means that your query has failed and the failure has not been handled gracefully.  echo out your $sqlinsert and I think you'll see the issue.

 

 

 

when i echo it out it comes up with this 

 

Warning: mysql_query() expects parameter 1 to be string, resource given in C:\xampp\htdocs\PhpProject1\New Review.php on line 27

error inserting new record

 

and when i switch them about all it says is 

 

'error inserting new record' but no warning codes.. whats the next thing i could do as it not inserting it into the database which i am trying to achieve