Cannot submit a form on php and mysqli_query

Question

I have a html form that i want to submit its data into a specific database in wamp using phpmyadmin, the connection is successfully done. However, the data cannot be submitted. I get this message after submitting the data in the form :

Successful connection
( ! ) Warning: mysqli_query() expects parameter 1 to be mysqli, resource given in C:\wamp\www\Ex\insert-data.php on line 11
Call Stack
#   Time    Memory  Function    Location
1   0.0005  136600  {main}( )   ..\insert-data.php:0
2   0.0023  144480  mysqli_query ( )    ..\insert-data.php:11
Error inserting new records!

My Code in 'insert-data.php' is:

<?php


if(isset($_POST['submitted'])){
 include('connect.php');
 $fname = $_POST['fname'];
 $lname = $_POST['lname'];

      $sqlinsert=
    "INSERTINTO`test`(`FName`,`LName`)VALUES('$fname','$lname')";

if(!mysqli_query($dbconn,$sqlinsert)){

 die('Error inserting new records!');
}
echo "1 record added to database";



}

?>


<!DOCTYPE html>      



<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<h1>Insert Data into DB</h1>
</head>

<body>
    <form method="post" action="insert-data.php" >
    <input type="hidden" name="submitted" value="true" />

    <label>First Name</label>
    <input type="text" name="fname"  />  
    <label>last Name</label>
    <input type="text" name="lname"  />

    <input type="checkbox" name="check" />
    <input type="radio" name="radios" />
    <input type="submit" value="submit"></button>

    </form>

  </body>
  </html>

Any idea? ....Thanks

Use MySQLi properly! Prepare your statements! http://php.net/manual/en/mysqli.prepare.php . Also, another point when showing an error that has appeared please make clear which line is number 11. However, in the case your query string needs to be spaced — Liam Sorsby, Jul 23 '15 at 16:24
No,it has a space in my code ..but here by mistake appears like that .. — Judy, Jul 23 '15 at 16:25
@LiamSorsby bro please read the error clearly. It says "Warning: mysqli_query() expects parameter 1 to be mysqli," which means that first parameter in mysqli_query() which is $dbconn has the problem in it. — William Francis Gomes, Jul 23 '15 at 16:28
sidenote: `` that will always be true and will repost on page reloading. You should set your conditional statement to the submit button along with using `!empty()` as part of that conditional. A header redirect is better and using sessions and tokens to avoid that. You also have a syntax error here `` — Funk Forty Niner, Jul 23 '15 at 16:29
@WilliamFrancisGomes I apologise. I hold my hands up, your right. I am sorry. I seem to have missed that section. The irony. :-) — Liam Sorsby, Jul 23 '15 at 16:29
@Judy are you by any chance running mysql_connect() instead of mysqli_connect()? — Liam Sorsby, Jul 23 '15 at 16:31
you need to post your connection codes, replacing credentials with `xxx`. Edit: @LiamSorsby or PDO for that matter. Who knows ;-) — Funk Forty Niner, Jul 23 '15 at 16:31
you've been asked what to do and you didn't. This, about your db connection codes. — Funk Forty Niner, Jul 23 '15 at 16:36
Follow this link: http://php.net/manual/en/function.mysqli-connect.php and fix your connect.php file. — William Francis Gomes, Jul 23 '15 at 16:39
@LiamSorsby [it was indeed `mysql_` after all](http://stackoverflow.com/questions/31592926/cannot-submit-a-form-on-php-and-myqli-query#comment51138385_31592926) Liam ;-) finally, *solved.* cheers — Funk Forty Niner, Jul 23 '15 at 17:15

score 1 · Accepted Answer · edited May 23 '17 at 12:14

1

you posted your connection codes in comments (which belongs in your question I might add) being mysql_ based.

You need to use mysqli

those different MySQL APIs do not intermix. You must use the same one from connection to query.

http://php.net/manual/en/function.mysqli-connect.php

Example pulled from the manual:

<?php 
//conection:
$link = mysqli_connect("myhost","myuser","mypassw","mybd") 
        or die("Error " . mysqli_error($link));

and remember to replace $link with $dbconn and your own credentials.

This doesn't help you:

die('Error inserting new records!');

this does:

or die(mysqli_error($dbconn));

Since you seem new to this, use prepared statements right away.

References:

Your present code is open to SQL injection.

Add error reporting to the top of your file(s) which will help find errors.

<?php 
error_reporting(E_ALL);
ini_set('display_errors', 1);

// rest of your code

Sidenote: Displaying errors should only be done in staging, and never production.

Just for argument's sake, put a space between INSERT and INTO:

$sqlinsert= "INSERT INTO `test` (`FName`,`LName`) VALUES ('$fname','$lname')";

You seem to have made a reference to that in comments that they are seperated, but I said it anyway.

Plus, try putting your connection/include on top of your conditional statement.

Connection:

Your connection should be this and replacing the xxx with your own credentials.

$db_host     = "xxx";
$db_username = "xxx";
$db_pass     = "xxx";
$db_name     = "xxx";

$dbconn = mysqli_connect("$db_host","$db_username","$db_pass","$db_name") 
or die("Error".mysqli_error($dbconn));

and nothing else. No instances of mysql_ at all.

Sidenote: @ symbols are error suppressors. You can add them back in once everything is working.

Closing notes:

Kudos to Liam (Sorsby).

edited May 23 '17 at 12:14

Community

1
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answered Jul 23 '15 at 16:37

Funk Forty Niner

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I corrected that .. I'm not getting the warning but still error inserting the records, I'm getting this instead : Successful connectionError inserting new records! – Judy Jul 23 '15 at 16:40
@Judy you need to find out why your query failed. consult http://php.net/manual/en/mysqli.error.php - add `or die(mysqli_error($dbconn))` to `mysqli_query()` – Funk Forty Niner Jul 23 '15 at 16:42
Thank you anyway...but there is no problem with the connection...the problem is about inserting the data . – Judy Jul 23 '15 at 16:47
@Judy reload my answer and look at error reporting and why your query is failing under **this does:** – Funk Forty Niner Jul 23 '15 at 16:48
Thanks a lot... this is what i wrote and it worked. $dbconn = @mysqli_connect("$db_host","$db_username","$db_pass","$db_name") or die("Error".@mysqli_error($dbconn)); @mysql_select_db("$db_name") or die("No database"); echo"Successful connection"; – Judy Jul 23 '15 at 16:57
@Judy you also have ` ` in your question but no related code for it in your SQL. If that is indeed part of it, you need to modify your question/code for it. I tested your code, no problems. Check to see if you haven't a restraint somewhere, that the column names have the right type and lengths. If your inputs have apostrophes, or anything else that MySQL will complain about, then that would be a contributing factor. I see no errors with what I posted as a solution. – Funk Forty Niner Jul 23 '15 at 17:03
@Judy the @ symbols in @mysqli_connect etc. **suppresses** errors. and you can't use `@mysql_select_db("$db_name") or die("No database");` - there should be **no** instances of `mysql_` whatsoever. Please go over the manual on connecting, as stated in my answer. – Funk Forty Niner Jul 23 '15 at 17:04
@Judy You're welcome Judy. I'm glad to see this was resolved, *cheers* – Funk Forty Niner Jul 23 '15 at 17:10
1

@Fred-ii- Growing another cactus here :)? – Rizier123 Jul 23 '15 at 17:16
1

@Rizier123 yes, God love 'em :-) – Funk Forty Niner Jul 23 '15 at 17:17
1

@Fred-ii- Nice answer! Finally Solved. +1 from me. This is why I don't get rep. I can't be bothered with answers like this. Haha – Liam Sorsby Jul 23 '15 at 17:22
1

@LiamSorsby Thank you Liam. I made a slight edit. See "Closing notes". *cheers* ;-) – Funk Forty Niner Jul 23 '15 at 17:25
Haha! Thank you very much ;-) haha – Liam Sorsby Jul 23 '15 at 17:25
@LiamSorsby You're welcome Liam. You were the *crystal ball* on this one ;-) – Funk Forty Niner Jul 23 '15 at 17:27
1

Now onto the next one. Ha – Liam Sorsby Jul 23 '15 at 17:28

score 0 · Answer 2 · edited Jul 24 '15 at 06:53

0

Use separated words like,

INSERT INTO `test` (`FName`,`LName`) VALUES ('$fname','$lname')";

edited Jul 24 '15 at 06:53

Rohan Kumar

40,431
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answered Jul 23 '15 at 16:22

lingo

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2

OP stated in comments: *"No,it has a space in my code ..but here by mistake appears like that .. – Judy 4 mins ago"* – Funk Forty Niner Jul 23 '15 at 16:30

Cannot submit a form on php and mysqli_query

2 Answers2