PHP script doesn't say there is an error but it won't send to MySQL

Question

I found a lot of these on the side but I don't understand them. Can someone just post the good code and point the wrong things out.

Code didn't want to be posted here is a foto of it

html form

php action

If it helps iam using byethost so I could be something byethost related

You will need to post your code to get help. There's not much we can do with this amount of information. — Asur, May 12 '16 at 06:57
post your code and your link also do not help out to know the problem — Niroj Adhikary, May 12 '16 at 06:57
Is a foto good too? I got 2 of those with the code, removed the Css of the html form though — einstein8612, May 12 '16 at 07:19
in the photo `[host]` `[username]` and `[mypassword]` are substituted intentionally or this is the actual code you are trying to run? — Velimir Tchatchevsky, May 12 '16 at 07:23
The host username password is cuz I don't want people to see my host username and password :) — einstein8612, May 12 '16 at 07:24
You shouldn't use mysql_* functions. You should use PDO or mysqli_* . http://stackoverflow.com/questions/12859942/why-shouldnt-i-use-mysql-functions-in-php — Michas, May 12 '16 at 12:09

score 0 · Answer 1 · answered May 12 '16 at 07:42

0

Your post indexing is wrong. You can not use value of id attrbute with POST. You have to use value of name attribute so change your code to

$_POST['firstname'] AND $_POST['lastname']

answered May 12 '16 at 07:42

DD77

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lhrec_106 · Answer 2 · 2016-05-12T11:51:08.610

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A few things: 1. form should be specified method.

<div> 
  <form action="login_action.php" method="post" >
  <label for="fname">First Name</label>
  <input type="text" id="fname" name="firstname" >

  <label for="lname">Last Name</label>
  <input type="text" id="lname" name="lastname" >
</div>

otherwise, the form will be using get method, and there is no data send to $_POST. refer to http://www.w3schools.com/tags/att_form_method.asp

As @User7 mentioned. The data index in $_POST is based on the input name, not id

 $sql ='INSERT INTO login_info(fname, lname) VALUES ("'.$_POST['firstname'].'","'.$_POST['lasttname'].'")';
 if(!mysql_query($sql, $con)){
  die('error:'.mysql_error());
 }

edited May 12 '16 at 11:51

answered May 12 '16 at 08:03

lhrec_106

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You must understand xD, iam such a web devenlopment nub can you put the hole code... – einstein8612 May 12 '16 at 09:03
What do you mean by good code? These are just minor changes to your original code. – lhrec_106 May 12 '16 at 11:51

PHP script doesn't say there is an error but it won't send to MySQL

2 Answers2