CODEWITHSUNDEEP
phphtml
Sam
Sam

Reputation: 2347

Confused why form is not submitting

I don't think there's anything I left out, but the form doesn't seem to be transmitting any data upon hitting submit. My code is as follows, I realize it's kind of long, but I wanted to include the entire form. All of the attached functions just check if the input was valid:

<form name="formname" id="formname" action="database.php" method = post onsubmit="return checker()">

Username: <input type='text' id='un' onchange="checkname(id)" onkeyup="checkempty(id)" ><div id="un1"></div><br>

Reporter Title:<br>

<select id="s2" onblur="checktype(id)" >

<option value="choose">Choose one</option>

<option value="Assistant Director">Assistant Director</option>

<option value="Director">Director</option>

<option value="DB Admin">DB Admin</option>

<option value="Systems Admin">Systems Admin</option>

</select><div id="s21"></div>
<br>

Password: <input type='password' id='pw' onchange="checkpass(id)" onkeyup="checkempty(id)"><div id="pw1"></div><br>

Email: <input type='text' id='eml'onchange="checkemail(id)" onkeyup="checkempty(id)"><div id="eml1"></div><br>

Description:<br> <textarea rows="6" cols="20" id="desc" onchange="checkdesc(id)" onkeyup="checkempty(id)" ></textarea><div id="desc1"></div><br>

Type of Incident:<br>

<select id="s1" onblur="checktype(id)" >

<option value="choose">Choose one</option>

<option value="afs">afs</option>

<option value="password">password</option>

<option value="hardware">hardware</option>

<option value="other">other</option>

</select><div id="s11"></div>
<br>

<?php

include("connect.php");

$ret = mysql_query("SELECT * FROM countries");

echo "Choose your location:<br>";

echo "<select id='countries' onblur='checktype(id)'>";

echo "<option value='choose'>Choose one</option>";

while($array = mysql_fetch_array($ret)){

echo "<option value='{$array['abbrevs']}'>{$array['full']}</option>";

}

echo "</select><br>";


?>

<div id="countries1"></div><br>

<p>
Would you like an email copy?


<select id="s3">

<option value="no">No</option>

<option value="yes">Yes</option>

</select>


<input type="submit" id = "sub" value= "Submit">
</p>


</form>

and the php I tried to receive it with

<?php

include("connect.php");

$username = $_GET['un'];
$password = $_GET['s2'];
$reporter = $_GET['pw'];
$email = $_GET['eml'];
$description = $_GET['desc'];
$type = $_GET['s1'];
$country = $_GET['countries'];
$emailopt = $_GET['s3'];




?>

the checker function:

function checker(){

if(isgood && isgood1 && isgood2 && isgood3 && isgood4 && isgood5 && isgood6)
{
return true;
}
else{

document.getElementById('subb').style.visibility="visible";
document.getElementById('subb').innerHTML = "You must fully complete the form.";
return false;


}

where the "isgoods" where just quick flags I made for validations, which was all working properly.

Upvotes: 1

Views: 186

Answers (3)

Sam
Sam

Reputation: 2347

Well, I feel stupid. The problem was that I didn't give my HTML elements names, but rather just IDs. I was so used to using ajax and submitting by just getting the value with javascript via the ID that I forgot about names O.o

Upvotes: 3

James Skidmore
James Skidmore

Reputation: 50338

You are submitting your form with method="POST", so in your PHP you need to read the POST values:

$username = $_POST['un'];
...

Had you submitted your form with method="GET", your PHP code would work. Read about the difference here.

Upvotes: 1

Clive
Clive

Reputation: 36965

Your form's method is POST so you should use $_POST instead of $_GET in your PHP script.

Also you should fix up your HTML so it's valid, you need to quote the method attribute properly:

<form name="formname" id="formname" action="database.php" method="post" onsubmit="return checker()">

Upvotes: 4

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