I've been tearing my hair out trying to figure out why the isset($_POST['Submit']) is not executing with my form. The data from the form is just not passing into the php code. Basically the code does not seem to be recognizing something like $ffname = $_POST["ffname"];
<?php
$ffname = $flname = $femail = $fcemail = $fpass = $fcpass = "";
if(isset($_POST['ffname'])){
$ffname = $_POST["ffname"];
$flname = $_POST["flname"];
$femail = $_POST["femail"];
$fcemail = $_POST["fcemail"];
$fpass = $_POST["fpass"];
$fcpass = $_POST["fcpass"];
echo "<p>Hello World<p>";
$con = mysqli_connect("localhost", "root", "") or die(mysqli_error());
mysqli_select_db($con, "userdata") or die(mysqli_error($con));
mysqli_query($con,"INSERT INTO tbluser (fname, lname, email, pass) VALUES('$ffname', '$flname', '$femail', '$fpass')")
or die (mysqli_error($con));
}
?>
<form method="post">
First Name: <input type="text" name="ffname" id="ffname" value="<?php echo $ffname;?>"><br>
Last Name: <input type="text" name="flname" value="<?php echo $flname;?>"><br>
E-mail: <input type="email" name="femail" value="<?php echo $femail;?>"><br>
Confirm E-mail: <input type="email" name="fcemail" value="<?php echo $fcemail;?>"><br>
Password: <input type="password" name="fpass" value="<?php echo $fpass;?>"><br>
Confirm Password: <input type="password" name="fcpass" value="<?php echo $fcpass;?>"><br>
<input type="submit" name="Submit" value="submit">
</form>
The other answer by #DerVO is correct. But there seems to be something else at play, since you say it still doesn't work.
A comment became too long, so I've built a full answer here.
Step 1:
Add a name to your input:
<input type="submit" name="Submit" value="submit">
However, relying on the submit in your $_POST is not the best plan. So I suggest watching a different form field - for example, ffname:
Step 2:
Improve your watch, using a different field:
if ( isset( $_POST['ffname'] ) ) {
// do your work
}
Lastly, you may be munging your form action attribute.
Step 3:
In order to keep things simple, if the form is supposed to submit to the same page, you can simply omit the form action.
<form method="post">
Betweeen these three items, the form will work, unless you have some problem with your server.
Step 4:
Clean up your form formatting. You've got odd spacing which is problematic. In an html element, the property="value" code needs to be without spaces, but spaces between properties. Example:
<!-- Your version -->
<input type = "text"name = "ffname"id = "ffname"value="<?php echo $ffname;?>"><br>
<!-- Clean / correct version -->
<input type="text" name="ffname" id="ffname" value="<?php echo $ffname;?>"><br>
Here's a "clean" version of your whole form:
<form method="post">
First Name: <input type="text" name="ffname" id="ffname" value="<?php echo $ffname;?>"><br>
Last Name: <input type="text" name="flname" value="<?php echo $flname;?>"><br>
E-mail: <input type="email" name="femail" value="<?php echo $femail;?>"><br>
Confirm E-mail: <input type="email" name="fcemail" value="<?php echo $fcemail;?>"><br>
Password: <input type="password" name="fpass" value="<?php echo $fpass;?>"><br>
Confirm Password: <input type="password" name="fcpass" value="<?php echo $fcpass;?>"><br>
<input type="submit" name="Submit" value="submit">
</form>
You need to give your input submit a name:
<input type="submit" name="Submit" value="Submit">
You have pass name of element in $_POST
try put name attribute in input submit
<input type = "submit" name="Submit" value = "1">
Related
I created a section for editing. When I edit the information and click the save button, the information is not saved and the header section does not display completely.
<?php
if (isset($_POST['submit_btn']))
{
$id = $_POST['id'];
$fn = trim($_POST['name']);
$ln = trim($_POST['lastname']);
$age = trim($_POST['age']);
$q = "UPDATE `users` SET `fn` = '$fn',
`ln` = '$ln',
`age` = '$age'
WHERE id = '$id'";
mysqli_query($dbconnect,$q);
if (mysqli_affected_rows($dbconnect) > 0)
redirect("?msg=ok&id=**$id**");
else
redirect("?msg=error&id=**$id**");
}
else
echo ("Not In If(isset)");
?>
<form action="" method="post">
<label for="name">FirstName:</label>
<input type="text" name="name" id="name" value="<?php echo $row['fn']?>">
<br>
<label for="lastname">LastName:</label>
<input type="text" name="lastname" id="lastname" value="<?php echo $row['ln']?>">
<br>
<label for="age">Age:</label>
<input type="text" name="age" id="age" value="<?php echo $row['age']?>">
<br>
<input type="submit" name="submit_btn" value="Save">
<a href="index2.php">
Back
</a>
</form>
</body>
Bold sections do not work here.
Below is a picture of the result:
In the link that I specified, after clicking on save the ID will not be displayed and all the information filled in the forms will be lost.
Sorry if the result is styleless and boring and I just created this page to practice php😁
Thank you for being responsive🙏🙏🙏
You are mistaking a POST request with a GET request.
Part, which appears in the URL is sent to the webserver in GET request.
Your form is submitting POST request to the webserver, logic in the code does the same, but you are trying to display information from url (GET).
Please check the examples in php.net:
POST variables: https://www.php.net/manual/en/reserved.variables.post.php
GET variables: https://www.php.net/manual/en/reserved.variables.get.php
You can take an example with GET request variable below, however, be careful with trusting the "end client" and always prepare your statements, which you send to your database to execute queries.
if (isset($_GET['submit']))
{
$number = $_GET['number'];
echo $number
? "Number which was submitted: $number <br>"
: 'Number is not set';
} else {
echo 'Form has not been yet submitted';
}
?>
<form action="" method="get">
<input type="number" name="number" placeholder="Number">
<input type="submit" name="submit" value="Save">
</form>
I created form on one page in my wordpress:
<form action="scores_send.php" method="post">
<strong>Date: </strong><input type="date" name="date" /><br />
<strong>TEAM 1:</strong>
Player 1.: <input type="text" name="team1_player1" /><br />
Player 2.: <input type="text" name="team1_player2" /><br />
<strong>TEAM 2:</strong>
Player 1.: <input type="text" name="team2_player1" /><br />
Player 2.: <input type="text" name="team2_player2" /><br />
<strong>SCORE:</strong>
Team 1. <input type="number" name="score_team1" min="0" max="5"/> : <input type="number" name="score_team2" min="0" max="5"/> Team 2.<br />
<input type="submit" value="Add" />
</form>
And also I wrote php code which connect to my external (no-wordpress) postgres database (connection parameters in code are not true) and insert information from form to one table:
<?php
$date = $_POST['date'];
$t1p1 = $_POST['team1_player1'];
$t1p2 = $_POST['team1_player2'];
$t2p1 = $_POST['team2_player1'];
$t2p2 = $_POST['team2_player2'];
$score1 = $_POST['score_team1'];
$score2 = $_POST['score_team2'];
if($date and $t1p1 and $t1p2 and $t2p1 and $t2p2 and $score1 and $score2) {
$connection = pg_connect("host=127.0.0.1 port=5432 dbname=scores user=user password=pass")
or die('Brak polaczenia z serwerem PostgreSQL');
$db = #pg_select_db('scores', $connection)
or die('Nie moge polaczyc sie z baza danych');
$team1 = #pg_query("INSERT INTO scores.games (score_team1, score_team2, datetime, team1_player1, team1_player2, team2_player1, team2_player2) VALUES ('$score1', '$score2', '$date', '$t1p1', '$t1p2', '$t2p1', '$t2p2'");
if($team1) echo "Scores added.";
else echo "ERROR! Scores not added";
pg_close($connection);
}
?>
I tried 2 ways:
1. to create template php file in folder of my theme and
2. to paste this code to content editor in wordpress (using this plugin: https://wordpress.org/plugins/insert-php/)
and both ways is not working. I changed part "action" in , but maybe wrong.
Just change your action of your form to : http://yoursite.com/scores_send.php
make sure that your scores_send.php is on the root of your installation.
So, your form look like :
<form action="http://yoursite.com/scores_send.php" method="post">
You should use http://example.com/page-slug/ in form action
Where page-slug is the slug of another page create by make scores_send.php as template.It will post all form value to score_send.php.From where you can proceed further to insert into external DB.
<form action="http://example.com/page-slug/" method="post">
I am learning PHP now, so pardon my silly question which I am not able to resolve.
I have created a simple web form where in I display the values entered by a user.
function submitform()
{
document.forms["myForm"].submit();
} // in head of html
<form action ="res.php" id="myForm" method="post" >
Name: <input type="text" name="name" size="25" maxlength="50" /> <br> </br>
Password:<input type="password" name="password" size="25" maxlength="50" />
Description: <textarea name="editor1"> </textarea>
<input type="submit" value="Submit" onclick="submitForm()" />
</form>
and res.php contains:
foreach($_POST as $field => $value)
{
echo "$field = $value";
}
When I click on the submit button, I just get a blank page without any values from the form. Can anyone please let me know what am I missing?
There's no need for the javascript. This should do:
<form action ="res.php" id="myForm" method="post" >
Name: <input type="text" name="name" size="25" maxlength="50" /> <br> </br>
Password:<input type="password" name="password" size="25" maxlength="50" />
Description: <textarea name="editor1"> </textarea>
<input type="submit" value="Submit" />
</form>
Let's start with fixing errors:
JavaScript is case-sensitive. I see that your function name is submitform and the form's onclick calls submitForm.
The javascript is not really necessary from what you've shown us, I would try this on a single php page and see if it works:
Create a test.php file for test purpose:
<?php
if($_POST){
foreach($_POST as $key=>$value){
echo "$key: $value<br />";
}
}else{
<form action="test.php" method="post">
<input type="text" value="1" name="name1" />
<input type="text" value="2" name="name2" />
<input type="submit" value="submit" name="Submit" />
</form>
}
?>
If it does work, slowly work your way into your current form setup to see what is breaking it. If it doesn't work, there's something larger at play.
There are 2 things you should do now.
Remove the JavaScript function to submit the form. It's not required (or necessary). The default behavior of a submit button is to... well... submit. You don't need to help it with JavaScript.
Enable error display by using error_reporting(E_ALL).
After you do both things, you should be able to debug and assess the problem much more easily.
Put your php code inside php tags!
<?php
foreach($_POST as $field => $value)
{
echo $field ." = ." $value.'<br />';
}
?>
If you do
<?php
print_r($_POST);
?>
what do you get?
If this still doesn't work, does your server parse php?
Create the file test.php and access it directly http://localhost/test.php or whatever your URL is
<?php
echo 'Hello';
?>
if this doesn't work..it's a whole diferent problem
You can submit an HTML form using PHP with fsubmit library.
Example:
require_once 'fsubmit.php';
$html = "<form action ='res.php' method='post'><input type='text' name='name'/></form>";
$form = new Fsubmit();
$form->url = 'http://submit_url_here.com';
$form->html = $html;
$form->params = ['name'=>'kokainom'];
$response = $form->submit();
echo $response['content'];
I'm trying to upload a form to the database, that uses the same field names, as part of a CMS.
This is the form (I have used JQuery to multiply the div the number of options per product):
<form method="post" enctype="multipart/form-data" action="testing.php">
<div class="OptExtra1">
<h3>Additional Option</h3>
<label for="RESAddType">File type (i.e. “CD” or “Download”)</label>
<input name="RESAddType[]" type="text" id="RESAddType" size="48" class="FW" />
<label for="RESAddTitle">File title (i.e. “Boxed Set of 4 CDs”)</label>
<input name="RESAddTitle[]" type="text" id="RESAddTitle" size="48" class="FW" />
<label for="RESAddFType">File format (As “MP3” // “WORD” // “PDF”)</label>
<input name="RESAddFType[]" type="text" id="RESAddFType" size="48" class="FW" />
<label for="RESAddPrice">File price (Enter as “6.99” – <strong>NO “£” SIGN!</strong>)</label>
<input name="RESAddPrice[]" type="text" id="RESAddPrice" size="48" class="FW" />
<label for="RESAddFName">File name</label>
<input name="RESAddFName[]" type="text" id="RESAddFName" size="48" class="FW" />
<label for="RESAddTxt">File text</label>
<textarea name="RESAddTxt[]" id="RESAddTxt" cols="70" rows="50" class="mceAdvanced"></textarea>
<label for="RESAddSample">File text</label>
<textarea name="RESAddSample[]" id="RESAddSample" cols="70" rows="50" class="mceVSimple"></textarea>
<input type="button" value="Add another option" class="SubmitButton" onclick="inserter()" />
<hr />
</div>
<p><input type="submit" name="submit" value="Add Resource" class="SubmitButton"/><input type="hidden" name="RESCatCode" value="2" /><input type="hidden" name="RESCatSubCode" value="5" /><input type="hidden" name="submitted" value="true" /></p>
and this is what I have so far for the PHP
if(isset($_POST['submitted'])){
$RESCode = 100;
$RESAddType = $_POST['RESAddType'];
$RESAddTitle = htmlentities($_POST['RESAddTitle'], ENT_QUOTES);
$RESAddFType = $_POST['RESAddFType'];
$RESAddPrice = $_POST['RESAddPrice'];
$RESAddFName = $_POST['RESAddFName'];
$RESAddTxt = mysql_real_escape_string($_POST['RESAddTxt']);
$RESAddSample = mysql_real_escape_string($_POST['RESAddSample']);
for ($i=0; $i < count($_POST['RESAddType']); $i++) {
$OptionQuery = mysql_query ("INSERT INTO ResAdd (RESAddCode, RESCode, RESAddType, RESAddTitle, RESAddFType, RESAddPrice, RESAddFName, RESAddTxt, RESAddSample) VALUES ('', '".$RESCode."', '".$RESAddType."', '".$RESAddTitle."', '".$RESAddFType."', '".$RESAddPrice."', '".$RESAddFName."', '".$RESAddTxt."', '".$RESAddSample."');");
}
header("Location: welcome.php");
exit;
}
It kind of worked, but is just putting in the word "array" into the database. Also the htmlentities and mysql_real_escape_string posts don't upload anything to the database.
Any ideas please?
You've forgotten something :)
$OptionQuery = mysql_query ("INSERT INTO ResAdd (RESAddCode, RESCode, RESAddType, RESAddTitle, RESAddFType, RESAddPrice, RESAddFName, RESAddTxt, RESAddSample) VALUES ('', '".$RESCode[$i]."', '".$RESAddType[$i]."', '".$RESAddTitle[$i]."', '".$RESAddFType[$i]."', '".$RESAddPrice[$i]."', '".$RESAddFName[$i]."', '".$RESAddTxt[$i]."', '".$RESAddSample[$i]."');");
Look at all these $i in the code above. If you don't use them, the script will try to use the entire array (and trying to print or save an array as a string always results in printing "Array").
P.S.I've edited my answer. Sorry for the previous one, I've misunderstood the code and posted a wrong answer.
You need to looping
$cnt=count($_POST['RESAddType']);
for ($counter=0; $counter < $cnt; $counter++)
{
$_POST['RESAddType'][$counter]// to access the value
$_POST['RESAddPrice'][$counter]//
//create your query here
}
I have a page which basically allows an admin user to create manager user types (basically a register function. So when the values are submitted, they are stored into the DB, very very basic stuff. However, I have a hidden variable type..reasons are I have 3 different user levels and I have declared they identification as an integer (e.g. 7 = manager, 8 =user etc.)
Can someone help me out with how to correctly pass this hidden value so it stores in the database...
Here is my form:
<form id="userreg" name="userreg" method="post" action="adduser-process.php">
<label>Full Name:</label> <input name="fullname" size="40" id="fullname" value="<?php if (isset($_POST['fullname'])); ?>"/>
<br />
<label>Username:</label> <input name="username" size="40" id="username" value="<?php if (isset($_POST['username'])); ?>"/> <br />
<label>Password:</label> <input name="password" size="40" id="password" value="<?php if (isset($_POST['password'])); ?>"/> <br />
<label>Email Address:</label> <input name="emailaddress" size="40" id="emailaddress" value="<?php if (isset($_POST['emailaddress'])); ?>"/>
<br />
<input name="userlevel" type="hidden" size="1" id="userlevel" value="<?php $_POST[5]; ?>" /> <br />
<input value="Add User" class="addbtn" type="submit" />
</form></div>
Next, here is the script that runs the query:
<?php
require_once "config.php";
$fullname = $_POST['fullname'];
$username = $_POST['username'];
$password = $_POST['password'];
$emailaddress = $_POST['emailaddress'];
$userlevel = $_POST[5];
$sql = "INSERT INTO users_tb VALUES('".$user_id."','".$fullname."','".$username."',MD5('".$password."'),'".$emailaddress."','".$userlevel."')";
$result = mysql_query($sql, $connection)
or die("MySQL Error: ".mysql_error());
header("Location: administratorfrontview.php");
exit();
?>
I'm basically trying to pass the hidden typem with a constant value of '5' just for this form, as it will not be changed...also while im here, for some reason, the 'fullname' is not stored in the DB either!!?? WTH?? all other fields are processed fine. Any help is much appreciated! Thank you.
Two things. One, $userlevel should equal $_POST['userlevel'] not 5 as POST data isn't always in that order. Two, your insert statement should be preceded with the column names (to prevent any data from going in the wrong order).
$sql = "INSERT INFO users_tb (id, name, username, password, email, userlevel) ".
"('".$user_id."','".$fullname."','".$username."',MD5('".$password."'),'".
$emailaddress."','".$userlevel."')";
Your PHP for outputting the value is wrong. Use:
<?= $_POST[5]; ?>
or
<?php echo $_POST[5]; ?>