How to display mysql data from php in html form.(NOT TABLE)

Question

I have been looking for a solution for this for a while but they all pertain to html tables. I have a simple form and have manually added values into the database using phpMyAdmin. I have a drop down menu at the top and whenever the admin selects a particular name from the drop down menu and presses the 'Display the fields' button, I want all the respective fields to be filled in with the values after which the admin can make changes onto any particular field and update. How can I get those values to be filled? I have tried multiple codes but keep getting errors such as undefined index, undefined variable etc. Can someone help me with that?

 <!doctype html>

<?php

$servername = "localhost";
$username = "root";
$password = "";
$dbname = "db_dealer_track";
$conn = new mysqli($servername, $username, $password, $dbname);

if($conn->connect_error){
die("Connection failed". $conn->connect_error);
}


if(isset($_POST['id1'])){

$sql = "SELECT * FROM tbl_dealer_info ";
$sql .= "WHERE $account_name = 'account_name' ";
$result = mysqli_query($conn, $sql);
while($row = mysqli_fetch_array($result)){


?>

<html>
<head>
<title>iMobile </title>
</head>
<body bgcolor = "#D6DFE3">
<center><b><h2>Please enter the following information: </h2></b></center>

<form action = "dealer_track.php" method = "post"> 
<strong><center> <u>Fields marked with an asterisk(*) are required.</u><br><br>
Name of the dealer:* // This is where the admin selects the user they would like to update

<?php 
$sql = "SELECT account_name FROM tbl_dealer_info ";
$result = mysqli_query($conn, $sql);
echo "<select name = 'account_name' id = 'id'>"; 
echo "<option value = ''>";
while($row = mysqli_fetch_array($result)){
    echo "<option value = '" .$row['account_name'] . "'>" . $row['account_name'] . "</option>";
}
echo "</select>";
?>
<br><br>
<input type = submit id = "id1" name = "id1" value = "Display the fields" /><br>
</center>
<hr>
<br><br>
</form> 

<form action = "dealer_track.php" method = "post">

Email:*<br>     
<input type = "email" name = "email" id = "id3" value = "<?php echo $row['email']?>" Required /><br><br>


RSM:*<br>
<?php
$sql = "SELECT rsm_val FROM tbl_rsm_drop_down ";
$result = mysqli_query($conn, $sql);
echo "<select name = 'rsm_val'>"; 
echo "<option value = ''></option>";
while($row = mysqli_fetch_array($result)){
    echo "<option value = '" .$row['rsm_val'] . "'>" . $row['rsm_val'] . "</option>";
}
echo "</select>";
?>
<br><br>
**// My radio buttons aren't getting checked though**

iPhone Boost Approved: 
<input type = "radio" name = "boost_app" <?php if(isset($boost_app)&& $boost_app =="Yes")?> value = "Yes" />Yes 
<input type = "radio" name = "boost_app" <?php if(isset($boost_app)&& $boost_app =="No")?> value = "No" />No<br><br>
</form>
<?php
}}  // While loop and if loop at the start
?>

</body>
</html>

Who on earth up-votes such a question ???? Please read the tooltip when you hover the up-arrow before clicking. — YvesLeBorg
– YvesLeBorg, Commented Sep 12, 2017 at 17:26
Your code is hard to understand. Can you post a clearer version of your code? — Mohamed Chaawa
– Mohamed Chaawa, Commented Sep 12, 2017 at 17:28
Some advices: fetch all db data that you need on the top (php) part of the page, in corresponding arrays. Then, in the html part, just loop through those arrays and build the required html elements. That way your code will look a lot cleaner and will be a lot more manageable. And start using mysqli (or - my recommendation - PDO) with the help of the OOP functions. And use "prepared statements" (in order to avoid mysql injection) and "exception handling" (for having the chance to discover eventual errors). — user7941334
– user7941334, Commented Sep 12, 2017 at 18:08
The value attribute is not working either. No errors, yet nothing is getting populated — Sukrit Jaie
– Sukrit Jaie, Commented Sep 12, 2017 at 21:33

Jay · Accepted Answer · 2017-09-12 17:44:22Z

1

I'm taking a wild guess here, so I'm assuming you want to select a user from a dropdown (That could be a bad idea if many people are in said database), but you would want to make a simple HTML form and name it somethign you will remember. Under the form put this?

<?php
if(isset($_POST['formnamehere'])) {
$sql = "SELECT * FROM (table name) WHERE accountname=" . $accountname;
$result = $conn->query($sql);
$row = $result->fetch_assoc();
echo $row['accountname'];
//put other things here, etc.
}
?>

Granted this code is not meant to be used exactly. but to give you a general idea.

answered Sep 12, 2017 at 17:44

Jay

708 bronze badges

Sign up to request clarification or add additional context in comments.

7 Comments

Sukrit Jaie Over a year ago

Nothing happens. No errors but the fields aren't filling up either

Jay Over a year ago

@SukritJaie You need JS to make it auto pop up.

Jay Over a year ago

Try $result = $conn->query($sql) or die($mysqli->error); Also personal preference of mine mysqli_query($conn, $sql); to $conn->query($sql);

Jay Over a year ago

If errors continue, I'll setup a subdomain on my vps and attempt to help you. :) This would have irritated me in the past

Sukrit Jaie Over a year ago

My database table has multiple columns. So I thought of using JS too, but the onchange event will only extract one column from the database. Thank you so much for helping me!

|

Abdoulie Kassama · Accepted Answer · 2017-09-12 17:45:33Z

1

You code is a bit messy but here is what you need to do generally. First query for the unique record:

$sqlQuery = "SELECT id, firstname, lastname FROM Table Where id = '$id'";

Then run the query:

$result = $connection->query($sqlQuery ); //nb: $connection is your connection variable

Then check if any result found:

if ($result->num_rows > 0) { ........ }

If any records found then put the fetched data in variables like this

while($row = $result->fetch_assoc()) {
  $firstname = $row["firstname"];
  $lastname = $row["lastname"];
  //and so on....
}
// You can display these variables any how you want in here, eg:
echo "<h2>$firstname</h2>";

or

<input type="text" id="firstname" name="firstname" value="<?php echo $firstname ?>" /> 
//nb: you must close the php tag before using html and re open it after

if "if ($result->num_rows > 0) {...} is false, just use an else {...} to display a message

edited Sep 12, 2017 at 17:45

answered Sep 12, 2017 at 17:38

Abdoulie Kassama

79011 silver badges20 bronze badges

1 Comment

Sukrit Jaie Over a year ago

I tried this but am still getting undefined variable. P.s I dont just want to display all the fields right away, only when the submit button with display the fields is pressed

Skully · Accepted Answer · 2021-03-11 15:55:10Z

0

You can run a query with your active connection to fetch your respective information from the table you want, along with a search clause for where the name is equal to a given value.

Query:

$result = mysqli_query($con, "SELECT `data` FROM `table` WHERE `name` = '$name';");

You can then display your data on your front end by outputting the result.

<?php
    if($row = mysqli_fetch_array($result)) {
        echo $row["data"];
    }
?>

edited Mar 11, 2021 at 15:55

Skully

3,1743 gold badges29 silver badges43 bronze badges

answered Mar 9, 2021 at 21:14

Vasyl Lukanych

33 bronze badges

Collectives™ on Stack Overflow

How to display mysql data from php in html form.(NOT TABLE)

3 Answers 3

7 Comments

1 Comment

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

7 Comments

1 Comment

Comments

Your Answer

Sign up or log in

Post as a guest

Related