Get value from MySQL database and populate dropdown box [duplicate] - php

Question

This question already has answers here:
Using PHP to populate a <select></select> dropdown? [duplicate]
(7 answers)
Closed 7 months ago.
I hope someone can help me figure this out as I'm new to PHP and MySQL. But basically, what I'm trying to do is to pull the data from the database and populate the dropdown box and textbox, and so I can update the value in the textbox.
This is the table structure:
id | hf_name | hf_price
------------------------
AI | test 1 | 123
AI | test 2 | 123
So let's say I wanna update "test 1" price, I would select "test 1" in the dropdown selection and update the price in the textbook.
This are my code:
<form class='data_form' action='data/food_update.inc.php' method='post'>
<select>
<?php while ($row = mysqli_fetch_array($result)):; ?>
<option><?php echo $row[1];?></option>
<?php endwhile;?>
</select>
<input type='number' name='hf_price' placeholder='$'>
<button type='submit'>UPDATE</button>
</form>
Include:
<?php
//CONNECTION TO MySQL
include '../dbh.php';
//VARIABLES
$hf_name = $_POST['hf_name'];
$hf_price = $_POST['hf_price'];
$by_user = $_SESSION['Fname'];
$up_date = date('Y-m-d');
$sql = "SELECT hf_name FROM hotfoods";
$result = mysql_query($conn, $sql);
?>
Thanks!

Answer 1

Try changing the option to
<option value="<?php echo $row[2];?>"><?php echo $row[1];?></option>
Where I'm assuming $row[2] variable contains the hf_price value.

Answer 2

$query = "select * from tablename";
$result = mysqli_query($query);
<select>
<?php
while($row = mysqli_fetch_assoc($result){
?>
<option value='<?php echo $row['hf_name']; ?>'><?php echo $row['hf_name']; ?></option>
<?php
}
?>
</select>

Answer 3

but for this you need to change your id datatype into int,
$query = "select * from tablename";
$result = mysqli_query();
<select id="nameID" name="hf_name">
<?php
while($row = mysqli_fetch_assoc($result){
?>
<option value='<?php echo $row['id']; ?>'><?php echo $row['hf_name']; ?></option>
<?php
}
?>
</select>
This will be fine when you fetch details but it does not update any value because for this you need to use Ajax or Jquery for update on Select, let me to try it.
Ajax
$("#nameID").onchange(function(){
var id = $("#id").val();
$.ajax(
url: "update.php",
type: "POST",
data: {'id'=id},
success: function(){
alert("ok");
});
});
Try this it might work.

Answer 4

You should assign a name attribute value (in this case, hf_name) on select element so that that become available in $_POST array.
<form class='data_form' action='data/food_update.inc.php' method='post'>
<select name="hf_name">
<?php while ($row = mysqli_fetch_array($result)):; ?>
<option value="<?php echo $row['hf_name']?>"><?php echo $row['hf_name'];?></option>
<?php endwhile;?>
</select>
<input type='number' name='hf_price' placeholder='$'>
<button type='submit'>UPDATE</button>
</form>
Now this is on which the form will be submitted. Where you can take user's submitted data through $_POST array.
<?php
//CONNECTION TO MySQL
include '../dbh.php';
//VARIABLES
$hf_name = $_POST['hf_name'];
$hf_price = $_POST['hf_price'];
$query = sprintf("UPDATE hotfoods SET hf_price='%d' WHERE hf_name='%s'",
mysql_real_escape_string($hf_price),
mysql_real_escape_string($hf_name));
$result = mysql_query($conn, $query);
if ($result) {
echo 'Price updated!';
} else {
echo 'Problem while updating price';
}
?>