PHP put data from database into form - php

Question

I'm relatively new to PHP and have been working on a basic survey as a practice project.
The user logs in, answers the questions and the data is sent to the database. So far so good.
However, since the only thing that the login is set up for is to do the survey, I want them to be able to go back and change their answers at any time.
I've set it up so that it checks if the user has already filled in the info on the database, and edits values rather than adding new row if they've filled it in before.
However, I now want them to be able to see the choices that they made and edit them.
I know how to get the data back and echo it to the page, what I don't know how to do is get it to display the form with the default values set at the users previous selections.
Apologies for any display problems, its being a bit erratic!
so for example:
<?php if(!isset($_SESSION['username'])){
// are logged in
header("location:login.php");
} ?>
<form name="questionnaire" action="submit_survey.php" method="post">
<input name="Q1" type="radio" method="post" value="1">1
<input name="Q1" type="radio" method="post" value="2">2
<input name="Q1" type="radio" method="post" value="3">3
<input name="Q1" type="radio" method="post" value="4">4
<input name="Q2" type="radio" method="post" value="1">1
<input name="Q2" type="radio" method="post" value="2">2
<input name="Q2" type="radio" method="post" value="3">3
<input name="Q2" type="radio" method="post" value="4">4
</form>
PHP sending to database:
include('db_login2.php');
$con = mysqli_connect("$host", "$username", "$password", "$db_name");
if(mysqli_connect_errno($con)) {
echo "failed to connect" . mysql_connect_error();
}else {
$username = $_SESSION['username'];
//Question 1.1
$Q1 = $_POST['Q1'];
$Q2 = $_POST['Q2'];
//INJECTION PREVENTION (Not shown)
$resulting = mysqli_query($con, "SELECT username FROM users WHERE username='$username'");
$counting = mysqli_num_rows($resulting);
if($counting != 0){
mysqli_query($con,"UPDATE users SET Public_Services='$Q1', Politicians='$Q2'
WHERE username='$username'");
$_SESSION['completed'] = true;
header("location:login.php");
mysqli_close($con);
} else {
$sql= "INSERT INTO users(username, Public_Services, Politicians)
VALUES ('$username','$Q1', '$Q2')";
if (!mysqli_query($con,$sql))
{
die('Error: ' . mysqli_error($con));
}
//if it works
$_SESSION['completed'] = true;
header("location:loggedin.php");
mysqli_close($con);
}
}
?>
PHP for getting data back from database
include('db_login2.php');
$con = mysqli_connect("$host", "$username", "$password", "$db_name");
if(mysqli_connect_errno($con)) {
echo "failed to connect" . mysql_connect_error();
}else {
$username = $_SESSION['username'];
$result = mysqli_query($con,"SELECT * FROM users WHERE username='$username'");
echo $row = mysqli_fetch_array($result);
mysqli_close($con);
}
?>
I was wondering if it was possible to do something like having inline php in the html form which comes to life depending on the value of the user's original selection.
Any thoughts?
Thanks for any help!

Answer 1

To populate the survey with the users previous answers, you would us the $row = mysqli_fetch_array($result);
<?php
... // all your other code
$row = mysqli_fetch_array($result);
?>
<form name="questionnaire" action="submit_survey.php" method="post">
<input name="Q1" type="radio" value="1" <?php if($row['Public_Services'] == 1) echo 'checked="checked"'?>>1
<input name="Q1" type="radio" value="2" <?php if($row['Public_Services'] == 2) echo 'checked="checked"'?>>2
<input name="Q1" type="radio" value="3" <?php if($row['Public_Services'] == 3) echo 'checked="checked"'?>>3
<input name="Q1" type="radio" value="4" <?php if($row['Public_Services'] == 4) echo 'checked="checked"'?>>4
<input name="Q2" type="radio" value="1" <?php if($row['Politicians'] == 1) echo 'checked="checked"'?>>1
<input name="Q2" type="radio" value="2" <?php if($row['Politicians'] == 2) echo 'checked="checked"'?>>2
<input name="Q2" type="radio" value="3" <?php if($row['Politicians'] == 3) echo 'checked="checked"'?>>3
<input name="Q2" type="radio" value="4" <?php if($row['Politicians'] == 4) echo 'checked="checked"'?>>4
</form>
or using a php for loop
<?php
... // all your other code
$row = mysqli_fetch_array($result);
echo '<form name="questionnaire" action="submit_survey.php" method="post">';
for($i=1;$i<=4;$i++){
echo '<input name="Q1" type="radio" value="'.$i.'"';
if($row['Public_Services'] == $i) echo 'checked="checked"';
echo " />".$i."<br />\n";
}
echo "<br />\n";
for($i=1;$i<=4;$i++){
echo '<input name="Q2" type="radio" value="'.$i.'"';
if($row['Politicians'] == $i) echo 'checked="checked"';
echo " />".$i."<br />\n";
}
echo "</form>";
?>

Answer 2

You could either output the form's results as values in the HTML itself, or you could create a small Javascript snippet via PHP that will set the appropriate values once the form is loaded. The second method is much cleaner, but your clients will need to support Javascript for this to work. This isn't usually a problem, unless they've explicitly disabled Javascript, in which case the page just won't show the form data.
Using jQuery, you could do something like:
$(function() {
<?php
if (isset($foo)) { // foo is a value from your database
echo '$("#fooField").val("' . $foo . '")';
}
// ....more fields here
?>
}
Thus, the aforementioned code will output a jQuery code on load which will set the appropriate values for all of the form elements

Answer 3

You can have tags inside the html and echo the form based on the values.
Or you simply echo every line.
Edit: You might have a look at this: http://php.net/manual/en/control-structures.alternative-syntax.php