如何根据数据库输出的内容显示输入字段

Question

I am trying to displaying the input fields depending upon the what outputs are coming from the database. Please share any idea or logic in this. I have more than 25 fields.

Page1.php

I have Name(Text type), Email(Email type), gender(Radio), country(Select dropdown), Address( Textarea) in the form. The user will click on check box whatever he needs from the form and click on submit then the value of the fields which he selects will store in the database.

Form page

Page2.php

Here I am displaying the fields which are selected from Page1.php but also I need input type related to which fields come from the database.

For example: In the page1 I choose Name, Email, Country and submitted the form then In page2 I have to display the output <input type="text" name="Name">,<input type="text" name="Name">,<select></select>.

I need to know what logic I have to use it. Can any one help me with the script?

Form code

<form action="" method="post">
<input type="checkbox" name="check-fields[]"  value="Name">
<label>Name(Text type)</label>
<input type="checkbox" name="check-fields[]"  value="Email">
<label>Email(Email type)</label>
<input type="checkbox" name="check-fields[]"  value="Gender">
<label>Gender(Radio type)</label>
<input type="checkbox" name="check-fields[]"  value="Select">
<label>Country(Drop down)</label>
<input type="checkbox" name="check-fields[]"  value="textarea">
<label>Address(Textare)</label>
<input type="submit" name="submit" value="submit">
</form>

Storing the value in the database

if (isset($_POST['submit'])) {
// prepare and bind
$a=$_POST['check-fields'];
$elements= implode(',',$a);
$sql_elements="INSERT INTO test (elements) VALUES (?)";
$stmt = $conn->prepare($sql_elements);
$stmt->bind_param("s",  $elements);
$stmt->execute();
echo "New records created successfully";
$stmt->close();
$conn->close();
}

Page 2

I am getting output but how can I set the input fields?

$sql_fields="SELECT elements FROM test WHERE id=3";
if ($stmt = $conn->prepare($sql_fields)) {
            $stmt->execute(); 
            $stmt->bind_result($elements);
        while ($stmt->fetch()) {
            $arr=explode(",", $elements);
            echo "<pre>";
            print_r($arr);
            echo "</pre>";
       }
   }

$stmt->close();
$conn->close();
?>

Answer 1

You're trying to create a personalized form for different users if I'm correct, meaning that you will allow a user to generate a form based on a selection of fields the user can choose from.

What you need to do is create a table called "forminputtypes" where you create a row for each different input field you can think of, and want to give the user as a choice to select from. Your table would looke like this:

Id | FieldName | Type  | etc ->
-------------------------------
1  | Name      | text  | ...
2  | Email     | email | ...
3  | Gender    | radio | ...

You can add more columns in the table to store more information (think of possible values in a radio input or a dropdown).

Now at in Page1.php you select all input types from this table, and display them like you already are. However, the value of these checkboxes will be the corresponding Id value of the record in the database like so:

<input type="checkbox" name="check-fields[]"  value="3">

Now when someone chooses the Gender field, you can see in your Page2.php that the user did so by matching his choice '3' to the record in the database. If you want to save this information, you can create another table called UserFormInputFields which will function as a couple table between your user table and the FormInputTypes table.

When you want to display the form in Page2.php, you simply get all the input fields the user chose by selecting on Id from the FormInputTypes table. Since you know the type of each input field (because it's a column in the table) you can display them all correctly.

Answer 2

If you want to do it in core PHP only.Just put if else around the html tags.If value is coming from Database then you show particular tag else nothing. You can modify your page 2 code like.

<?php
$sql_fields="SELECT elements FROM test WHERE id=3";
if ($stmt = $conn->prepare($sql_fields)) {
            $stmt->execute(); 
            $stmt->bind_result($elements);
        while ($stmt->fetch()) {
            $arr=explode(",", $elements);
            echo "<pre>";
            print_r($arr);
            echo "</pre>";
       }
   }

$stmt->close();
$conn->close();
?>
<form action="" method="post">
<?php if(in_array("Name",$arr)) { ?>
 <input type="text" name="name"  >
 <label>Name</label>
<?php } ?>
<?php if(in_array("Email",$arr)) { ?>
 <input type="email" name="email">
 <label>Email</label>
<?php } ?>
<?php if(in_array("Gender",$arr)) { ?>
 <input type="radio" name="gender"  value="Male">
 <input type="radio" name="geneer"  value="Female">
<label>Gender(Radio type)</label>
<?php } ?>
//.............write other fiels like this.
<input type="submit" name="submit" value="submit">
</form>

I hope this helped.