Trying to create variables with name of input field from an array but not working

Question

I am working on a project and trying to create a generic section to save all kinds of form data to the database. I wrote down the following code to send all the data to php field and hence send it to the database. But the issue is, its giving me an error.

if(isset($_POST['data_for']) && $_POST['data_for']=='save') {
    $data = $_POST['formdata'];
    print_r($data); // This is showing proper array as an output
    foreach ($data as $key => $value) {     
        echo $value['name']; //This gives the key (index value) of the form "Eg. email"
        echo $value['value']; //This gives the value of the user input "eg. [email protected]"
        
        $$value['name'] = $value['value']; //This line gives error as "Array to string conversion"
    }
    echo $email; //This is just a test to print a variable created in runtime 
   //The insertion to database code goes here.

}

The above code is getting values from the below jquery

$(document).on('submit','form.cat1', function(e){
    e.preventDefault();
    var forum = $(this).attr('forum');
    var method = $(this).attr('method');
    var nonce = $(this).attr('nonce');
    var data_for = $(this).attr('data-for');
    var formdata = $(this).serializeArray();
    //alert(formdata);
    $.ajax({
        url:'formSubmitPoint.php',
        method:method,
        data:{formdata:formdata, nonce:nonce, forum:forum, data_for:data_for},
        //processData: false,
        //contentType: false, 
        success:function(data){
            console.log(data);
            if (data['result']=='success') {
                if (data['action']=='redirect') {
                    window.location.href=data['location'];
                }
                if (data['action']=='show') {
                    $(data['location']).html(data['message']);
                }
            }
            if (data['result']=='error') {
                if (data['action']=='show') {
                    $(data['location']).html(data['message']);
                }
            }
        },
        error:function(data){
            console.log(data);
        }
    });
})

And the jquery pulls data from the below html

<form class="was-validated cat1" method="post" forum='feedback' data-for="save" nonce="{$nonce}">
  <div class="form-group">
    <label for="newPass">Name</label>
    <input type="text" class="form-control" id="name" placeholder="Your Name" name="name" required>
    <div class="valid-feedback">Valid.</div>
    <div class="invalid-feedback">Please fill out this field.</div>
  </div>
  <div class="form-group">
    <label for="newPass">Email</label>
    <input type="email" class="form-control" id="email" placeholder="Your Email Address" name="email" required>
    <div class="valid-feedback">Valid.</div>
    <div class="invalid-feedback">Please fill out this field.</div>
  </div>
  <div class="form-group">
    <label for="newPass">Contact Number</label>
    <input type="number" class="form-control" id="contact" placeholder="Your contact number" name="contact" required>
    <div class="valid-feedback">Valid.</div>
    <div class="invalid-feedback">Please fill out this field.</div>
  </div>
  <div class="form-group">
    <label for="newPass">Message</label>
    <textarea class="form-control" id="message" name="message" required></textarea>
    <div class="valid-feedback">Valid.</div>
    <div class="invalid-feedback">Please fill out this field.</div>
  </div>
  <button type="submit" name="submit" class="btn btn-primary btn-sm">Submit</button>
</form>

$$value['name'] = $value['value'[;, fix the syntax $$value['name'] = $value['value']; — Ericgit
– Ericgit, Commented Jul 24, 2020 at 2:49
Sorry it was a typo here... I have written it correctly in my code, still its not working. — Vikash Mishra
– Vikash Mishra, Commented Jul 24, 2020 at 2:52
@Nikkorian The $$var (double dollar) is a reference variable that stores the value of the $variable inside it. — Ericgit
– Ericgit, Commented Jul 24, 2020 at 2:54

nagyl · Accepted Answer · 2020-07-24 03:47:26Z

1

$$value['name'] Will give me $email when the value of $value['name'] will be email

There is no possible way of doing that. You can store it's value, or a reference to that object by doing

$email = $value['value'];  //this is a copied object
$email = &$value['value']; //this is a reference

EDIT

You can do

foreach ($data as $key => $value) {     
    echo $value['name'];
    echo $value['value'];

    $text  = $value['name'];
    $$text = $value['value'];

    echo $email;
}

You can't create a Variable Variable from an array, because you would convert an array into a string. You must create a string type variable to help it.

foreach ($data as $key => $value) {     
    $text  = $key;
    $$text = $value;

    echo $email;
}

edited Jul 24, 2020 at 3:47

answered Jul 24, 2020 at 3:11

nagyl

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10 Comments

Vikash Mishra Over a year ago

@nagyi, it works in other projects with the same way. I have been working on the same set of code for long. This time there is something that I am missing which leads to the error.

nagyl Over a year ago

@VikashMishra can you please tell me the difference between $email, and it's value 'email', if not a reference/or a new object? Or can you tell me what $$value['name'] does?

Vikash Mishra Over a year ago

@nagyi I am just trying to create a new variable with the value of the variable. Eg $$value['name'] Will give me $email when the value of $value['name'] will be email when it will be something else the variable name will be something else. I am not hardcoding the name of the variable as it will be generated in runtime.

nagyl Over a year ago

That's ok, but why do you need that? Why can't you just store the value in a normal variable? xd Why do you benefit from doing that instead of a logical way?

Vikash Mishra Over a year ago

It helps me in multiple aspects, like I do not have to hardcode the name of the variable for each and every form that I submit using this method.

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