Insert Only One Record In MYSQL database using php

i am working on a project where i want to insert only one record in a specific time.

For Example: in the project client rate a contractor only once.(when the construction is in working) once he rated he cannot rate second time whatsoever.

I checked online but didn’t find what i am looking for.

Now I have 2 Ideas.

1- Either insert checkboxes and make them disable permanently, once the rating is done(with server side validation).

2- Rate according to numbers (like 7/10) and display them.

which one is batter strategy to opt.

Code Part:

CheckBox Solution Code(no solution available on google on serverside validation):

Checkbox: <input type="checkbox" id="myCheck">

<p>Click the "Try it" button to disable the checkbox.</p>

<button onclick="myFunction()">Try it</button>

<script>
function myFunction() {
  document.getElementById("myCheck").disabled = true;
}
</script>

Also checked this on stackoverflow (but this is not i am looking for):

[an example serverside solution link][1]

Number Solution:

<table>
  <tr><td>Rate Your Contractor (Out Of 10)</td><td><input id="intLimitTextBox"></td></tr>
</table>

<script>

function setInputFilter(textbox, inputFilter) {
  ["input", "keydown", "keyup", "mousedown", "mouseup", "select", "contextmenu", "drop"].forEach(function(event) {
    textbox.addEventListener(event, function() {
      if (inputFilter(this.value)) {
        this.oldValue = this.value;
        this.oldSelectionStart = this.selectionStart;
        this.oldSelectionEnd = this.selectionEnd;
      } else if (this.hasOwnProperty("oldValue")) {
        this.value = this.oldValue;
        this.setSelectionRange(this.oldSelectionStart, this.oldSelectionEnd);
      } else {
        this.value = "";
      }
    });
  });
}

// Install input filters.
setInputFilter(document.getElementById("intLimitTextBox"), function(value) {
  return /^d*$/.test(value) && (value === "" || parseInt(value) <= 10); });

</script>

Is There any php way(or both ways) to implement the solution of my problem?

PS: i am not very expert in web devolepment, sorry if any mistake i am doing.

Answer