//html

 <form name="useraddcooperativestructureform" class="useraddcooperativestructureclass" id="useraddcooperativestructureformid"> <input type="file" class="cooperativestructureclass" name="cooperativestructurename" id="imagenameid" placeholder="Cooperative Structure"><br> <button type="button" class="btn btn-primary" id="coopuploadbtnid">Upload</button> </form>

//php

 <?php require('../database/connection.php'); $pimage = $_FILES['cooperativestructurename']['name']; $pimagetmp = $_FILES['cooperativestructurename']['tmp_name']; $folder = "css/"; move_uploaded_file($pimagetmp, $folder.$pimage); $query = "INSERT INTO tbl_coopstructure VALUES(null, '$pimage')"; $sqlquery = mysqli_query($connection,$query); ?>

//javascript

 var coopbtn = document.getElementById('coopuploadbtnid'); if(coopbtn){ coopbtn.addEventListener('click',function(){ var filename = document.forms['useraddcooperativestructureform']['cooperativestructurename'].value; $.post("../controller/admincoopstructureinsert.php",{"name":filename},function(){ setTimeout(function(){ $('.message-validation-upload-structure').css('display','flex'); $('.message-validation-upload-structure').css('background-color','#02ff2a'); $('.insertclass').html("User successfully uploaded image"); },1) setTimeout(function(){ $('.message-validation-upload-structure').css('display','none'); },3000) }) }) }

Cuando inserté un archivo correctamente, pero no hay ningún valor en la base de datos, solo la identificación única tiene un valor que se inserta. alguien me puede decir cual es la solucion? #principiante