PHP에서 자바 스크립트를 넣을 때 폼을 실행할 수 없습니다

Question

여기 내 PHP 코드와 자바 스크립트를 넣은 파일은 같지만 자바 스크립트 코드를 입력하면 PHP의 폼이 데이터베이스에 추가되지 않습니다. 여기

echo '<form action="$_SERVER["REQUEST_URI"];" method="POST">'; 

     echo '<div class="col-sm-12">'; 

      $t = mysqli_query($connect, $q); 

       echo '<div class="col-sm-6">'; 
        echo "<select name='start' class='form-control'required >"; 
          echo '<option value="">Select Location</option>'; 
         while ($row = $result->fetch_assoc()) { 
          unset($loc_name); 
          $loc_name = $row['loc_name']; 

          echo '<option value="'.$loc_name.'">'.$loc_name.'</option>'; } 
        echo "</select><br>"; 
       echo '</div>'; 

       echo '<div class="col-sm-6">'; 
        echo '<select name="status" class="form-control" required >'; 
         echo '<option value="">Status</option>'; 
         echo '<option>Priority</option>'; 
         echo '<option>Not Priority</option>'; 
        echo '</select><br>'; 
       echo '</div>'; 

       echo '<div class="col-sm-6">'; 
        echo '<input type="number" name="pieces" placeholder="Piece(s) of Item" class="form-control" required />'; 
       echo '</div>'; 

       echo '<div class="col-sm-6">'; 
        echo '<input type="date" name="date" class="form-control" required />'; 
       echo '</div>'; 

       echo '<center><button type="submit" name="submit" class="btn btn-default">Submit</button></center>'; 
       echo '</form>'; 

(나는. 여기에 삽입 코드를 삽입하지 않는 것)은 PHP 당신은 당신에게 자바 스크립트 내부에 스크립트 태그를 묶어야합니다

   echo ' 
        <div class="form-group col-lg-6"> 
         <label for="exampleInputText">Total Price</label> 
         <input type="text" name="totalval" class="form-control" id="totalval" onchange="updateDue()"> 
        </div> 
        <div class="form-group col-lg-6"> 
         <label for="exampleInputText">Initial Deposit</label> 
         <input type="text" name="inideposit" class="form-control" id="inideposit" onchange="updateDue()"> 
        </div> 
        <div class="form-group col-lg-6"> 
         <label for="exampleInputText">Outstanding Dues</label> 
         <input type="text" name="remainingval" class="form-control" id="remainingval"> 
        </div>'; 


echo ' 
<script type="text/javascript"> 
function updateDue() { 

    var total = parseInt(document.getElementById("totalval").value); 
    var val2 = parseInt(document.getElementById("inideposit").value); 

    // to make sure that they are numbers 
    if (!total) { total = 0; } 
    if (!val2) { val2 = 0; } 

    var ansD = document.getElementById("remainingval"); 
    ansD.value = total - val2; 
} 
</script>'; 

Answer 1

내 자바 스크립트 코드입니다. 예

echo '<script> 
function updateDue() { 

    var total = parseInt(document.getElementById("totalval").value); 
    var val2 = parseInt(document.getElementById("inideposit").value); 

    // to make sure that they are numbers 
    if (!total) { total = 0; } 
    if (!val2) { val2 = 0; } 

    var ansD = document.getElementById("remainingval"); 
    ansD.value = total - val2; 
    } 
</script> ';