ajax 调用的 Php 文件未回显 javascript 代码

Question

I'm trying to create a page where a user can upload a file and select the people they want to email it to. Once they click submit, I prevent page refresh and reset their inputs in the form. Now I want to place their previously entered information into a table on the same page (different section of the page).

Answer 1

If you did want to proceed with this concept, you would capture the output of the PHP script in a done() function, insert it in an element on the page, and run eval() . Like this...

$.ajax({
    type: "POST",
    url: "../FileDrop/dbSystem.php",
    data: {tags: JSON.stringify(tags), file: 
          $('input[name=fileName]').val()};
}).success(function(result) {
  $( '#element' ).html(result);
  $( '#element script' ).each( () => { $(this).html().eval() }) 
});

But it would make alot more sense to return data that you use to execute the javascript in your page - that is, keep all that logic together rather than splitting some of it off into a PHP file.

<?php 
// php process that checks for 'valid'...
// create a json response
$output = array('valid' => 1);
echo json_encode($output);
?>

.... and in your JS

.success(function(result) {
  // result is a stringified JSON object. We need to convert it into an actual JSON object with parse()
  result = JSON.parse(result);
  // might not matter, but result.valid will be a number which JSON converts into a string. By adding the + right before the variable, that tells JS to use it as a number for the comparison
  if (+result.valid == 1) {
    $(".outputDiv").show();
  }
});