Where do PHP echos go when you are posting to a page?-CodePudding

This might be a dumb question. I'm fairly new to PHP. I am trying to get a look at some echo statements from a page I'm posting to but never actually going to. I can't go directly to the page's url because without the post info it will break. Is there any way to view what PHP echos in the developer console or anywhere else?

Here is the Ajax:

function uploadImage(image) {
  
      var data = new FormData();
      data.append("image", image);

      imgurl = 'url';
      filepath = 'path';
      $.ajax({
        url: imgurl,
        cache: false,
        contentType: false,
        processData: false,
        data: data,
        type: "post",
        success: function(url) {
          var image = $('<img >').attr('src', path   url);
          $('#summernote').summernote("insertNode", image[0]);
        },
        error: function(data) {
          console.log(data);
        }
      });
    }

And here is the php file:

<?php
  $image = $_FILES['image']['name'];
  $uploaddir = 'path';
  $uploadfile = $uploaddir . basename($image);
  if( move_uploaded_file($_FILES['image']['tmp_name'],$uploadfile)) {
    echo $uploadfile;
  } else {
    echo "Unable to Upload";
  }
?>

So this code runs fine but I'm not sure where the echos end up and how to view them, there is more info I want to print. Please help!

CodePudding user response：

Here is a basic example...

HTML (Form)

<form action="script.php" method="POST">
    <input name="foo">
    <input type="submit" value="Submit">
</form>

PHP Script (script.php)

<?php

    if($_POST){
        echo '<pre>';
        print_r($_POST); // See what was 'POST'ed to your script.
        echo '</pre>';
        exit;
    }

    // The rest of your PHP script...

Another option (rather than using a HTML form) would be to use a tool like POSTMAN which can be useful for simulating all types of requests to pages (and APIs)

CodePudding user response：

You already handle the response from PHP (which contains all the outputs, like any echo)

In the below code you have, url will contain all the output.

To see what you get, just add a console.log()

$.ajax({ 
    ...
    success: function(url) {
        // Output the response to the console
        console.log(url);

        var image = $('<img >').attr('src', path   url);
        $('#summernote').summernote("insertNode", image[0]);
    },
    ...
}

One issue with the above code is that if the upload fails, your code will try to add the string "Unable to upload" as the image source. It's better to return JSON with some more info. Something like this:

// Set the header to tell the client what kind of data the response contains
header('Content-type: application/json');

if( move_uploaded_file($_FILES['image']['tmp_name'],$uploadfile)) {
    echo json_encode([
        'success' => true,
        'url'     => $uploadfile,
        // add any other params you need
    ]);
} else {
    echo json_encode([
        'success' => false,
        'url'     => null,
        // add any other params you need
    ]);
}

Then in your Ajax success callback, you can now check if it was successful or not:

$.ajax({ 
    ...
    dataType: 'json', // This will make jQuery parse the response properly
    success: function(response) {
        if (response.success === true) {
            var image = $('<img >').attr('src', path   response.url);
            $('#summernote').summernote("insertNode", image[0]);
        } else {
            alert('Ooops. The upload failed');
        }
    },
    ...
}

If you add more params to the array in your json_encode() in PHP, you simply access them with: response.theParamName.