當您發布到頁面時，PHP回聲會去哪里？

這可能是一個愚蠢的問題。我對 PHP 相當陌生。我正在嘗試從我要發布但實際上從未去過的頁面中查看一些回聲陳述句。我不能直接轉到頁面的網址，因為沒有帖子資訊它會中斷。有沒有辦法在開發者控制臺或其他任何地方查看 PHP 的回聲？

這是阿賈克斯：

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);
        }
      });
    }

這是php檔案：

<?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";
  }
?>

所以這段代碼運行良好，但我不確定回聲最終在哪里以及如何查看它們，我想列印更多資訊。請幫忙！

uj5u.com熱心網友回復：

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.

uj5u.com熱心網友回復：

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)

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