What does ajax do with the echo statements in a php function called from javascript_问答_开发者

What does ajax do with the echo statements in a php function called from javascript

开发者 https://www.devze.com 2023-03-29 03:34 出处：网络

I have a script called at the end of an html page like this. <body> <script type=\"text/javascript\">

相关专题：echo php

I have a script called at the end of an html page like this.

<body>

<script type="text/javascript">

 $(document).ready(function() {
               //define php info and make ajax call to recreate divs from XML data
               $.ajax({
                   url: "get_nodes.php",
                   type: "POST",
                   data: { },
                   cache: false,
                   success: function (response) {

                       if (response != '') 
                       {
                          alert(response);

                       }
                   }
               });
 });

</script>

</body>

The php script has echo statements that write javascript. The echo statements are not printed in the document. They are being sent somewhere, I don't know where. If I look at the response from the ajax call to php with the alert function, all the echo statements are printed in the alert box as I would expect them to be printed in the document itself.

My questions are:

What does ajax do with the php echo statements?
What is the way to have them printed in the document?

Here is the PHP file get_nodes.php called through ajax.

<?php

function get_nodes() {
// load SimpleXML
$nodes = new SimpleXMLElement('communities.xml', null, true);

foreach($nodes as $node) // loop through 
{

        echo "\n<script type='text/javascript'>\n";

        echo "  var newdiv = document.createElement('div');\n";
        echo "  newdiv.id = '".$node['ID']."';\n";
        echo "  newdiv.className ='comdiv ui-widget-content';\n";
        echo "  newdiv.style.top = '".$node->TOP."'\n";
        echo "  newdiv.style.left = '".$node->LEFT."';\n";
        echo "  newdiv.style.width = '".$node->WIDTH."';\n";
        echo "  newdiv.style.height = '".$node->HEIGHT."';\n";
        echo "  $( '#page' ).append(newdiv);\n";
        echo "  var heading = document.createElement('p');\n";
        echo "  heading.innerHTML = '".$nod开发者_如何学Goe->NAME."';\n";
        echo "  heading.className ='comhdr editableText ui-widget-header';\n";
        echo "  $('#".$node['ID']."').append(heading);\n";
        echo "  $('#".$node['ID']."').resizable();";
        echo "  $('#".$node['ID']."').draggable();";
        echo "  $('#".$node['ID']."').draggable('option', 'handle', '.comhdr');";

        printf('$("#%s").append("<a href=\\"#\\" onClick=\\"delete_div(\'%s\');\\">Delete</a>&nbsp;&nbsp;");', $node

['ID'], $node['ID']);

        printf('$("#%s").append("<a href=\\"#\\" onClick=\\"add_url(\'%s\');\\">Add URL</a>&nbsp;&nbsp;");', $node

['ID'], $node['ID']);

        echo "\n</script>\n"; 
}

   return;
}

echo get_nodes();

?>

As Incognito pointed out, your ajax call does not care about what your PHP function is doing - all that matters in this case is the content returned from the server, not how the content is created.

Try adding the dataType option to your ajax call to specify a response type of script, and use the append or appendTo function to add the script to the document body:

$.ajax({
    url: "get_nodes.php",
    type: "POST",
    data: { },
    dataType: 'script',
    cache: false,
    success: function (response) {
        if (response != '') 
        {
            $(document.body).append(response);
        }
    }
});

Also, a suggestion: In your server-side function, why not echo HTML markup instead of a javascript function which creates markup?