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2015-12-18 18:21 浏览 195

错误ajax和PHP“Undefined index”

I'm a new to AJAX so I'm sorry if the solution is obvious. I must send the dates to a MySQL table without refreshing the page and I currently have a problem doing this.

The code gives me the following error:

" Notice: Undefined index: post in inserisci.php on line 6"

FORM

 <form id="form" class="form-horizontal shadow-z-1" action="includes/inserisci.php" method="post">
        <fieldset>
            <legend style="text-align: center; position: relative; top: 8px;">Inserisci un Post</legend>
            <hr>
            <div class="form-group">
                <label for="inputTitle" class="col-lg-2 control-label">Titolo</label>
                <div class="col-lg-10">
                    <input type="text" class="form-control" id="inputText" name="titolo" placeholder="Scrivi qui il Titolo">
                </div>
            </div>
            <div class="form-group">
                <label for="textArea" class="col-lg-2 control-label" name="testo">Messaggio</label>
                <div class="col-lg-10">
                    <textarea id="messaggio"></textarea>
                </div>
                </div>

               <div class="form-group is-empty is-fileinput">
                    <label for="inputFile" class="col-md-2 control-label">File</label>

                    <div class="col-md-10">
                      <input type="text" readonly="" class="form-control" placeholder="Browse..." pmbx_context="19E61A0C-3526-4E51-8535-935982C4C335">
                      <input type="file" id="inputFile" multiple="" pmbx_context="1D2BCAEA-08CC-476A-8F4A-EF6BD51B9102">
                    </div>
                  <span class="material-input"></span></div>

            <div class="col-md-10 col-md-offset-2">
                      <button type="button" class="btn btn-default" onclick="document.getElementById('modalposta').style.display = 'none';">Cancel</button>
                      <button type="submit" class="btn btn-primary" onClick='send(); return false;' >Submit</button>
                    </div>
        </fieldset>
    </form>

INSERISCI.JS

  function send(){
var content = $.ajax({
        type: "POST",
        url: "index.php",
        data: {titolo:titolo & post:testo}           
  })
  .done(function() {
    alert( "success" );
  })                                                 
  .fail(function() {
    alert( "error" );
  })                                                
  .always(function() {
    alert( "complete" );
  });
}

INSERISCI.PHP

<?php

include('../core.php');

$titolo=$_REQUEST ['titolo'];
$post=$_REQUEST ['post'];

$sql = mysql_query("");

mysql_close();

?>

Other question. I need to reload the content of this without refresh the page.

INDEX.PHP

<?php $sql = mysql_query("")  or die ("Nessun errore");
    if (mysql_num_rows($sql) > 0)
    {


        while ($row = mysql_fetch_assoc($sql))
        {
            echo '

            <article class="flipper white-panel">
        <a class="flipcard flip" style="
        position: absolute;
        right: 0px;
        top: -10px;
        z-index:1;
    "><i class="material-icons" style="font-size: 40px;">bookmark</i></a>
        <div class="articolo">
        <img src="http://i.imgur.com/sDLIAZD.png" alt="">

            <h4><a href="#">'.$row ['titolo'].'</a></h4>
            <p>'.Markdown($row ['contenuto']).'</p>
             <div class="pull-right">
                            <span class="label label-default">alice</span>
                            <span class="label label-primary">story</span>
                            <span class="label label-success">blog</span>
                            <span class="label label-info">personal</span>
                            <span class="label label-warning">Warning</span>
                            <span class="label label-danger">Danger</span>
                          </div>
                          <hr>
                          </div>
        <div class="commenti">
            <ul class="comment-list">
                        <li>
                            <div class="comment-img">

                            </div>
                            <div class="comment-text">

                            </div>
                        </li>
                    </ul>
                    <input type="text" class="form-control inputcomment" placeholder="Lascia un Commento.."/>
                  </div>
      </article>
         ';
        }


    }

Thanks everyone.

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3条回答 默认 最新

  • 已采纳
    douzhuozhu9544 douzhuozhu9544 2015-12-18 18:24

    Notes:

    1. Please do not use mysql_* functions. They are deprecated.
    2. Your code is vulnerable to SQL Injection Attacks. Please make sure you address them.

    That's a syntax error. That's not a valid JavaScript Object. Change your data object to:

    data: {titolo:titolo, post:testo}
    

    Also you need to close the $.ajax before you are using .done.

      var content = $.ajax({
            type: "POST",
            url: "index.php",
            data: {titolo:titolo & post:testo}           // Should be closed here.
      })
      .done(function() {
        alert( "success" );
      })                                                 // Should not be closed. Remove ;
      .fail(function() {
        alert( "error" );
      })                                                 // Should not be closed. Remove ;
      .always(function() {
        alert( "complete" );
      });
      // });                                                Remove this!
    

    If nothing works out, please change these lines:

    $titolo = $_POST['titolo'];
    $post = $_POST['post'];
    

    And wait, did you get the values for those two?

    var titolo = $('input[name="titilo"]').val();
    var testo = $('input[name="testo"]').val();
    
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  • duancai9010 duancai9010 2015-12-18 18:25

    This is because the data you are passing in your AJAX request is not a valid javascript object and you are not getting the variables from the form:

    data: {titolo:titolo & post:testo}
    

    You should instead do something like:

    var titolo = $('form input[name="titolo"]').val();
    var testo = $('form input[name="testo"]').val();
    data: {
       titolo:titolo,
       post:testo
    }
    

    Also, as a side note, when accessing array elements, its not good form to put a space between the variable and the bracket. I'm surprised that even works.

    $_REQUEST ['titolo']
    

    Should be

    $_REQUEST['titolo']
    
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  • douxia2137 douxia2137 2015-12-18 19:27

    If you want to use ajax form submit, Update your form definition with

    <form id="form" class="form-horizontal shadow-z-1" action="#" method="post">
    

    use following updated javascript

    function send() {
        var content = $.ajax({
            type: "POST",
            url: "process.php",
            data: {
                titolo:  "Value for titolo",
                post: "Value for post"
            }
        }).done(function() {
            alert( "success" );
        }).fail(function() {
            alert( "error" );
        }).always(function() {
            alert( "complete" );
        });
    }
    
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