2013-03-09 15:53
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php + ajax +评论系统错误[关闭]

i am creating a comment system using php mysql and ajax but keep geting a syntax error and i can not find the dam error this is the comments.php code if anyone help me a will appreciate that



mysql_connect("localhost","root",""); //connect to the database: change user and pass to your own DB's user and pass
mysql_select_db("ajaxComment"); //change this to the name of your mysql database
if(isset($_POST['type'] == "addcomment")) {
    $author = mysql_real_escape_string($_POST['author']);
    $message = mysql_real_escape_string($_POST['message']);
    //the mysql_real_escape_string() is to prevent sql injections and such.
    if($author == "" || $message == "") { //one of the fields was empty...
        die("<p><font color='red'>Error: Please fill out BOTH fields.</font></p>"); //...tell them
    $q1 = mysql_query("INSERT INTO `comments` (`author`, `message`) VALUES ('$author', '$message')") or die("<p>Mysql Error: <b>".mysql_error()."</b></p>"); //if there was a mysql error, let them know
    echo "<p><font color='green'>Comment added successfully. You should see it in a few seconds.</font></p>";
} elseif(isset($_POST['type'] == "getcomments")) { //ajax wants the comments.
    $q1 = mysql_query("SELECT * FROM `comments`");
    $n1 = mysql_num_rows($q1);
    if($n1 == 0) { //no comments found
        die("<p><font color='red'>No comments were found.</font></p>");
    echo "<table border=1>";
    echo "<tr><td><b>Author</b></td><td><b>Comment</b></td></tr>";
    while($r1 = mysql_fetch_assoc($q1)) { //if there were any comments, loop through them
        $a = $r1['author'];
        $m = $r1['message'];
        echo "<tr><td>$a</td><td>$m</td></tr>";
    echo "</table>";
} else {
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "">
<html xmlns="">
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Ajax Comment Tutorial</title>
<script src="" type="text/javascript"></script>
<!--The above code will get the jquery AJAX library and include it into the page-->
<script type="text/javascript">
(document).ready(function() {
    $("submit").click(function(){ //this will execute code if the submit button was clicked
        var user = $("#author").val(); //receive the contents of the field with the ID "author" and put it into the user var
        var comment = $("#message").val(); //same as above, but with the message field instead.
        $.post("comments.php",{type: "addcomment",author: user,message: comment},function(data){$("#return").html(data)});
        //the first parameter of the above, is the page to send the post data too. In this example, this page is called comments.php, so we send it there.
        //the second parameter is the post data, it goes: {pn: pd} where pn is the name you use in PHP's $_POST and pd is the data it will contain.
        //the third parameter is for the return data, in this case we sent it to the "return" div we made below our form. The first two parameters are not required.
        return false; //this will stop the page submitting, so it won't refresh.
    function getComments() { //this function will collect the comments
        $.post("comments.php",{type: "getcomments"},function(data){$("#comments").html(data)});
        //for an explanation on this, simply read the click function above.
        setInterval("getComments()",10000); //gets the comments every 10 seconds.

    <div style="font-size: 18px;">
        <p>Current Comments: </p>
        <div id="comments"></div> <!--This div will be filled with AJAX-->
        <hr />
        <p>Add a comment:</p>
        <form action="comments.php" method="post">
            <p>Username: <input type="text" name="author" id="author" /></p>
            <p>Comment: <textarea name="message" cols="70" rows="10" id="message"></textarea></p>
            <p><input type="submit" name="submit" value="Add Comment" id="submit" /></p>
        <div id="return"></div> <!--This div will be filled with the return data from the comment processing script at the top-->
<?php }  ?>
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1条回答 默认 最新

  • doyhq66282 2013-03-09 15:57

    PHP error is in this line:

    if(isset($_POST['type'] == "addcomment")) {

    you probably want it to be:

    if (isset($_POST['type']) && $_POST['type'] == "addcomment") {
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