Question

我正在尝试将输入字段中的值插入到数据库中，并将ajax作为会话系统的一部分。我正在使用输入表单，如下所示。

<input data-statusid="' .$statuscommentid. '" id="reply_'.$statusreplyid.'" class="inputReply" placeholder="Write a comment..."/>

使用以下jquery我在用户按下回车键时执行一个功能。

  $(document).ready(function(){ 
$('.inputReply').keyup(function (e) {
    if (e.keyCode === 13) {
       replyToStatus($(this).attr('data-statusid'), '1',$(this).attr("id"));
    }
  });

  });

在这个函数中是我遇到问题的地方，我用jquery调用函数没有问题，但我对ajax做错了什么而且我不知道是什么？

$.ajax({ type: "POST", url: $(location).attr('href');, data: dataString, cache: false, success: function(){ $('#'+ta).val(""); } });

此外，这是我用来插入数据库的php

    <?php //status reply input/insert
//action=status_reply&osid="+osid+"&user="+user+"&data="+data
if (isset($_POST['action']) && $_POST['action'] == "status_reply"){
    // Make sure data is not empty
    if(strlen(trim($_POST['data'])) < 1){
        mysqli_close($db_conx);
        echo "data_empty";
        exit();
    }
    // Clean the posted variables
    $osid = preg_replace('#[^0-9]#', '', $_POST['sid']);
    $account_name = preg_replace('#[^a-z0-9]#i', '', $_POST['user']);
    $data = htmlentities($_POST['data']);
    $data = mysqli_real_escape_string($db_conx, $data);
    // Make sure account name exists (the profile being posted on)
    $sql = "SELECT COUNT(userid) FROM user WHERE userid='$userid' LIMIT 1";
    $query = mysqli_query($db_conx, $sql);
    $row = mysqli_fetch_row($query);
    if($row[0] < 1){
        mysqli_close($db_conx);
        echo "$account_no_exist";
        exit();
    }
    // Insert the status reply post into the database now
    $sql = "INSERT INTO conversation(osid, userid, postuserid, type, pagetext, postdate)
            VALUES('$osid','$userid','$postuserid','b','$pagetext',now())";
    $query = mysqli_query($db_conx, $sql);
    $id = mysqli_insert_id($db_conx);
    // Insert notifications for everybody in the conversation except this author
    $sql = "SELECT authorid FROM conversation WHERE osid='$osid' AND postuserid!='$log_username' GROUP BY postuserid";///change log_username
    $query = mysqli_query($db_conx, $sql);
    while ($row = mysqli_fetch_array($query, MYSQLI_ASSOC)) {
        $participant = $row["postuserid"];
        $app = "Status Reply";
        $note = $log_username.' commented here:<br /><a href="user.php?u='.$account_name.'#status_'.$osid.'">Click here to view the conversation</a>';
        mysqli_query($db_conx, "INSERT INTO notifications(username, initiator, app, note, date_time) 
                     VALUES('$participant','$log_username','$app','$note',now())");
    }
    mysqli_close($db_conx);
    echo "reply_ok|$id";
    exit();
}

＆GT;

提前感谢任何帮助，我们将不胜感激

Answer 1

为什么不为Ajax调用设置正确的URL而不是使用location.href？

var ajax = ajaxObj("POST", location.href);

另外，我猜ajaxObj没有定义或编码良好。您正在使用jQuery，为什么不尝试jQuery ajax？

http://api.jquery.com/jquery.ajax/

Answer 2

顺便说一下，你有一个很酷的ajax POST和DB crud操作的例子： jQuery Ajax POST example with PHP

Answer 3

var ajax = ajaxObj("POST", location.href);
    ajax.onreadystatechange = function() {
        if(ajaxReturn(ajax) == true) {
            var datArray = ajax.responseText.split("|");
            if(datArray[0] == "reply_ok"){
                var rid = datArray[1];
                data = data.replace(/</g,"&lt;").replace(/>/g,"&gt;").replace(/\n/g,"<br />").replace(/\r/g,"<br />");
                _("status_"+sid).innerHTML += '<div id="reply_'+rid+'" class="reply_boxes"><div><b>Reply by you just now:</b><span id="srdb_'+rid+'"><a href="#" onclick="return false;" onmousedown="deleteReply(\''+rid+'\',\'reply_'+rid+'\');" title="DELETE THIS COMMENT">remove</a></span><br />'+data+'</div></div>';
                _("replyBtn_"+sid).disabled = false;
                _(ta).value = "";
                alert("reply ok!");
            } else {
                alert(ajax.responseText);
            }
ajax.send("action=status_reply_ok&sid="+sid+"&user="+user+"&data="+data);
        }
    }

Ajax数据库插入不起作用

3 个答案: