Question

当我运行ajax操作时，返回文本的第一个字符为返回字符。但是，用于发送的回显文本是固定的，没有空格或返回。如果我警告退货文本，则没有退货。

我正在从Godaddy主机运行代码，并在PHP5上运行MSQL

innerHTML that crates the button - _("comment_"+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 = "";

js代码-

function deleteComment(commentid,commentbox){
    var conf = confirm("Press OK to confirm deletion of this status and its replies");
    if(conf != true){
        return false;
    }
    var ajax = ajaxObj("POST", "php_parsers/comments_system.php");
    ajax.onreadystatechange = function() {
        if(ajaxReturn(ajax) == true) {
            var deleteOk = ajax.responseText;  //This is retuning a return carraige
            if(deleteOk = "1"){
                _(commentbox).style.display = 'none';
                _("replytext_"+commentid).style.display = 'none';
                _("replyBtn_"+commentid).style.display = 'none';
            } else {
                alert(ajax.responseText);
            }
        }
    }
    ajax.send("action=delete_comment&commentid="+commentid);
}

这是PHP代码

if (isset($_POST['action']) && $_POST['action'] == "delete_comment"){
    if(!isset($_POST['commentid']) || $_POST['commentid'] == ""){
        mysqli_close($db_conx);
        echo "comment id is missing";
        exit();
    }
    $commentid = preg_replace('#[^0-9]#', '', $_POST['commentid']);
    // Check to make sure this logged in user actually owns that comment
    $sql = "SELECT author_name FROM comments WHERE parent_id=$commentid LIMIT 1";
    $query = mysqli_query($db_conx, $sql); 
    while ($row = mysqli_fetch_array($query, MYSQLI_ASSOC)) { 
        $author_name  = $row[author_name];
    }
    if ($author_name == $log_username) {
        mysqli_query($db_conx, "DELETE FROM comments WHERE parent_id=$commentid");
        mysqli_close($db_conx);
        echo "1";
        exit();
    }else{
        echo "$sql - $author_name";
    }
}

所以我现在必须编辑我的代码以使用int值，因为看起来IF语句仍然与返回值匹配，但是如果我使用文本return则它不匹配。我有几乎相同的代码来创建注释，但是按预期工作。

AJAX返回文字中的回车

0 个答案: