我正在使用AJAX通过SQL在表中创建新行。我的代码相当简单,但是我注意到它有时会在数据库中创建重复的条目(有时甚至是3)。有办法防止这种情况吗?这是我的代码(通过php echo发送)。

更新的主页面

echo "
    <script language='javascript'>


$('#newreportBTN').live(\"click\", function() {

 //Get variables from boxes
        var data = $(\"#newreport\").serialize();

    $.ajax({
            type: 'POST',
            url: \"report_new.php\",
            data: data,
           success: function (html) {
                //if process.php returned 1/true (send mail success)
                if (html==1) {
                    //Send notification to user
                     $('#adminReportInfo').fadeOut(300);

                //if process.php returned 0/false (send mail failed)
                } else alert('Sorry, unexpected error. Please try again later.');
            }
        });

        //cancel the submit button default behaviours
        return false;
    });


    </script>
    ";


report_new.php

<?php
//Start session
session_start();

//Include database connection details
require_once('config.php');

//Array to store validation errors
$errmsg_arr = array();

//Validation error flag
$errflag = false;

//Connect to mysql server
$link = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD);
if(!$link) {
    die('Failed to connect to server: ' . mysql_error());
}

//Select database
$db = mysql_select_db(DB_DATABASE);
if(!$db) {
    die("Unable to select database");
}

//Function to sanitize values received from the form. Prevents SQL injection
function clean($str) {
    $str = @trim($str);
    if(get_magic_quotes_gpc()) {
        $str = stripslashes($str);
    }
    return mysql_real_escape_string($str);
}



$company = $_POST['company'];
$claimnumber = $_POST['claimnumber'];
$fullname = $_POST['fullname'];
$dateofloss = $_POST['dateofloss'];
$foruser = $_POST['foruser'];
$status = $_POST['status'];


//Input Validations
if($company == '') {
    $errmsg_arr[] = 'Company Missing';
    $errflag = true;
}
if($claimnumber == '') {
    $errmsg_arr[] = 'Claim # Missing';
    $errflag = true;
}
if($fullname == '') {
    $errmsg_arr[] = 'Full name missing';
    $errflag = true;
}


//If there are input validations, redirect back to the registration form
if($errflag) {
    $_SESSION['ERRMSG_ARR'] = $errmsg_arr;
    session_write_close();
    header("location: user_error.php");
    exit();
}

//Create INSERT query
$qry = "INSERT INTO documents (company, claimnumber, fullname, dateofloss, foruser, status) VALUES ('$company', '$claimnumber','$fullname','$dateofloss','$foruser','$status')";
$result = @mysql_query($qry);

//Check whether the query was successful or not
if($result) {
    //header("location: admin.php?status=new");
    echo '1';
    exit();
}else {
    die("Query failed");
    echo mysql_error();
}
?>

最佳答案

检查以下内容:


确保仅绑定一次该事件。
您可以取消绑定特定选择器的所有事件,然后再次重新绑定。
您还可以添加preventDefault方法(http://api.jquery.com/event.preventDefault/


示例:http://jsfiddle.net/FZ7Dk/

$(function() {

    $('#newreportBTN').live('click', function(e) {
        e.preventDefault();

        /*Code here*/

        /*on success*/
        $('#newreportBTN').unbind();
        $('#newreportBTN').bind('click');
    });

});

关于javascript - AJAX:重复的条目,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/7307219/

10-10 23:18