提交從jQuery來PHP

Question

我已經發送數據到我的jQuery的文件，我測試的數據串，一切都存儲在哪裏它應該是，但是當我發佈到我的PHP文件中的所有變量都爲空。有人能幫我嗎？

$(document).ready(function() { 
    $(".button").click(function() { 
    //$('.error').hide(); 
    var firstname = $("input#First_Name").val(); 
    var lastname = $("input#Last").val(); 
    var areacode = $("input#area_code").val(); 
    var phonenumber = $("input#Phone_Number").val(); 
    var emailaddress = $("input#emailaddress").val(); 
    var confirmemail = $("input#confirm_email").val(); 
    var password = $("input#Password_Input").val(); 
    var confirmpassword = $("input#ConfirmPassword").val(); 
    var streetaddress = $("input#StreetAddress_input").val(); 
    var streetaddress2 = $("input#StreetAddress2_input").val(); 
    var city = $("input#City_Input").val(); 
    var state = $("input#StateInput").val(); 
    var zipcode = $("input#ZipCode").val(); 
    var month = $("input#month_input").val(); 
    var day = $("input#day_input").val(); 
    var year = $("input#year_input").val(); 
    var services = $("input#services_input").val(); 
    var agreement = $("input#agreement").val(); 

    var dataString = 'firstname=' + firstname + '&lastname=' + lastname + '&areacode=' + areacode + '&phonenumber=' + phonenumber + '&emailaddress=' + emailaddress + '&confirmemail=' + confirmemail + '&password=' + password + '&streetaddress=' + streetaddress + '&streetaddress2=' + streetaddress2 + '&city=' + city + '&state=' + state + '&zipcode=' + zipcode + '&month=' + month + '&day=' + day + '&year=' + year + '&services=' + services + '&agreement=' + agreement; 
    alert(dataString); 
    $.ajax({ 
     type: "POST", 
     url: "http://www.vectorcreditsolution.com/js/process.php", 
     data: dataString, 
      success: function() { 
      alert("Yay it was sent"); 
    } 
     }); 
     return false; 
    }); 

，然後我的PHP文件

<?php 
    $FirstName = $_POST["firstname"]; 
    $LastName = $_POST['lastname']; 
    $AreaCode = $_POST['areacode']; 
    $PhoneNumber = $_POST['phonenumber']; 
    $EmailAddress = $_POST['emailaddress']; 
    $Password = $_POST['password']; 
    $StreetAddress = $_POST['streetaddress']; 
    $StreetAddress2 = $_POST['streetaddress2']; 
    $City= $_POST['city']; 
    $State = $_POST['state']; 
    $ZipCode = $_POST['zipcode']; 
    $Month = $_POST['month']; 
    $Day = $_POST['day']; 
    $Year= $_POST['year']; 
    $Service=$_POST['services']; 

    var_dump($_POST["firstname"]); 
    var_dump($_POST['firstname']); 
    var_dump($_POST[firstname]); 

Answer 1

你是不是輸出從AJAX返回的數據，並沒有使用工具來查看。也許你期待var轉儲出現在頁面上。他們不會，他們將返回到成功的功能，你可以使用這些數據，你覺得合適。

jQuery的AJAX方法管從你的PHP文件的輸出到成功函數內部的「RET」變量。例如，如果您希望返回HTML成功消息，則可以將該HTML添加到頁面而不刷新它。或者任何一件事。

您應該閱讀AJAX。

更改此：

$.ajax({ 
type: "POST", 
    url: "http://www.vectorcreditsolution.com/js/process.php", 
    data: dataString, 
    success: function() { 
     alert("Yay it was sent"); 
    } 
}); 

這樣：

$.ajax({ 
type: "POST", 
    url: "http://www.vectorcreditsolution.com/js/process.php", 
    data: dataString, 
    success: function(ret) { 
     alert(ret); 
    } 
}); 

，然後單擊與jQuery的頁面上提交按鈕。

不要將此與其他答案結合在此頁面上。一個將會改變你的數組鍵（serialize）。

Answer 2

嘗試張貼的數據是JavaScript字典，而不是作爲一個大的字符串：

$.ajax({ 
    // ... 
    data: { 
     firstname: firstname, 
     lastname: lastname, 
     // etc. 
    } 
}); 

Answer 3

我會使用序列化（）jQuery的功能。而不是尋找每個領域的價值序列化整個窗體

$("#form-id").serialize(); 


$.ajax({ 
    type: "POST", 
    url: "http://www.vectorcreditsolution.com/js/process.php", 
    data: $("#form-id").serialize(), 
     success: function() { 
     alert("Yay it was sent"); 
} 
    }); 
    return false; 
});