发送数据使用Ajax和PHP，但没有MySQL来发生

我想发出一个简单的数据，如姓名，代码，日期...等我的数据库，我得到了使用JavaScript中的数据，并把它交给我的PHP文件使用Ajax，但没有任何反应，我不知道发生了什么，但Ajax什么都不做！因此，请任何人查看代码，并告诉发生了什么事。谢谢发送数据使用Ajax和PHP，但没有MySQL来发生

P.S.我的脚本在我的HTML文件的结尾处，就在结束标记之前。

<script type="text/javascript"> 
 
debugger; 
 
window.onload = function() { 
 
    var platform = $("input:text[id=platform]").val(); 
 
    var source = $("input:text[id=source]").val(); 
 
    var code = $("input:text[id=code]").val(); 
 
    var thedate = $("input:text[id=date]").val(); 
 
    var payment_type = $("input:text[id=payment_type]").val(); 
 
    var change_for = $("input:text[id=change_for]").val(); 
 
    var order_comment = $("#order_comment").val(); 
 
    var decline_reason = $("input:text[id=decline_reason]").val(); 
 
    var order_reference_by_vendor = $("input:text[id=order_reference_by_vendor]").val(); 
 
    var deliveryprovider_title = $("input:text[id=deliveryprovider_title]").val(); 
 
    var social_reason = $("input:text[id=social_reason]").val(); 
 
    var government_identification = $("input:text[id=government_identification]").val(); 
 
    
 
     if(platform == ""){platform = "NULL"} 
 
     if(source == ""){source = "NULL"} 
 
     if(code == ""){code = "NULL"} 
 
     if(thedate == ""){thedate = "NULL"} 
 
     if(payment_type == ""){payment_type = "NULL"} 
 
     if(change_for == ""){change_for = "NULL"} 
 
     if(order_comment == ""){order_comment = "NULL"} 
 
     if(decline_reason == ""){decline_reason = "NULL"} 
 
     if(order_reference_by_vendor == ""){order_reference_by_vendor = "NULL"} 
 
     if(deliveryprovider_title == ""){deliveryprovider_title = "NULL"} 
 
     if(social_reason == ""){social_reason = "NULL"} 
 
     if(government_identification == ""){government_identification = "NULL"} 
 

 
    alert ("before ajax"); 
 

 
     $.ajax({ 
 
     type:"post", 
 
     url:"savedata.php", 
 
     data:{"platform": platform, 
 
      "source": source, 
 
      "code": code, 
 
      "thedate": thedate, 
 
      "payment_type": payment_type, 
 
      "change_for": change_for, 
 
      "order_comment": order_comment, 
 
      "decline_reason": decline_reason, 
 
      "order_reference_by_vendor": order_reference_by_vendor, 
 
      "deliveryprovider_title": deliveryprovider_title, 
 
      "social_reason": social_reason, 
 
      "government_identification": government_identification}, 
 
     success: function(msg){ 
 
     alert("Success Insert data"); 
 
     } 
 
     }); 
 

 
    alert ("after ajax"); 
 

 
     } 
 
</script>

和这里的PHP代码

<?php 
 

 
function db_connect() 
 
{ 
 
\t $servername = "localhost"; 
 
\t $username = "root"; 
 
\t $password = ""; 
 
\t $dbname = "integrationdb"; 
 

 
    @mysql_connect($servername,$username,$password) or die ("error in host connection"); 
 
    @mysql_select_db($dbname) or die("error in db connection"); 
 
} 
 

 
db_connect(); 
 

 
$platform = $_POST['platform']; 
 
$source = $_POST['source']; 
 
$code = $_POST['code']; 
 
$thedate = $_POST['thedate']; 
 
$payment_type = $_POST['payment_type']; 
 
$change_for = $_POST['change_for']; 
 
$order_comment = $_POST['order_comment']; 
 
$decline_reason = $_POST['decline_reason']; 
 
$order_reference_by_vendor = $_POST['order_reference_by_vendor']; 
 
$deliveryprovider_title = $_POST['deliveryprovider_title']; 
 
$social_reason = $_POST['social_reason']; 
 
$government_identification = $_POST['government_identification']; 
 

 
$qry = "INSERT INTO orderinformation VALUES ('$platform','$source','$code','$thedate','$payment_type','$change_for','$order_comment','$decline_reason','$order_reference_by_vendor','$deliveryprovider_title','$social_reason','$government_identification')"; 
 

 
$result=mysql_query($qry); 
 
if(isset($result)) { 
 
    echo "YES"; 
 
} else { 
 
    echo "NO"; 
 
} 
 
?>

编辑原谅我，如果我要问... ...它在JSON做区别，如果我使用（'），而不是（“）??！

来源

2017-07-22 Hossam

你看到这个警示（“成功插入数据”响应数据）;正在显示？ –

你是否在控制台中发现任何错误？ –

请参阅https://stackoverflow.com/questions/12859942/why-shouldnt-i-use-mysql-functions-in-php并停止使用这些MySQL的funtions插件PHP。 – BenRoob

尝试通过chrome toolbar查看网络，它可以帮助您跟踪的要求，读它的头，还要检查其可能包含错误信息

来源

2017-07-22 20:12:46

我得到了问题，我没有调用jQuery的正确CDN，现在它运行成功..使用'contentType：“application/json; charset = utf-8“'，** savadata。PHP，\t 200，\t XHR，\t jquery.min.js：4，15.5 KB，\t 19.76小号**，但没有数据在MySQL数据库中添加!!!!!!!!!!!!! – Hossam

发送数据使用Ajax和PHP，但没有MySQL来发生

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