Senden von Daten an MySQL mit Ajax und PHP, aber nichts passiert

Question

Ich möchte eine einfache Daten, wie Name, Code, Datum usw. in meine Datenbank senden, ich habe die Daten mit JavaScript und schickte es an meine PHP-Datei mit Ajax, aber nichts passiert und ich weiß nicht, was passiert, aber Ajax nichts tun !! Also, bitte kann jemand den Code überprüfen und erzählen, was los ist. danke

P.S. mein Skript im ENDE meiner HTML-Datei, direkt vor dem End-Tag.

<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>

und hier ist der PHP-Code

<?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"; 
 
} 
 
?>

• BEARBEITEN Entschuldigung, wenn ich frage ... wird es einen Unterschied machen, wenn ich (') statt (") in JSON verwende ??!

Answer 1

Versuchen Sie das Netzwerk über chrome toolbar es sehen können Sie die Anforderung helfen verfolgen, dessen Kopf lesen, auch die Antwortdaten überprüfen, die die Fehlermeldung enthält möglicherweise