Einfügen von Datensätzen in mysql Datenbank mit php unter Verwendung von Ajax

, Wie zu tun Kodierung dieser Code Mit der Verwendung von Ajax.Bitte Helfen Sie.
Ich bin Bignner hier und ich habe geschrieben, dieser code funktioniert es, aber ich will mit ajax, weil diese nicht wollen, um die Seite neu laden...?

PHP-Datei

    //Code For Making Form And getting Data…..
    <html>
    <body>
    Fill -ID,NAME,EMAIL_ID,PASSWORD,CREDITS,
    <form action="Form_Data.php" method="post"> 

ID: <input type="text" name="ID"><br><br>
NAME: <input type="text" name="NAME"><br><br>
PASSWORD: <input type="text" name="PASSWORD"><br><br>
CREDITS: <input type="text" name="CREDITS"><br><br>
E_mail: <input type="text" name="EMAIL_ID"><br><br>
CREATED_ON:<input type="text" name="CREATED_ON"><br><br>
MODIFIED_ON:<input type="text" name="MODIFIED_ON"><br><br>
<input type="submit">
</form>

</body>
</html>

//code für Daten aus Formular-Daten.

<html>
<?php
include 'connnect.php';
   mysql_set_charset('utf8');
   //query for insert data into tables
   $ID = $_POST['ID'];
   $NAME =$_POST['NAME'];
   $EMAIL_ID =$_POST['EMAIL_ID'];
   $PASSWORD =$_POST['PASSWORD'];
   $CREDITS =$_POST['CREDITS'];
   $CREATED_ON=$_POST['CREATED_ON'];
   $MODIFIED_ON=$_POST['MODIFIED_ON'];

   $query = "INSERT INTO `user_table` 
         (`ID`,`NAME`,`EMAIL_ID`,`PASSWORD`,`CREDITS`,`CREATED_ON`,`MODIFIED_ON`)
         VALUES
         ('$ID','$NAME','$EMAIL_ID','$PASSWORD','$CREDITS','$CREATED_ON','$MODIFIED_ON')";
         $query_run= mysql_query($query);
         $retval=mysql_query($query,$conn);
           if ($query_run)
   { echo 'It is working';
}
   mysql_close($conn);
?>
</html>

Habe ich Versucht, Aber ... Ist Blewo...

file for html and ajax
    <html>
     <HEAD>    
       <script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.7.1/jquery.min.js"></script>
       </HEAD>
    <body>
    <div id="status_text">
    Fill -ID,NAME,EMAIL_ID,PASSWORD,CREDITS,
    <form onsubmit="return false" method="post">
    ID: <input type="text" id="ID" name="ID"><br><br>
    NAME: <input type="text" id="NMAE" name="NAME"><br><br>
    PASSWORD: <input type="text"  id= "PASSWORD"name="PASSWORD"><br><br>
    CREDITS: <input type="text" Id= "CREDITS"name="CREDITS"><br><br>
    Email_ID: <input type="text" id="Email_ID"name="EMAIL_ID"><br><br>
    CREATED_ON:<input type="text" id="CREATED_ON" name="CREATED_ON"><br><br>
    MODIFIED_ON:<input type="text" id="MODIFIED_ON" name="MODIFIED_ON"><br><br>
    <input type="submit"  id="btn_submit" name="submit" value="Send">
    </div>

<script>

//on the click of the submit button 
$("#btn_submit").click(function(){
    //get the form values
 var ID = $('#ID').val();     
 var NAME = $('#NAME').val();     
 var PASSWORD = $('#PASSWORD').val();
 var CREDITS = $('#CREDITS').val();
 var EMAIL_ID = $('#EMAIL_ID').val();
 var CREATED_ON = $('#CREATED_ON').val();
 var MODIFIED_ON = $('#MODIFIED_ON').val();
 //make the postdata
 var postData = '&ID='+ID+'&NAME='+NAME+'&PASSWORD='+PASSWORD+'&REDITS'+CREDITS+'&EMAIL_ID'+EMAIL_ID+'&CREATED_ON'+CREATED_ON+'&MODIFIED_ON'+MODIFIED_ON;
 //call your .php script in the background, 
 //when it returns it will call the success function if the request was successful or 
 //the error one if there was an issue (like a 404, 500 or any other error status)
});
 $.ajax({
    url : "Form_Data.php",
    type: "POST",
    data : postData,
    success: function(data,status,  xhr)
     {
        //if success then just output the text to the status div then clear the form inputs to prepare for new data
        $("#status_text").html(data);
        $('#ID').val();
        $('#NAME').val('');
        $('#PASSWORD').val('');
        $('#EMAIL_ID').val('');
        $('#CREATED_ON').val('');
        $('#MODIFIED_ON').val('');
         }

});  
</script>


</form>

</body>
</div>
</html>

code für die Abfrage...

<html>
<?php
include 'connnect.php';
   mysql_set_charset('utf8');
   //query for insert data into tables
   $ID = $_POST['ID'];
   $NAME =$_POST['NAME'];
   $EMAIL_ID =$_POST['EMAIL_ID'];
   $PASSWORD =$_POST['PASSWORD'];
   $CREDITS =$_POST['CREDITS'];
   $CREATED_ON=$_POST['CREATED_ON'];
   $MODIFIED_ON=$_POST['MODIFIED_ON'];


$query = "INSERT INTO `user_table` 
         (`ID`,`NAME`,`EMAIL_ID`,`PASSWORD`,`CREDITS`,`CREATED_ON`,`MODIFIED_ON`)
         VALUES
         ('$ID','$NAME','$EMAIL_ID','$PASSWORD','$CREDITS','$CREATED_ON','$MODIFIED_ON')";
         $query_run= mysql_query($query);
         $retval=mysql_query($query,$conn);
           if ($query_run)
   { echo 'It is working';
}
   mysql_close($conn);
?>
</html>

Was ist, wenn mein Passwort ist pas'word123? Werfen Sie einen Blick auf PDO oder wie zu vermeiden, mysql-injection stackoverflow.com/questions/60174/...
Könnten Sie uns zeigen, was Sie bisher ausprobiert haben? Wir sind hier um zu helfen und nicht schreiben Sie Ihren code für Sie
Was planen Sie für AJAX? jQuery? Haben Sie versucht, alles so weit?
doch ich habe getan, einige Codierung zeigt, bearbeitet...
nicht immer das, was Fehler ...?

InformationsquelleAutor Ashish Kumar | 2015-03-18

Ich habe Es Gelöst ... Wie man mithilfe von Ajax und MYSQL...
PHP-CODE

<?php
include 'connnect.php';
   mysql_set_charset('utf8');
   //query for insert data into tables

    $ID = $_POST['ID'];
   $NAME =$_POST['NAME'];
   $EMAIL_ID =$_POST['EMAIL_ID'];
   $PASSWORD =$_POST['PASSWORD'];
   $CREDITS =$_POST['CREDITS'];
   $CREATED_ON=$_POST['CREATED_ON'];
   $MODIFIED_ON=$_POST['MODIFIED_ON'];


$query = "INSERT INTO `user_table` 
         (`NAME`,`EMAIL_ID`,`PASSWORD`,`CREDITS`,`CREATED_ON`,`MODIFIED_ON`)
         VALUES
         ('$NAME','$EMAIL_ID','$PASSWORD','$CREDITS','$CREATED_ON','$MODIFIED_ON')";
         $query_run= mysql_query($query);
        //$retval=mysql_query($query,$conn);
          if ($query_run)
          { 
                echo 'It is working';
          }

   mysql_close($conn);
?>

HTML-DATEI....

<html>
 <HEAD>    
   <script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.7.1/jquery.min.js"></script>
   </HEAD>
<body>
<div id="status_text">
Fill -ID,NAME,EMAIL_ID,PASSWORD,CREDITS,

ID: <input type="text" id="ID" name="ID"><br><br>
NAME: <input type="text" id="NAME" name="NAME"><br><br>
PASSWORD: <input type="text"  id= "PASSWORD"name="PASSWORD"><br><br>
CREDITS: <input type="text" Id= "CREDITS"name="CREDITS"><br><br>
Email_ID: <input type="text" id="EMAIL_ID"name="EMAIL_ID"><br><br>
CREATED_ON:<input type="text" id="CREATED_ON" name="CREATED_ON"><br><br>
MODIFIED_ON:<input type="text" id="MODIFIED_ON" name="MODIFIED_ON"><br><br>
<input type="submit"  id="btn_submit" name="submit" value="Send"/>
</div>

<script>
<!-- 
//Browser Support Code
 function ajaxFunction(){
 var ajaxRequest;  //The variable that makes Ajax possible!

 try{
   //Opera 8.0+, Firefox, Safari
   ajaxRequest = new XMLHttpRequest();
 }catch (e){
   //Internet Explorer Browsers
   try{
      ajaxRequest = new ActiveXObject("Msxml2.XMLHTTP");
   }catch (e) {
      try{
         ajaxRequest = new ActiveXObject("Microsoft.XMLHTTP");
      }catch (e){
         //Something went wrong
         alert("Your browser broke!");
         return false;
      }
   }
 } }
//on the click of the submit button 
$("#btn_submit").click(function(){
    //get the form values
 var ID = $('#ID').val();     
 var NAME = $('#NAME').val();     
 var PASSWORD = $('#PASSWORD').val();
 var CREDITS = $('#CREDITS').val();
 var EMAIL_ID = $('#EMAIL_ID').val();
 var CREATED_ON = $('#CREATED_ON').val();
 var MODIFIED_ON = $('#MODIFIED_ON').val();
 //make the postdata
 //var postData = '&ID='+ID+'&NAME='+NAME+'&PASSWORD='+PASSWORD+'&CREDITS'+CREDITS+'&EMAIL_ID'+EMAIL_ID+'&CREATED_ON'+CREATED_ON+'&MODIFIED_ON'+MODIFIED_ON;
 //alert(postData);
 var myData={"ID":ID,"NAME":NAME,"PASSWORD":PASSWORD,"CREDITS":CREDITS,"EMAIL_ID":EMAIL_ID,"CREATED_ON":CREATED_ON,"MODIFIED_ON":MODIFIED_ON};
 //call your .php script in the background, 
 //when it returns it will call the success function if the request was successful or 
 //the error one if there was an issue (like a 404, 500 or any other error status)
 $.ajax({
    url : "Form_Data.php",
    type: "POST",
    data : myData,
    success: function(data,status,xhr)
     {
        //if success then just output the text to the status div then clear the form inputs to prepare for new data
        $("#status_text").html(data);
        $('#ID').val();
        $('#NAME').val('');
        $('#PASSWORD').val('');
        $('#EMAIL_ID').val('');
        $('#CREATED_ON').val('');
        $('#MODIFIED_ON').val('');
         }

}); 
}); 
</script>



</body>
</div>
</html>

InformationsquelleAutor Ashish Kumar

ändern Sie Ihr Skript, weil Ihre ajax war außerhalb der Klick-Funktion

//on the click of the submit button 
$("#btn_submit").click(function(){

    //get the form values
 var ID = $('#ID').val();     
 var NAME = $('#NAME').val();     
 var PASSWORD = $('#PASSWORD').val();
 var CREDITS = $('#CREDITS').val();
 var EMAIL_ID = $('#EMAIL_ID').val();
 var CREATED_ON = $('#CREATED_ON').val();
 var MODIFIED_ON = $('#MODIFIED_ON').val();
 //make the postdata
 var postData = '&ID='+ID+'&NAME='+NAME+'&PASSWORD='+PASSWORD+'&REDITS'+CREDITS+'&EMAIL_ID'+EMAIL_ID+'&CREATED_ON'+CREATED_ON+'&MODIFIED_ON'+MODIFIED_ON;
 //call your .php script in the background, 
 //when it returns it will call the success function if the request was successful or 
 //the error one if there was an issue (like a 404, 500 or any other error status)

  $.ajax({
    url : "Form_Data.php",
    type: "POST",
    data : postData,
    success: function(data,status,  xhr)
     {
        //if success then just output the text to the status div then clear the form inputs to prepare for new data
        $("#status_text").html(data);
        $('#ID').val();
        $('#NAME').val('');
        $('#PASSWORD').val('');
        $('#EMAIL_ID').val('');
        $('#CREATED_ON').val('');
        $('#MODIFIED_ON').val('');
         }

}); 

});

</script>

und ändern Sie Ihre php-code zu diesem

<?php
include 'connnect.php';
   mysql_set_charset('utf8');
   //query for insert data into tables
if(isset($_POST['NAME'])){
   $ID = $_POST['ID'];
   $NAME =$_POST['NAME'];
   $EMAIL_ID =$_POST['EMAIL_ID'];
   $PASSWORD =$_POST['PASSWORD'];
   $CREDITS =$_POST['CREDITS'];
   $CREATED_ON=$_POST['CREATED_ON'];
   $MODIFIED_ON=$_POST['MODIFIED_ON'];


$query = "INSERT INTO `user_table` 
         (`ID`,`NAME`,`EMAIL_ID`,`PASSWORD`,`CREDITS`,`CREATED_ON`,`MODIFIED_ON`)
         VALUES
         ('$ID','$NAME','$EMAIL_ID','$PASSWORD','$CREDITS','$CREATED_ON','$MODIFIED_ON')";
         $query_run= mysql_query($query);
         $retval=mysql_query($query,$conn);
           if ($query_run)
   { echo 'It is working';
}
}
     mysql_close($conn);
?>

InformationsquelleAutor Tariq hussain

In diesem code bin ich gerade Absenden Ihrer zwei input-Felder, die Ruhe, die Sie hinzufügen können, indem Sie sich. Versuchen Sie dies:

 <html>
    <body>
    Fill -ID,NAME,EMAIL_ID,PASSWORD,CREDITS,
    <form action="Form_Data.php" method="post"> 


NAME: <input id="name" type="text" name="NAME"><br><br>
PASSWORD: <input id="password" type="text" name="PASSWORD"><br><br>


<input type="submit" id="submit">
</form>

</body>
</html>

$("#submit").click(function() {
            var name= $("#name").val();
            var password= $("#password").val();

            $.ajax({
                type: "POST",
                url: "your_php_path.php",
                data: 'name=' + name+ '&password=' + password,
                success: function(result) {
                   alert(result);
                }
            });


        });

Es ist Nicht Arbeiten Chef...
stellen Sie sicher, dass Sie geben-index-name korrekt in der php-Datei $name =$_POST['name']; $PASSWORT =$_POST['Passwort']
sollte ich XMLHttpRequest-Objekt hier..
mailen Sie mir Ihre html-Seite und php-Seite .. werde ich beheben [email protected]
check meine neue Antwort ..seine arbeiten an meinem Ende .. können Sie überprüfen, auch durch das hinzufügen einer Zeile direkt nach if(isset(---)){ print_r($_POST); und alert-Ergebnis im Skript

InformationsquelleAutor Tariq hussain

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