使用ajax和php多次将值插入数据库

Question

inserting values into database more then one with ajax and php, using this code inserting only one value i dont know why please help me out.

PHP coding

$content=$_POST['content'];
$text=$_POST['text'];
mysql_query("insert into ajax(ajax,text) values ('$content','$text')");

ajax coding

$(function() {
    $(".comment_button").click(function() {

        var test = $("#content").val();
        var text = $("#text").val();

        var dataString = 'content='+ test;
        var datastring = 'text='+ text;

        if(test=='') {
            alert("Please Enter Some Text");        
        } else {
            $("#flash").show();
            $("#flash").html('<img src="loading.gif" align="absmiddle">&nbsp;<span class="loading">Wait..!</span>');            
            $.ajax({
                type: "POST",
                url: "demo_insert.php",
                data: dataString + datastring,
                cache: false,
                success: function(html) {
                    //document.getElementById('content').value='';
                    $("#flash").hide();
                    alert("Values Inserted Successfully");
                    //document.write("values inserted Successfully..."); // to show message as string           
                }
            });
        }

        return false;
    });
});

Answer 1

var dataString = 'content='+ test;
var datastring = 'text='+ text;

In the above code you are re-declaring the variable which overrides the first variable's data

The best way for you to do this is

$.ajax({
  type: "POST",
  url: "demo_insert.php",
  data: {content:test, text: text},
  cache: false,
  success: function(html){
 //document.getElementById('content').value='';
 $("#flash").hide();
 alert("Values Inserted Successfully");
 //document.write("values inserted Successfully..."); // to show message as string
  }
});