inserting values into database more than once with ajax and php

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;
    });
});

At least try that: var datastring = '&text='+ text; or read some basic tutos how to send data to server using $.ajax — A. Wolff
– A. Wolff, Commented Oct 13, 2014 at 16:03
Please, don't use mysql_* functions, They are no longer maintained and are officially deprecated. Learn about prepared statements instead, and use PDO or MySQLi. This article will help you decide. — Jay Blanchard
– Jay Blanchard, Commented Oct 13, 2014 at 16:13

Dharam · Accepted Answer · 2014-10-13 16:07:54Z

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
  }
});

answered Oct 13, 2014 at 16:07

Dharam

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3 Comments

Mani Over a year ago

Dear @Dharam thanks for your kind attention, my query is solved with your guide line.... but i have one more question..if i use var dataString = 'content='+ test; var string = 'text='+ text; then how i will use the code...plz guide me..thx

cyberwombat Over a year ago

Javascript is case sensitive so these are 2 different vars. he's just missing the ampersand if he wanted a query string format

Dharam Over a year ago

You can use the following

Var dataString = 'content='+test;  dataString += '&text='+text; //Note when you use var - it redeclares the variable, and note that in this line we are concatenating the earlier declared variable. Wrote this without testing, but this should work.

Also If all these data are coming from a form you can also use $('#formid').serialize(); and pass this directly

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