Member Avatar for mani2604

Hi everyone...


Well i do have a mysql query in one php page(php_1) & I want to submit the variables to the query in different php page(php_2) via form action but how am I supposed to do it without redirecting to php_1..

All I need is to post the data to the first php page query so that It performs some action over there & thats all... So hoping sumone who knows ajax might helpme out with a sample code or point me in the right direction...

Thanks,

php_1 page :-

$query = ("SELECT * FROM state4 WHERE (LONG_HI<'".$_POST['Ymax']."')");

php_2 page:-

Ymax<input type="text" name="Ymax" size="15">
	      <input type=SUBMIT name="submit" VALUE="Submit"></td>
  • 7 Contributors
  • 6 Replies
  • 3K Views
  • 2 Years Discussion Span
  • Latest Post Latest Post by Bachu
Member Avatar for Thirusha

I dont know much about php, but first change the submit button to a normal button

<input type="button" name="submit" VALUE="Submit" onclick="sendInfo()">

and use the onclick property to call your ajax function, there are many examples of a simple ajax function on the internet.

Member Avatar for ivan3510

Hi!
This is the codes:
(page php_2):

<html>
<script type="text/javascript">
function sendData(){
if(window.XMLHttpRequest){
xmlhttp=new XMLHttpRequest();
}else{
xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange=function(){
if(xmlhttp.readyState==4 && xmlhttp.status==200){
var Ymax=document.getElementById('Ymax').value;
result=xmlhttp.responseText;
//in the html element <div id="forResult"></div>
//will be writen what you write with "echo" in your php file
//in this example, "Thank you for submiting Ymax"
document.getElementById('forResult').innerHTML=result;
}
}
xmlhttp.open("GET","php_1.php?Ymax="+Ymax,true);
xmlhttp.send();
}
</script>
<body>
Ymax: <input type="text" id="Ymax" size="15"><br>
<input type="button" value="Submit" onclick="sendData()">
<div id="forResult"></div>
</body>
</html>

php_1:

<?php
$Ymax=$_GET['Ymax'];
//......
//query.....
echo "Thank you for submiting Ymax";
?>
Edited by ivan3510 because: n/a
Member Avatar for ragziraghu

The var Ymax in line 11 is out of scope for xmlhttp.open in line 19.
Hence it uses in "Ymax" input box in line 24 and prints [object HTMLInputElement]

Member Avatar for AleMonteiro

I would use JQuery:

    <script>
        $(function() // on DOM loaded
            $("#btnSubmit").click(function(){ // Click handler 
                var yMax = $("#txtYmax").val(); // Get YMax value
                $.ajax(
                    'MyPhpPage.php', // Page to handle the request
                    { yMax : yMax }, // Parameters to send
                    function(result) { // Function executed when the request is completed
                        $("#forResult").html(result);
                    },
                    function(erro) { // Function executed when the request throws an error
                        //do something
                    }
                );
            });
        });
    </script>
    <body>
        Ymax: <input type="text" id="txtYmax" size="15">
              <input type="button" value="Submit" id="btnSubmit">
        <div id="forResult"></div>
</body>
Member Avatar for fobos

AleMonteiro, this post is dead, please dont bring it back from the grave.

Member Avatar for Bachu 
Ymax<input type="text" name="Ymax" id="Ymax" size="15">
          <input type="button" name="submit" class="submitBtn" VALUE="Submit">
<script  type="text/javascript">
$(function(){
    $('.submitBtn').click(function(){

        $.post('otherPhpPage.php',{Ymax: $('#Ymax').val()},function(data){
            alert(data);
        });
    });
});
</script>

<!--##########  In otherPhpPage.php ########  -->

<?php
echo $yMax=$_REQUEST['Ymax']; 
?>

Result of this code will show an alert with your entered Ymas value

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