how to pass a value to php variable by ajax

Question

this is my javascript code :

function  category(row){
    dataparam = "oper=delete&row="+row;
    $.ajax({
        type: "POST",
        url: "multiupload.php",
        data: dataparam,
        error:function() {
                alert("sorry")
        },
        success: function(html) {
            alert(html);
        }
    });
}

and my php script :

$opers = (isset($_REQUEST['opers']) and $_REQUEST['opers'] != '' ) ? $_REQUEST['opers'] : '';
if($opers == "delete") {
        $row=$_REQUEST['row'];
        echo $row;
}

This is not working... I don't know what problem is. Please help me to get the $row variable in my php script.

Possible duplicate of http://stackoverflow.com/questions/6130662/passing-javascript-variable-to-php-using-ajax/? — Liam, May 26 '11 at 13:58

score 1 · Answer 1 · answered May 26 '11 at 13:57

1

In your dataparam variable you have "oper=delete&row="+row; and in the PHP code you test for $_REQUEST['opers']), since oper <> opers, the failure is perfectly normal, just add or remove the s somewhere.

answered May 26 '11 at 13:57

krtek

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martynthewolf · Answer 2 · 2011-05-26T14:12:46.597

It looks like you have this wrong, $_REQUEST['opers'] it should be $_REQUEST['oper']

$opers = (isset($_REQUEST['oper']) and $_REQUEST['oper'] != '' ) ? $_REQUEST['oper'] : '';  

if($opers == "delete") {
    $row=$_REQUEST['row'];
    echo $row;
}

I would also recommend that as you are expecting the values to come via URL you use the appropriate super global which is $_GET. There is a very small chance that a $_COOKIE could screw you over. If you use them and hapen to give it the value of 'oper'.

score 0 · Answer 3 · edited May 26 '11 at 13:58

0

You want $_REQUEST['opers'] while you pass oper. Notice the additional "s".

edited May 26 '11 at 13:58

krtek

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5
56
84

answered May 26 '11 at 13:57

RRStoyanov

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23

score 0 · Answer 4 · answered May 26 '11 at 14:00

$opers = (isset($_REQUEST['oper']) and $_REQUEST['oper'] != '' ) ? $_REQUEST['oper'] : '';  
if($opers == "delete")
{
     $row=$_REQUEST['row'];
     echo $row;
}

Hopefully the problem is with the extra curly brackets. Check it. I have corrected the code. Let me know if it works. And there is an extra s in opers.

how to pass a value to php variable by ajax

4 Answers4

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