JS function passing an empty string back to php, but has a value?

Question

So I am trying to pass back a couple values to a PHP page like this.

function showAccountInfo(obj){

    var value = obj.value;
    var content = obj.querySelector("option:checked").textContent;

    alert("value: " + value + " content: " + content);

    if(obj == ""){
        return;
    }
    else {
        if (window.XMLHttpRequest) {
            // code for IE7+, Firefox, Chrome, Opera, Safari
            xmlhttp = new XMLHttpRequest();
        } else {
            // code for IE6, IE5
            xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
        }
        xmlhttp.onreadystatechange = function() {
            if (xmlhttp.readyState == 4 && xmlhttp.status == 200) {
                document.getElementById("facilities").innerHTML = xmlhttp.responseText;
            }
        }
        xmlhttp.open("GET","getinfo.php?=q"+value+"&c="+content, true);
        xmlhttp.send(); 
    }
}

When the alert goes off it shows me the correct values for each variable. However when I pass off that value to getinfo.php and use the variable value to complete a query and echo it out to the page it shows me it as empty.

$q = ($_GET['q']);

$sql = "SELECT *, account.account_name FROM facility "
 . "INNER JOIN account ON account.account_id = facility.account_id "
 . "WHERE facility.account_id = '".$q."'";

echo $sql;

Result: SELECT *, account.account_name FROM facility INNER JOIN account ON account.account_id = facility.account_id WHERE facility.account_id = ''

I had everything functioning properly prior my previous question.

Have you checked the generated URL is what you expect in the Net tab of your developer tools? Is the data in the response correct? — Quentin
– Quentin, Commented Mar 5, 2015 at 17:29
@Quentin just checked, the URL is giving me the correct data. — BRBT
– BRBT, Commented Mar 5, 2015 at 17:32

charlietfl · Accepted Answer · 2015-03-05 17:35:55Z

1

Your url is incorrectly formatted

Change

xmlhttp.open("GET","getinfo.php?=q"+value+"&c="+content, true);
                                ^

To

xmlhttp.open("GET","getinfo.php?q="+value+"&c="+content, true);
                                 ^

Note change of first =

answered Mar 5, 2015 at 17:35

charlietfl

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1 Comment

BRBT Over a year ago

Yep, that was it..should have noticed that, thank you!

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