PHP, JSON not displaying correctly on delete

Question

I have the following HTML:

<html>
<head>
    <script src="results.js" type="text/javascript"></script>
    <title>Results</title>
</head>
<body>
    <div id="resultsDiv"></div>
</body>
</html>

The following JavaScript:

window.onload = function() {
    init();
}

var xmlhttp;
var intervalHandle;

function init() {   
    getAjaxData();
    intervalHandle = setInterval(getAjaxData, 2000);
}

function getAjaxData() {
    if (window.XMLHttpRequest) {
        // IE7+. Firefox, Chrome, Opera, Safari
        xmlhttp=new XMLHttpRequest();
    } else {
        // IE6, IE5
        xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
    }
    xmlhttp.onreadystatechange = showJSONData;
    xmlhttp.open("GET", "results2Json.php", true); // GET results from DB
    xmlhttp.send();
}

function showJSONData() {
    if (xmlhttp.readyState==4 && xmlhttp.status==200) {
        var data = JSON.parse(xmlhttp.responseText);

        if (data.results.length > 0) { // Delete this line and it's closing brace, below, and the code does what expected
            var output = '<table border=1>';
            output += '<tr><th>Name</th></tr>';

            for (var i = 0; i < data.results.length; i++) {
                output += '<tr><td>' + data.results[i].name + '</td></tr>';
            }

            output += '</table>';
            document.getElementById("resultsDiv").innerHTML = output;
        } //Closing brace
    }
}

And finally the following PHP:

<?php
$con = mysqli_connect("localhost","root","");
mysqli_select_db($con,"random");
    $result = mysqli_query($con, "select name from random_name");

$rs = array();
while($rs[] = mysqli_fetch_assoc($result)) {
    // do nothing ;-)
}
mysqli_close($con);

unset($rs[count($rs)-1]);  //removes a null value
print("{ \"results\":" . json_encode($rs) . "}");
?>

all just pulling a single column from a DB, and automatically refreshing the results every 2 seconds. If I add to the DB, the results are refreshing fine. If I delete items from the DB all works fine also, until I attempt to delete the last item. This then remains on screen. I'm aware that if I take out the if (data.results.length > 0) line then the final item will delete OK but what I want to achieve is that the table header is not displayed either if the SQL query in the php file returns no results. Any help would be great, thanks.

Side note: that last line of your PHP code is bad practice, you should use echo json_encode(array('results' => $rs)); so that your entire output is controlled by json_encode — scrowler
– scrowler, Commented May 8, 2014 at 22:55
@scrowler Thanks, I'll bare that in mind and change in my actual code. — eldubleu
– eldubleu, Commented May 8, 2014 at 22:57

bittids · Accepted Answer · 2014-05-08 23:14:22Z

1

If there are no results in function showJSONData, your program has no output, so the final result does not update. Change function showJSONData to

function showJSONData() {
    if (xmlhttp.readyState==4 && xmlhttp.status==200) {
        var data = JSON.parse(xmlhttp.responseText);

        if (data.results.length > 0) { // Delete this line and it's closing brace, below, and the code does what expected
            var output = '<table border=1>';
            output += '<tr><th>Name</th></tr>';

            for (var i = 0; i < data.results.length; i++) {
                output += '<tr><td>' + data.results[i].name + '</td></tr>';
            }

            output += '</table>';
            document.getElementById("resultsDiv").innerHTML = output;
        } //Closing brace

        else{
            document.getElementById("resultsDiv").innerHTML = "no results";
       }
    }

answered May 8, 2014 at 23:14

bittids

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1 Comment

eldubleu Over a year ago

Thanks, that's tidier than the answer above

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