CODEWITHSUNDEEP
javascriptphpmysqlajaxinternal-server-error
Skin99
Skin99

Reputation: 91

XMLHTTP: Internal server error - issue with GET?

I'm trying to retrieve some data from a MySQL database that I have stored on a server. I'm using PHP, Javascript and AJAX to get the data.

When I run the HTML file (New.html) in Chrome and look at the code using developer tools, it says;

GET http://example.net/Example/getuser.php?q=2 500 (Internal Server Error)

showUser @ New.html:31onchange @ New.html:52

I think this refers to xmlhttp.onreadystatechange, and there is a red X beside the .send() line.

<html>
<head>
    <title>New</title>
    <meta charset="UTF-8">
                              //Javascript Code
<script> 
        function showUser(str) {
            if (str == " ") {
                document.getElementById("txtHINT").innerHTML = " ";
                return;
            } else {
                if (window.XMLHttpRequest) {
                    // code for IE7+, Firfox, Chrome, Opera, Safari
                    xmlhttp = new XMLHttpRequest();  
            } else {
                   // code for IE6, IE5
            xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
            }
            xmlhttp.onreadystatechange = function() {
                if (this.readyState == 4 && this.status == 200) {
                    document.getElementById("txtHint").innerHTML = this.reponseText;
                }
            };
            xmlhttp.open("GET","getuser.php?q="+str,true);
            xmlhttp.send(); //LINE 31
        }
    }
</script>
                      //CSS for HTML table
<style> 
    table         {
        width: 100%;
        border-collapse: collapse;
    }
    table, td, th {
        border: 1px solid black;
        padding: 5px;
    }   
    th            {
       text-align: left;
    }
</style>
</head>
                   <!-- Code for Form -->
<body>
    <form>
        <select name ="users" onchange="showUser(this.value)"> //LINE 51
            <option value=" ">Select a person:</option>
            <option value="1">Peter Griffin</option>
            <option value="2">Lois Griffin</option>
            <option value="3">Joseph Swanson</option>
            <option value="4">Glenn Quagmire</option>
        </select>
    </form>
    <br>
    <div id="txtHint"><b>MySQL Data should go here</b></div>
</body>
</html>

Does anyone know how to solve these issues? Perhaps the .open() needs to be placed earlier in the code? Or maybe a handler of some kind is needed?

PHP File:

<html>
<head>
    <title>Latest Attempt</title>
</head>
<?php
    $q = intval($_get['q']);
    // put your connection code here
    $servername = 'localhost';
    $username = 'user';
    $password = '12345678';
    $dbname = 'ajax_demo';
    
    // create connection
    $conn = new mysqli($servername, $username, $password, $dbname);
    
    // check connection
    if ($conn->connect_error) {
        die("Connection failed: " . $conn->connect_error);
    }
    echo "Connected successfully";
    
    mysqli_select($conn,"ajax_demo");
    $sql="SELECT * FROM my_DB WHERE id = '".$q."'";
    $result = mysqli_query($conn,$sql);
    
    echo "<table>
    <tr>
    <th>FirstName</th>
    <th>LastName</th>
    <th>Age</th>
    <th>Hometown</th>
    <th>Job</th>
    </tr>";
    while ($row = mysqli_fetch_array($result)) {
        echo "<tr>";
        echo "<td>" . $row['FirstName'] . "</td>";
        echo "<td>" . $row['LastName'] . "</td>";
        echo "<td>" . $row['Age'] . "</td>";
        echo "<td>" . $row['Hometown'] . "</td>";
        echo "<td>" . $row['Job'] . "</td>";
        echo "</tr>";
    }
    echo "</table>";
    mysqli_close($conn);
?>
</html>

Upvotes: 0

Views: 576

Answers (1)

EhsanT
EhsanT

Reputation: 2085

You have a few misspells in your code which if you fix them your code will work perfectly:

In you JavaScript

At this line

document.getElementById("txtHint").innerHTML = this.reponseText;

You have misspelled reponseTextand it should be responseText so This line will be like:

document.getElementById("txtHint").innerHTML = this.responseText;

In your PHP

You have to remove all these extra html tags from the top of the page:

<html>
<head>
    <title>Latest Attempt</title>
</head>

and this tag from the bottom of the page:

</html>

Then you in this line:

$q = intval($_get['q']);

You have to type $_GET in capital so it will be like:

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

And finally mysqli does not have mysqli_select() function, so you have to remove this line completely:

mysqli_select($conn,"ajax_demo");

And now you are good to go :)

Upvotes: 1

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