Using Ajax to query a database

Question

I want to check my database from inside a JavaScript function. I have never done this before, and I can't seem to figure it out. I don't know if this is the right way, but it's the one I found. This is my simple HTML :

<!doctype html>
<html>
<head>
    <script src="Test.js"></script>
</head>
<body >
<p><a href="#" onClick="start();">Click me</a></p>
</body>

The JavaScript:

function start(){
    $.ajax({
        type: "POST",
        url: "checkDatet.php",
        datatype: "html",
        data: {functionname: 'Name', arguments: ['John']},
        success: function(data) {
            alert(data);
        }
    });
}

And the PHP :

<?php
function familyName($name1) {
    $username = "user";
    $password = "pass";
    $hostname = "localhost";
    $output ="AAA";
    $dbhandle = mysql_connect($hostname, $username, $password)
        or die("Unable to connect to MySQL");
    $selected = mysql_select_db("databasee", $dbhandle)
        or die("Could not select database 'databasee'");
    $names = mysql_query("SELECT name WHERE name1 = '$name'");
    if($row = mysql_fetch_assoc($names)){
        $data = $row{'name'};
        if($data <> null)
            $output= "found it";
        else
            $output= "havent found it";
    }
    echo $output;
}
?>

better use dataType of text. change `datatype: "html",` to `datatype: "text",` — guradio, Oct 10 '15 at 11:54
I am getting no errors. It's just that it doesn't show that alert. — Ann, Oct 10 '15 at 11:58
`familyName()` is a PHP function, but nothing is calling it. Do you know PHP? — CodeGodie, Oct 10 '15 at 12:00
Possible duplicate of [using jquery $.ajax to call a PHP function](http://stackoverflow.com/questions/2269307/using-jquery-ajax-to-call-a-php-function) — CodeGodie, Oct 10 '15 at 12:02
Also, should i have the second line in my html ? I am getting ' Cannot resolve file' on that line — Ann, Oct 10 '15 at 12:02
I know a bit of PHP. What do you mean by nothing is calling it ? Is that $.ajax block doing ? // edit: Now it shows an alert saying: The page at localhost says: — Ann, Oct 10 '15 at 12:04
the ajax block is sending a POST request to your PHP file. But your PHP code if not grabbing the sent POST data, nor calling that function. You need to learn about POST in PHP — CodeGodie, Oct 10 '15 at 12:05
Is checkDatet.php in the same directory of the html file? You must specify a relative URL. Besides, where is called familyName($name)? — Luciano, Oct 10 '15 at 12:08
Yes it's in the same folder. I don't know how to call my function... — Ann, Oct 10 '15 at 12:10

minychillo · Accepted Answer · 2015-10-10T12:24:20.787

2

Nothing is actually calling your function ! Do this:

function start(){
    $.get('checkDatet.php?name=John', function(data) {
        alert(data);
    }
});

Then in "checkDatet.php" you need to call your function:

<?php
familyName($_GET['name']);

function familyName($name1) {
$username = "user";
$password = "pass";
$hostname = "localhost";
$output ="AAA";
$dbhandle = mysql_connect($hostname, $username, $password)
    or die("Unable to connect to MySQL");
$selected = mysql_select_db("databasee", $dbhandle)
    or die("Could not select database 'databasee'");
$names = mysql_query("SELECT name WHERE name1 = '$name'");
if($row = mysql_fetch_assoc($names)){
    $data = $row{'name'};
    if($data <> null)
        $output= "found it";
    else
        $output= "havent found it";
}
echo $output;
}
?>

edited Oct 10 '15 at 12:24

answered Oct 10 '15 at 12:12

minychillo

395
4
17

It tells me : "undefined index 'name' " on this line : familyName($_GET['name']); – Ann Oct 10 '15 at 12:21
Ups, sorry, it should have been familyName($_POST['name']); – minychillo Oct 10 '15 at 12:23
In the alert box that pops up – Ann Oct 10 '15 at 12:25
I figured it out now. Thanks for all your help guys ! – Ann Oct 10 '15 at 12:28

score 2 · Answer 2 · answered Oct 10 '15 at 12:14

You call calling a file, and in that you are not calling the function, so just change your php code like this

<?php
    $username = "user";
    $password = "pass";
    $hostname = "localhost";
    $output ="AAA";
    $name = $_POST['name'];
    $dbhandle = mysql_connect($hostname, $username, $password)
        or die("Unable to connect to MySQL");
    $selected = mysql_select_db("databasee", $dbhandle)
        or die("Could not select database 'databasee'");
    $names = mysql_query("SELECT name WHERE name1 = '$name'");
    if($row = mysql_fetch_assoc($names)){
        $data = $row{'name'};
        if($data <> null)
            $output= "found it";
        else
            $output= "havent found it";
    }
    echo $output;

?>

Using Ajax to query a database

2 Answers2