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I created this small script, from what I have learned about Ajax + Javascript so far. I am running this in Chrome btw, so XMLHttpRequest(); should be working

PHP, 

    $conn = new PDO('mysql:host=localhost; dbname=test', 'root', '')

    $stmt = $conn->query("SELECT text FROM ajax");
    foreach($stmt as $each){

?>

Ajax,

<script type='text/javascript'>

    var request = new XMLHttpRequest();
    request.open('GET', <?php echo $each['text']; ?>, false);
    request.send();
    console.log(request);

</script>
<?php } ?>

Now, in database test where tbl name is ajax I have rows id & text, in which text has three filled-in rows, but the ajax code is not showing me anything. It is not echoing anything at all, let alone update instantly, when I am adding texts in my rows. What am I doing wrong?

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3
  • 1
    I would recommend you to use jQuery for ajax requests.. Commented May 9, 2013 at 15:07
  • I would love to @Dremp but, I have no clue of jQuery Commented May 9, 2013 at 15:07
  • that code.. can get changed into $.get('<?php echo $each['text']; ?>', function(data) { /* console.log(data); */}); Commented May 13, 2013 at 16:54

2 Answers 2

Reset to default
1

out.php

<?php
     $conn = new PDO('mysql:host=localhost; dbname=test', 'root', '')

     $stmt = $conn->query("SELECT text FROM ajax");
     echo json_encode($stmt);
?>

test.html

<script>
    var request = new XMLHttpRequest();
    request.open('GET', "http://mysite.com/out.php", false);
    request.onreadystatechange=function()
    {
        if (request.readyState==4 && request.status==200)
        {
             var data = JSON.stringify(request.responseText); //The data from the server will be in responseText
             //data now contains an array of JSON objects of your data
             for(i=0;i<data.length;i++) {
                 console.log(data[i].text);   //.text is a variable based on your MYSQL field
             }
        }
    }
    request.send();
    console.log(request);
</script>

First you need to make a php script that outputs the data straight out nothing else (as in the out.php section above) and then have the javascript script in a whole other page (as in the test.html section) and give the request.open call the url to out.php in the example above http://mysite.com/out.php, but replace that with your actual url to the out.php file, and since you are on localhost something like http://localhost/Portal/Test/out.php

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8 Comments

I tried your code, although I undestood some of the mistakes I've made, but I did not get the script to work. Nothing is being desplayed: btw: I replaced all my ajax script with mine.
I just reedited my answer a moment ago i had xmlhttp.readyState==4 && xmlhttp.status==200 when it needed to be request.readyState==4 && request.status==200 to match your variable name, did you copy the older version?
And is your text field a url or just text? if it is just text that wont work, the ajax call needs a url to where the data it is trying to retrieve is being outputted.
It is pure text, so how do I assing the url? Sorry, for the noob question. Just show me that one, and I'll stop bothering you
I am on localhost btw: localhost/Portal/Test/ajax.php
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First I would remove the foreach loop. Change it to the following:

echo json_encode($stmt);

That will give your AJAX request variable a JSON encoded object to work with. Currently your result variable from your AJAX request is only displaying the results within the console.

Maybe try changing console.log(request) to the follwing:

alert(JSON.stringify(request));
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2 Comments

Changed it, but How do I manupilate the the object inside the ajax code?
JSON objects are the equivalent of an associative array in PHP. In other words instead of using a numerical value for the values key, a string can be used. Research JS & JSON objects as there are tons of examples of looping through the object.

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