Need help learning ajax, grabbing data from mysql

I have an ajax script, which I kinda understand, but still need some extra help.

$('.images').click(function(){
    var imageId = $(this).attr('id');
    alert(imageName);
    $.ajax({
            type: "get",
            url: "imageData.php",
            dataType: "json",
            data: {getImageId: imageId},
            error: function() {
                alert("error");
            },
            success: function(data){
     开发者_StackOverflow社区           alert(imageId);
                $("#images_"+imageId).html(data);
            }
        });
    //$('#images_'+imageId).toggle();

});

I have that code, it goes to this imageData.php file

<?php
if(isset($_GET)){
    $images = "";
    $path = 'img/';
    $imageId = $_GET['getImageId'];
    $sql = mysql_query("SELECT * FROM images WHERE iID = '".$imageId."'");
    while($row = mysql_fetch_array($sql)){
        $images .= $path.$row['images'];
}
    $json = json_encode($images);
?>
<img src='<?php echo $json;?>'/>

    <?php
}
    ?>

Why does it output error when I try to echo a string from $images, but it outputs correctly when I do echo $imageId;? I'm trying to output something from mysql, but not trying to output just the id.

Need help please, thank you

---

You don't need use json_encode here, there is not data that needs to be in JSON format. There is also no reason to loop over the result set, if the query only returns one image.

Try this:

<?php
if(isset($_GET['getImageId'])) {
    $path = '';
    $imageId = mysql_real_escape_string($_GET['getImageId']); // SQL injection!
    $result = mysql_query("SELECT images FROM images WHERE iID = '".$imageId."'");
    $row = mysql_fetch_array($result);
    if($row) {
        $path = 'img/' . $row['images'];
    }
}
?>
<?php if($path): ?>
    <img src='<?php echo $path;?>'/>
<?php endif; ?>

If the iID is actually an integer, you need to omit the single quotes in the query.

You also have to change the dataType from json to html, as you are returning an image tag (HTML) and not JSON:

$.ajax({
    type: "get",
    url: "imageData.php",
    dataType: "html",
    data: {getImageId: imageId},
    error: function() {
        alert("error");
    },
    success: function(data){
        $("#images_"+imageId).html(data);
    }
});

---

Another option is to return only text (the link) and create the images on the client side:

<?php
if(isset($_GET['getImageId'])) {
    $path = '';
    $imageId = mysql_real_escape_string($_GET['getImageId']); // SQL injection!
    $result = mysql_query("SELECT images FROM images WHERE iID = '".$imageId."'");
    $row = mysql_fetch_array($result);
    if($row) {
        echo 'img/' . $row['images'];
    }
}
?>

And in JavaScript:

$.ajax({
    type: "get",
    url: "imageData.php",
    dataType: "text",
    data: {getImageId: imageId},
    error: function() {
        alert("error");
    },
    success: function(data){
        $("#images_"+imageId).html('<img src="' + data + '" />');
    }
});

---

As you may get many images because you use while loop you probably want to do this like so:

in php:

$x = 0;
$another = array();
while($row = mysql_fetch_array($sql)){
    $another[$x] = $path.$row['images'];
    $x++;
}

echo json_encode($another);

and in jquery (in your success callback):

$.each(data, function(i, v){
    // Do the image inserting to the DOM here v is the path to image
    $('#somelement').append('<img src="'+v+'"');
});

---

For outputing an image you must set src attribute of the image tag, if you already have one, or you can create it on the fly. See here how to do that > jQuery document.createElement equivalent?