Display data from SQL database in html: returned data object empty

Question

I want to retrieve data from my SQL database and use it on my HTML page.

I have a php script get_score.php like this:

<?php
$con = mysqli_connect( "localhost", "xxx1", "xxx2", "xxx3");
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$sql = "SELECT `name`, `score` FROM crossstreet ORDER BY `score` DESC LIMIT 5";
if (!mysqli_query($con, $sql))
{
die('Error: ' . mysqli_error($con));
}
$result = mysqli_query($con, $sql);
while ($row = mysqli_fetch_assoc($result)) {
    $output = ($row["name"]);
}
mysqli_free_result($result);
echo json_encode($result);

mysqli_close($con);
?>

And then I want to retrieve the data I via JQuery. I have read about the cross-origin issue but the data I retrieve and the html/Jquery are on the same server.

$.get("get_score.php", function( data ) {
  console.log(window[data]);
}, "json" );

This returns undefined. If I make it $(data) it returns an object like the one on this screenshot. enter image description here

Where is my mistake; how can I just get the data from the server to use in html?

You need to json_encode($output) not $result

Narendrasingh Sisodia
– Narendrasingh Sisodia

2015-05-26 11:08:11 +00:00
Commented May 26, 2015 at 11:08 — Narendrasingh Sisodia
– Narendrasingh Sisodia, Commented May 26, 2015 at 11:08

Barmar · Accepted Answer · 2015-05-26 11:10:01Z

2

You need to make $output an array so you get all the rows.

$output = array();
while ($row = mysqli_fetch_assoc($result)) {
    $output[] = $row['name'];
}

Then you should echo this:

echo json_encode($output);

And in your jQuery code, you should do console.log(data);

answered May 26, 2015 at 11:10

Barmar

789k57 gold badges554 silver badges669 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

Aneesh R S · Accepted Answer · 2015-05-26 11:09:11Z

0

mysqli_free_result($result);
echo json_encode($result);

i think this is the problem. Try to reverse the order. You are making the result free before using it. Do like this, instead of the above .

echo json_encode($result);
mysqli_free_result($result);

answered May 26, 2015 at 11:09

Aneesh R S

3,8174 gold badges25 silver badges35 bronze badges

1 Comment

Barmar Over a year ago

$result is not useful to send to the client.

Collectives™ on Stack Overflow

Display data from SQL database in html: returned data object empty

2 Answers 2

Comments

1 Comment

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

Comments

1 Comment

Your Answer

Sign up or log in

Post as a guest

Related