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Asked: 2026-06-11T02:41:42+00:00

I own a image hosting website and I capture image views using php and

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I own a image hosting website and I capture image views using php and mysql.

I use the following code to count the views.

include 'mysql.php';

$result = mysql_query("SELECT * FROM DB WHERE ID='$id'");
$row = mysql_fetch_array($result);
$views=$row['views'];

$query = "UPDATE DB SET views=$views+1 WHERE ID='$id'";
$result2 = mysql_query($query);

mysql_close($con);

views is mediumint(9) type field.

I noticed that the views get decreased day by day.can anyone say what is the problem and offer a solution.

Thanks.

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1 Answer

  1. Editorial Team

    You should use this to update instead:

    $query = "UPDATE DB SET views=views+1 WHERE ID='$id'";

    If a page takes a long time to execute, you can have one query overwrite another. Also using this, you might not need to even run the first query – unless you want other info from it.

    The reason you are getting an error is that one script is reading the data and grabbing the value, then updating it – based on the value it is storing – but in the meantime other scripts could be updating the row. You could avoid it by using transactions, but that seems utter overkill for what you are doing.

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