PHP code defining variable sqlshowvalue
$sqlshowvalue = 5;
if(isset($_POST['showmore'])) {
$sqlshowvalue += 5;
}
So I connect to my database and then when I run this SQL query below using the variable that I just defined above,
$result = mysqli_query($conn,"SELECT * FROM comments ORDER BY id DESC limit '$sqlshowvalue'");
So I am using mysqli as the method to connect to my DB and it gives me the following error:
Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given in ..
The reason it gives me this error is because something in my query is wrong and what it has to do is $sqlshowvalue, because if I replace sqlshowvalue with with just the number 5 (like shown below), it works fine:
$result = mysqli_query($conn,"SELECT * FROM comments ORDER BY id DESC limit 5");
So I am just wondering what I can do to make it so that the value for the limit is a PHP variable that I can be changed and the page updated.
Have you tried making
$result = mysqli_query($conn,"SELECT * FROM comments ORDER BY id DESC limit '$sqlshowvalue'");
to
$result = mysqli_query($conn,"SELECT * FROM comments ORDER BY id DESC limit ".$sqlshowvalue);
Remove ''
use only $sqlshowvalue
:-
$result = mysqli_query($conn,"SELECT * FROM comments
ORDER BY id DESC limit $sqlshowvalue");
for integer value no need of ''
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