[Solved] Why does this piece of code not work, when I pass value through input text?

by


I’d wrap $update with single quotes (notice that I flipped the quotations) and changed id1 into $id1:

mysqli_query($conn , "update insert1 set  ".$name." = '".$update."'
where id-1 = ".$id1 );

If id-1 is a string column type in the database then I’d wrap $id1 with single quotes. like this:

mysqli_query($conn , "update insert1 set  ".$name." = '".$update."'
where id-1 = '".$id1."'" );

Notes:

  • I’d double check if id-1 is intended in the WHERE condition, because it checks if the value in that column is 2 rather than 1. WHERE id - 1 = 1 is equivalent to WHERE id = 2 but more confusing to the reader (thanks to FirstOne for pointing that out).
  • As mentioned in another answer, your code is vulnerable for SQL injection, I’d check this: https://stackoverflow.com/a/16282269/4283725

1

solved Why does this piece of code not work, when I pass value through input text?