SQL UPDATE with LIKE

Question

I was trying to update approx 20,000 records in a table by using statements like this one, however, I got the message say 0 row(s) affected so it didn\'t work.<

Answer 1

was looking for same but didn't work for me.... but this works perfect:

$title = "Worklog details";
$log_name = 'PC01';
$update = $connection -> prepare("UPDATE nc_files 
SET title = ? 
WHERE log_name = ?");
$update->bind_param('ss',$title,$log_name);
$update->execute();
$update->close();

Answer 2

I had a similar trouble. The problem are the quotations marks " I Fixed my code as follow.

UPDATE Table SET Table.Field = "myreplace" WHERE (((Table.Field) Like '%A-16%'));

Regards, Alexwin1982

Answer 3

"log_name" should not be in quotes

Answer 4

By default MySQL allows double quoted names to be understood as either identifiers (like column names) or as string literals.

This is meant as a convenience, but I find the semantic ambiguity frustrating. MySQL must resolve the ambiguity, and cannot magically always guess the coder's intention, as you discovered.

-- In default sql_mode under 5.5
--
SELECT "foo"           -- "foo" is a *column* if tbl.foo exists, otherwise a string 
  FROM "tbl"           -- Oops!  ER_PARSE_ERROR (1064)  Can't do this in default mode.
 WHERE "foo" = "foo";  -- Both of these are strings

So, the way around it is to force unambiguous interpretation of identifiers:

1. Do not quote simple identifiers
2. Use MySQL-specific backticks for quoting
   (This is ODBC's SQL_IDENTIFIER_QUOTE_CHAR)
3. Always override the change the sql_mode to include ANSI_QUOTES (or a superset of it)
   Double quotes are then exclusively for identifiers, single quotes for strings.

#3 is my personal favorite, for readability and portability. The problem is it tends to surprise people who only know MySQL, and you have to remember to override the default.

Answer 5

this is because your column name log_name should not be in ' quotes.

"log_name" LIKE "%PC01%" condition will always fail and zero rows will get updated, try this:

UPDATE nc_files 
SET title ="Worklog details" 
WHERE log_name LIKE "%PC01%";

Answer 6

The log_name is a field name, remove literal quotes from it -

UPDATE nc_files SET title ="Worklog details" WHERE log_name LIKE "%PC01%"