Asked  7 Months ago    Answers:  5   Viewed   38 times

I would like to search my table having a column of first names and a column of last names. I currently accept a search term from a field and compare it against both columns, one at a time with

    select * from table where first_name like '%$search_term%' or 
    last_name like '%$search_term%';

This works fine with single word search terms but the result set includes everyone with the name "Larry". But if someone enters a first name then a space, then a last name, I want a narrower search result. I've tried the following without success.

    select * from table where first_name like '%$search_term%' or last_name 
    like '%$search_term%' or concat_ws(' ',first_name,last_name) 
    like '%$search_term%';

Any advice?

EDIT: The name I'm testing with is "Larry Smith". The db stores "Larry" in the "first_name" column, and "Smith" in the "last_name" column. The data is clean, no extra spaces and the search term is trimmed left and right.

EDIT 2: I tried Robert Gamble's answer out this morning. His is very similar to what I was running last night. I can't explain it, but this morning it works. The only difference I can think of is that last night I ran the concat function as the third "or" segment of my search query (after looking through first_name and last_name). This morning I ran it as the last segment after looking through the above as well as addresses and business names.

Does running a mysql function at the end of a query work better than in the middle?



What you have should work but can be reduced to:

select * from table where concat_ws(' ',first_name,last_name) 
like '%$search_term%';

Can you provide an example name and search term where this doesn't work?

Wednesday, March 31, 2021
answered 7 Months ago

You're escaping the $ in the variable by doing $. Try:

$query = "SELECT * FROM `cats` WHERE name='$name'";


From the discussion below.

The problem with the undefined index is the fact that you are using $row['age'] when really, the column name in the database is Age. Therefore you must use $row['Age'] when referring to the item. The same goes for name.

Wednesday, March 31, 2021
answered 7 Months ago

you can't print the result from mysqli_query, it is mysqli_resource and for dumping the error you need to change mysql_error() to mysqli_error()

$username = "bob";
$db = mysqli_connect("localhost", "username", "password", "user_data");
$sql1 = "select id from user_information where username='$username'";
$result = mysqli_query($db, $sql1) or die(mysqli_error());
while ($row = mysqli_fetch_array($result, MYSQLI_ASSOC)) { 
    echo $row['id'].'<br>'; 
Saturday, May 29, 2021
answered 5 Months ago

You can only use column aliases in GROUP BY, ORDER BY, or HAVING clauses.

Standard SQL doesn't allow you to refer to a column alias in a WHERE clause. This restriction is imposed because when the WHERE code is executed, the column value may not yet be determined.

Copied from MySQL documentation

As pointed in the comments, using HAVING instead may do the work. Make sure to give a read at this question too: WHERE vs HAVING.

Tuesday, June 1, 2021
answered 5 Months ago

yo need create the user "pma" in mysql or change this lines(user and password for mysql):

/* User for advanced features */
$cfg['Servers'][$i]['controluser'] = 'pma'; 
$cfg['Servers'][$i]['controlpass'] = '';

Linux: /etc/phpmyadmin/

Tuesday, July 13, 2021
answered 4 Months ago
Only authorized users can answer the question. Please sign in first, or register a free account.
Not the answer you're looking for? Browse other questions tagged :