Fetch values from different rows in one MySqli query

Question

How do I fetch and echo in php some data from a database called "somename", table "options", column "value" from all the rows which have the "name" starting with "dt_"

database name: "somename" database table: "options" database existing columns: "name", "value" database existing rows: "something1", "something2", "dt_option1", "dt_option2", "dt_option3"

database column i want: "value" database rows i want: "dt_option1", "dt_option2", "dt_option3"

I want the result to look like:

dt_option1 - 3432

dt_option2 - on

dt_option3 - off

dt_option_something - value

I tried and googled everywhere, I can't figure it out cause I am new to this.. Thank you for any help..

P.S. I just need the query and the echo..

bish · Accepted Answer · 2013-06-08 13:58:59Z

1

To fetch the value of the rows: select name, value from options where name like "dt_%"

In PHP build a normal mysqli qry and then use a while-loop

// Note: Don't have the mysqli_* syntax 100% in mind, perhaps you have to lookup the right syntax in PHP manual, but for the idea you should understand it

$result = mysqli_query("select name, value from options where name like 'dt_%'");

while($row = mysqli_fetch_assoc($result))
{
  echo $row['name'] . " - " . $row['value'];
}

edited cause asker changed question and also wants to see the name not only the value

answered Jun 8, 2013 at 13:58

bish

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1 Comment

Adrian M. Over a year ago

I tried but I get this error: Warning: mysqli_query() expects at least 2 parameters, 1 given in /home/x/public_html/api.php on line 15 Warning: mysqli_fetch_assoc() expects parameter 1 to be mysqli_result, null given in /home/x/public_html/api.php on line 17 Warning: mysql_fetch_row() expects parameter 1 to be resource, null given in /home/x/public_html/api.php on line 22 null

David Wilkins · Accepted Answer · 2013-06-08 13:54:52Z

1

Did you try anything like this? SELECT name, value FROM options WHERE name LIKE 'dt_%'

Edited since the asker changed the question

answered Jun 8, 2013 at 13:54

David Wilkins

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Comments

markdwhite · Accepted Answer · 2013-06-08 13:55:59Z

1

Do you mean something like:

SELECT name, value FROM somename.options WHERE name LIKE "dt_%"

answered Jun 8, 2013 at 13:55

markdwhite

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