php mysql query - output value of variable in the query

Question

I have the following mysql query in php:

$results = $wpdb->get_results("SELECT * FROM $wpdb->postmeta WHERE meta_key = '_wp_attachment_metadata' AND meta_value LIKE '%details%'");

However I want the query to be dynamic by changing the LIKE section of the query. Instead of:

LIKE '%details%'

I want to put a variable in there:

LIKE '% $format %'

where $format is a string.

Everything I have tried thus far has failed.

Whats the proper way to do this?

Answer 1

Try wrapping in braces:

LIKE '%{$format}%'

Answer 2

Since you are using double quotes you could simply do:

$results = $wpdb->get_results("SELECT * FROM $wpdb->postmeta WHERE meta_key = '_wp_attachment_metadata' AND meta_value LIKE '%$format%'");

Or simply concatenate the string:

$results = $wpdb->get_results("SELECT * FROM $wpdb->postmeta WHERE meta_key = '_wp_attachment_metadata' AND meta_value LIKE '%" . $format . "%'");

Answer 3

Your string is already in double quotes, so just surround your variable with curly braces and you should be good to go.

"SELECT * FROM $wpdb->postmeta WHERE meta_key = '_wp_attachment_metadata' AND meta_value LIKE '%{$format}%'"

Answer 4

Before passing the variable do this

$format = '%' . $format . '%';
now simply put it in the query.

$results = $wpdb->get_results("SELECT * FROM $wpdb->postmeta WHERE meta_key = '_wp_attachment_metadata' AND meta_value LIKE '$format'");

Answer 5

You already got the answer. I can do it exactly like you want.

$results = $wpdb->get_results("SELECT * FROM $wpdb->postmeta WHERE meta_key = '_wp_attachment_metadata' AND meta_value LIKE '%$details%'");

_{Since the query is wrapped in double quotes, you dont need to escape anything.}