MySQL – Cannot Select All Data from Table in PHP

MySQL

$sql = $conn->prepare('SELECT * FROM users WHERE user_id != null');
if($sql->rowCount() > 0){
}
else{
echo "there are no users"
}

I want to select all data from a table however the code above does not work. It echoes/prints out "there are no users." No errors are shown too.

Best Answer

user_id != NULL

always fails. Instead, say

user_id IS NOT NULL

And you need to ->execute it.

Related Solutions

Mysql – Calculate MATCH() AGAINST() scores from UNIFIED QUERY not FOR EACH TABLE

Perhaps you could record the following from the three(3) tables

table name
the column from the table name
FULLTEXT index on the column

Here is the code:

DROP TABLE IF EXISTS combined_data;
CREATE TABLE combined_data
(
    source_table VARCHAR(64),
    source_id INT NOT NULL,
    data TEXT NOT NULL,
    FULLTEXT (data)
) ENGINE=MyISAM;
--
ALTER TABLE combined_data DISABLE KEYS;
--
INSERT INTO combined_data (source_table,source_id,data) VALUES
SELECT 'table1',id,`result` FROM table1 WHERE MATCH(`result`) AGAINST('keyword');
--
INSERT INTO combined_data (source_table,source_id,data) VALUES
SELECT 'table2',id,`content` FROM table1 WHERE MATCH(`content`) AGAINST('keyword');
--
INSERT INTO combined_data (source_table,source_id,data) VALUES
SELECT 'table3',id,`text` FROM table1 WHERE MATCH(`text`) AGAINST('keyword');
--
ALTER TABLE combined_data ENABLE KEYS;

Now you can run a single query against the one table

SELECT *,MATCH(`data`) AGAINST('keyword') as `good_score`
FROM combined_data
ORDER BY `good_score` DESC;

Give it a Try !!!

Mysql – Replacing a PHP loop with a MySQL Query

Sure, just use a subquery

select distinct u.*
from 
  user_hashes uh
  inner join users u on uh.user_id = u.id
WHERE 
  uh.hash IN
  (
    select hash
    from user_hashes buh
    where user_id = ?
  )

Best Answer

Related Solutions

Mysql – Calculate MATCH() AGAINST() scores from UNIFIED QUERY not FOR EACH TABLE

Mysql – Replacing a PHP loop with a MySQL Query

Related Question