show tables get only the last name of table - mysql

Question

I'm trying to show the name of table in my database. I write this code :
function affiche_liste()
{
$db=new PDO('mysql:host=localhost;dbname=testf','root','');
$result = $db->query("SHOW TABLES");
foreach($result->fetch(PDO::FETCH_NUM) as $data) {
$tableList = $data[0];
}
return $tableList;
}
It give to me only the last table ?

Answer 1

For a simple query without parameters and the SQL hard coded you can use a generic function passing the connection and SQL to the function.
The following function () returns an array containing all rows in the result set.
function queryAll($db,$query){
$sth = $db->query($query);
$result = $sth->fetchAll(PDO::FETCH_NUM);
return $result;
}
For a simple query without parameters and the SQL hard coded you can use a generic function passing the connection and SQL to the function.
The following function () returns an array containing all rows in the result set.
function queryAll($db,$query){
$sth = $db->query($query);
$result = $sth->fetchAll(PDO::FETCH_NUM);
return $result;
}
$db=new PDO('mysql:host=localhost;dbname=testf','root','');
$db->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
$query = "SHOW TABLES";
$tables = queryAll($db,$query);
print_r($tables);

Answer 2

Each time you are looping through the results, your are overwriting the variable tableList. Instead, you need to append to an array of results.
function affiche_liste()
{
$db=new PDO('mysql:host=localhost;dbname=testf','root','');
$result = $db->query("SHOW TABLES");
$tableList = array();
foreach($result->fetch(PDO::FETCH_NUM) as $data) {
array_push($tableList, $data[0]);
}
return $tableList;
}

Answer 3

try this
`
function affiche_liste()
{$db=new PDO('mysql:host=localhost;dbname=testf','root','');
$result = $db->query("SHOW TABLES");
foreach($result->fetch(PDO::FETCH_NUM) as $data) {
$tableList[] = $data[0];
}
return $tableList;
}
`
i hope it'll work...
here your array is overwitting every time...that's the reason you were getting the last table name....so you need to append to the existing array...