MySQL check if a table exists without throwing an exception

Question

What is the best way to check if a table exists in MySQL (preferably via PDO in PHP) without throwing an exception. I do not feel like parsing the results of "SHOW TABLES LIKE" et cetera. There must be some sort of boolean query?

Answer 1

Querying the information_schema database using prepared statement looks like the most reliable and secure solution.

$sql = "SELECT 1 FROM information_schema.tables 
        WHERE table_schema = database() AND table_name = ?";
$stmt =  $pdo->prepare($sql);
$stmt->execute([$tableName]);
$exists = (bool)$stmt->fetchColumn();

Answer 2

If you're using MySQL 5.0 and later, you could try:

SELECT COUNT(*)
FROM information_schema.tables 
WHERE table_schema = '[database name]' 
AND table_name = '[table name]';

Any results indicate the table exists.

From: http://www.electrictoolbox.com/check-if-mysql-table-exists/

Answer 3

Using mysqli I've created following function. Assuming you have an mysqli instance called $con.

function table_exist($con, $table){
    $table = $con->real_escape_string($table);
    $sql = "show tables like '".$table."'";
    $res = $con->query($sql);
    return ($res->num_rows > 0);
}

Hope it helps.

Warning: as sugested by @jcaron this function could be vulnerable to sqlinjection attacs, so make sure your $table var is clean or even better use parameterised queries.

Answer 4

This is posted simply if anyone comes looking for this question. Even though its been answered a bit. Some of the replies make it more complex than it needed to be.

For mysql* I used :

if (mysqli_num_rows(
    mysqli_query(
                    $con,"SHOW TABLES LIKE '" . $table . "'")
                ) > 0
        or die ("No table set")
    ){

In PDO I used:

if ($con->query(
                   "SHOW TABLES LIKE '" . $table . "'"
               )->rowCount() > 0
        or die("No table set")
   ){

With this I just push the else condition into or. And for my needs I only simply need die. Though you can set or to other things. Some might prefer the if/ else if/else. Which is then to remove or and then supply if/else if/else.

Answer 5

Here is the my solution that I prefer when using stored procedures. Custom mysql function for check the table exists in current database.

delimiter $$

CREATE FUNCTION TABLE_EXISTS(_table_name VARCHAR(45))
RETURNS BOOLEAN
DETERMINISTIC READS SQL DATA
BEGIN
    DECLARE _exists  TINYINT(1) DEFAULT 0;

    SELECT COUNT(*) INTO _exists
    FROM information_schema.tables 
    WHERE table_schema =  DATABASE()
    AND table_name =  _table_name;

    RETURN _exists;

END$$

SELECT TABLE_EXISTS('you_table_name') as _exists

Answer 6

As a "Show tables" might be slow on larger databases, I recommend using "DESCRIBE " and check if you get true/false as a result

$tableExists = mysqli_query("DESCRIBE `myTable`");

Answer 7

$q = "SHOW TABLES";
$res = mysql_query($q, $con);
if ($res)
while ( $row = mysql_fetch_array($res, MYSQL_ASSOC) )
{
    foreach( $row as $key => $value )
    {
        if ( $value = BTABLE )  // BTABLE IS A DEFINED NAME OF TABLE
            echo "exist";
        else
            echo "not exist";
    }
}

Answer 8

Zend framework

public function verifyTablesExists($tablesName)
    {
        $db = $this->getDefaultAdapter();
        $config_db = $db->getConfig();

        $sql = "SELECT COUNT(*) FROM information_schema.tables WHERE table_schema = '{$config_db['dbname']}'  AND table_name = '{$tablesName}'";

        $result = $db->fetchRow($sql);
        return $result;

    }

Answer 9

If the reason for wanting to do this is is conditional table creation, then 'CREATE TABLE IF NOT EXISTS' seems ideal for the job. Until I discovered this, I used the 'DESCRIBE' method above. More info here: MySQL "CREATE TABLE IF NOT EXISTS" -> Error 1050

Answer 10

Why you make it so hard to understand?

function table_exist($table){ 
    $pTableExist = mysql_query("show tables like '".$table."'");
    if ($rTableExist = mysql_fetch_array($pTableExist)) {
        return "Yes";
    }else{
        return "No";
    }
}