MySQL check if a table exists without throwing an exception

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?

Answers:

Thank you for visiting the Q&A section on Magenaut. Please note that all the answers may not help you solve the issue immediately. So please treat them as advisements. If you found the post helpful (or not), leave a comment & I’ll get back to you as soon as possible.

Method 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();

Method 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 = '<div class="su-table su-table-alternate"></div>';

Any results indicate the table exists.

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

Method 3

Using mysqli i’ve created following function. Asuming you have an mysqli instance called $con.

function table_exist($table){
    global $con;
    $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.

Method 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.

Method 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

Method 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`");

Method 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";
    }
}

Method 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;

    }

Method 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

Method 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";
    }
}


All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0

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