MySQL 检查表是否存在而不抛出异常

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"; } }