php MySQL检查表是否存在而不抛出异常
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MySQL check if a table exists without throwing an exception
提问by clops
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?
在不抛出异常的情况下检查 MySQL 中是否存在表（最好通过 PHP 中的 PDO）的最佳方法是什么。我不想解析“SHOW TABLES LIKE”等的结果。必须有某种布尔查询？
回答by nickf
I don't know the PDO syntax for it, but this seems pretty straight-forward:
我不知道它的 PDO 语法，但这看起来很简单：
$result = mysql_query("SHOW TABLES LIKE 'myTable'");
$tableExists = mysql_num_rows($result) > 0;
回答by Michael Todd
If you're using MySQL 5.0 and later, you could try:
如果您使用的是 MySQL 5.0 及更高版本，您可以尝试：
SELECT COUNT(*)
FROM information_schema.tables
WHERE table_schema = '[database name]'
AND table_name = '[table name]';
Any results indicate the table exists.
任何结果都表明该表存在。
From: http://www.electrictoolbox.com/check-if-mysql-table-exists/
来自：http: //www.electrictoolbox.com/check-if-mysql-table-exists/
回答by Falk
Using mysqli i've created following function. Asuming you have an mysqli instance called $con.
使用 mysqli 我创建了以下函数。假设您有一个名为 $con 的 mysqli 实例。
function table_exist($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.
希望能帮助到你。
Warning:as sugested by @jcaron this function could be vulnerable to sqlinjection attacs, so make sure your $tablevar is clean or even better use parameterised queries.
警告：正如@jcaron 所建议的，此函数可能容易受到 sqlinjection 攻击，因此请确保您的$tablevar 是干净的，甚至更好地使用参数化查询。
回答by Esoterica
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.
如果有人来寻找这个问题，这只是张贴。即使它被回答了一点。一些答复使它比需要的更复杂。
For mysql* I used :
对于 mysql*，我使用了：
if (mysqli_num_rows(
mysqli_query(
$con,"SHOW TABLES LIKE '" . $table . "'")
) > 0
or die ("No table set")
){
In PDO I used:
在 PDO 中，我使用了：
if ($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.
有了这个，我只是将 else 条件推入 or。为了我的需要，我只需要死。虽然你可以设置或其他东西。有些人可能更喜欢 if/else if/else。然后删除或然后提供 if/else if/else。
回答by erandac
Here is the my solution that I prefer when using stored procedures. Custom mysql function for check the table exists in current database.
这是我在使用存储过程时更喜欢的解决方案。用于检查当前数据库中是否存在表的自定义 mysql 函数。
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
回答by Martin Lisicki
As a "Show tables" might be slow on larger databases, I recommend using "DESCRIBE " and check if you get true/false as a result
由于“显示表”在较大的数据库上可能会很慢，我建议使用“DESCRIBE”并检查结果是否为真/假
$tableExists = mysqli_query("DESCRIBE `myTable`");
回答by gilcierweb
Zend framework
Zend 框架
public function verifyTablesExists($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;
}
回答by Robin Andrews
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
如果想要这样做的原因是有条件的表创建，那么“CREATE TABLE IF NOT EXISTS”似乎是这项工作的理想选择。在我发现这一点之前，我使用了上面的“DESCRIBE”方法。更多信息：MySQL“如果不存在则创建表”-> 错误 1050
回答by Namkeen Butter
$q = "SHOW TABLES";
$res = mysql_query($q, $con);
if ($res)
while ( $row = mysql_fetch_array($res, MYSQL_ASSOC) )
{
foreach( $row as $key => $value )
{
if ( $value = BTABLE ) // BTABLE IS A DEFINED NAME OF TABLE
echo "exist";
else
echo "not exist";
}
}
回答by Hamed
Why you make it so hard to understand?
你为什么这么难理解？
function table_exist($table){
$pTableExist = mysql_query("show tables like '".$table."'");
if ($rTableExist = mysql_fetch_array($pTableExist)) {
return "Yes";
}else{
return "No";
}
}
