blocks|key|1347993|text|SELECT+SCHEMA_NAME
++FROM+INFORMATION_SCHEMA.SCHEMATA
+WHERE+SCHEMA_NAME+=+'DBName'|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1347994|如果您只需要知道数据库是否存在，以便在尝试创建它时不会出现错误，只需使用(来自here)：|unstyled|offset|length|1347995|CREATE+DATABASE+IF+NOT+EXISTS+DBName;|1347996|entityMap|0|LINK|mutability|MUTABLE|url|http://forums.mysql.com/read.php?101,213455,213496^0|0|13|4|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|T|8|@]|9|@$G|U|H|V|1|W]]|A|$]]|$1|I|3|J|5|6|7|X|8|@]|9|@]|A|$B|C]]|$1|K|3|-4|5|F|7|Y|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|R]]]]

<pre><code>SELECT SCHEMA_NAME
 FROM INFORMATION_SCHEMA.SCHEMATA
 WHERE SCHEMA_NAME = 'DBName'
</code></pre>

If you just need to know if a db exists so you won't get an error when you try to create it, simply use (From <a href="http://forums.mysql.com/read.php?101,213455,213496" rel="noreferrer">here</a>):

<pre><code>CREATE DATABASE IF NOT EXISTS DBName;
</code></pre>

blocks|key|1563638|text|检查数据库是否存在的一种简单方法是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1563639|SHOW+DATABASES+LIKE+'dbname';|code-block|syntax|javascript|1563640|如果名为“dbname”的数据库不存在，则会得到一个空集。如果它确实存在，您将得到一行。|1563641|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

A simple way to check if a database exists is:

<pre><code>SHOW DATABASES LIKE 'dbname';
</code></pre>

If database with the name 'dbname' doesn't exist, you get an empty set. If it does exist, you get one row.

blocks|key|1348109|text|来自像bash这样的shell|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1348110|if+[[+!+-z+"`mysql+-qfsBe+"SELECT+SCHEMA_NAME+FROM+INFORMATION_SCHEMA.SCHEMATA+WHERE+SCHEMA_NAME='db'"+2>&1`"+]];
then
++echo+"DATABASE+ALREADY+EXISTS"
else
++echo+"DATABASE+DOES+NOT+EXIST"
fi|code-block|syntax|javascript|1348111|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

From the shell like bash

<pre><code>if [[ ! -z "`mysql -qfsBe "SELECT SCHEMA_NAME FROM INFORMATION_SCHEMA.SCHEMATA WHERE SCHEMA_NAME='db'" 2&gt;&amp;1`" ]];
then
 echo "DATABASE ALREADY EXISTS"
else
 echo "DATABASE DOES NOT EXIST"
fi
</code></pre>

blocks|key|1347903|text|如果你正在寻找一个php脚本，请看下面。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1347904|$link+=+mysql_connect('localhost',+'mysql_user',+'mysql_password');
if+(!$link)+{
++die('Not+connected+:+'+.+mysql_error());
}

//+make+foo+the+current+db
$db_selected+=+mysql_select_db('foo',+$link);
if+(!$db_selected)+{
++die+('Cannot+use+foo+:+'+.+mysql_error());
}|code-block|syntax|javascript|1347905|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

If you are looking for a php script see below.

<pre><code>$link = mysql_connect('localhost', 'mysql_user', 'mysql_password');
if (!$link) {
 die('Not connected : ' . mysql_error());
}

// make foo the current db
$db_selected = mysql_select_db('foo', $link);
if (!$db_selected) {
 die ('Cannot use foo : ' . mysql_error());
}
</code></pre>

blocks|key|1348231|text|下面是一个用于检查数据库是否存在的bash函数：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1348232|function+does_db_exist+{
++local+db="${1}"

++local+output=$(mysql+-s+-N+-e+"SELECT+schema_name+FROM+information_schema.schemata+WHERE+schema_name+=+'${db}'"+information_schema)
++if+[[+-z+"${output}"+]];+then
++++return+1+#+does+not+exist
++else
++++return+0+#+exists
++fi
}+++++++++++|code-block|syntax|javascript|1348233|另一种选择是只尝试使用数据库。请注意，这也会检查权限：|1348234|if+mysql+"${db}"+>/dev/null+2>&1+</dev/null
then
++echo+"${db}+exists+(and+I+have+permission+to+access+it)"
else
++echo+"${db}+does+not+exist+(or+I+do+not+have+permission+to+access+it)"
fi|1348235|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Here is a bash function for checking if a database exists:


<pre class="lang-sh prettyprint-override"><code>function does_db_exist {
 local db="${1}"

 local output=$(mysql -s -N -e "SELECT schema_name FROM information_schema.schemata WHERE schema_name = '${db}'" information_schema)
 if [[ -z "${output}" ]]; then
 return 1 # does not exist
 else
 return 0 # exists
 fi
} 
</code></pre>

Another alternative is to just try to use the database. Note that this checks permission as well:

<pre class="lang-sh prettyprint-override"><code>if mysql "${db}" &gt;/dev/null 2&gt;&amp;1 &lt;/dev/null
then
 echo "${db} exists (and I have permission to access it)"
else
 echo "${db} does not exist (or I do not have permission to access it)"
fi
</code></pre>

blocks|key|1348528|text|一个非常简单的BASH-one代码：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1348529|mysqlshow+%7C+grep+dbname|code-block|syntax|javascript|1348530|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

A very simple BASH-one-liner:

<pre><code>mysqlshow | grep dbname
</code></pre>

blocks|key|1563785|text|检查PHP中是否存在数据库的一个很好的方法是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1563786|$mysql+=+mysql_connect("<your+host>",+"root",+"");

if+(mysql_select_db($mysql,+'<your+db+name>'))+{
++++echo+"Database+exists";
}+else+{
++++echo+"Database+does+not+exist";
}|code-block|syntax|javascript|1563787|这是我一直使用的方法。|1563788|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

A great way to check if a database exists in PHP is:

<pre><code>$mysql = mysql_connect("&lt;your host&gt;", "root", "");

if (mysql_select_db($mysql, '&lt;your db name&gt;')) {
 echo "Database exists";
} else {
 echo "Database does not exist";
}
</code></pre>

That is the method that I always use.

blocks|key|1348481|text|对于那些使用php和mysqli的人来说，这就是我的解决方案。我知道答案已经得到了回答，但我认为将答案作为mysqli准备好的语句也会很有帮助。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1348482|$db+=+new+mysqli('localhost',username,password);
$database="somedatabase";
$query="SELECT+SCHEMA_NAME+FROM+INFORMATION_SCHEMA.SCHEMATA+WHERE+SCHEMA_NAME=?";
$stmt+=+$db->prepare($query);
$stmt->bind_param('s',$database);
$stmt->execute();
$stmt->bind_result($data);
if($stmt->fetch())
{
++++echo+"Database+exists.";
}
else
{
++++echo"Database+does+not+exist!!!";
}
$stmt->close();|code-block|syntax|javascript|1348483|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

For those who use php with mysqli then this is my solution. I know the answer has already been answered, but I thought it would be helpful to have the answer as a mysqli prepared statement too.

<pre><code>$db = new mysqli('localhost',username,password);
$database="somedatabase";
$query="SELECT SCHEMA_NAME FROM INFORMATION_SCHEMA.SCHEMATA WHERE SCHEMA_NAME=?";
$stmt = $db-&gt;prepare($query);
$stmt-&gt;bind_param('s',$database);
$stmt-&gt;execute();
$stmt-&gt;bind_result($data);
if($stmt-&gt;fetch())
{
 echo "Database exists.";
}
else
{
 echo"Database does not exist!!!";
}
$stmt-&gt;close();
</code></pre>

blocks|key|1563811|text|CREATE+SCHEMA+IF+NOT+EXISTS+`demodb`+DEFAULT+CHARACTER+SET+utf8+;|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1563812|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>CREATE SCHEMA IF NOT EXISTS `demodb` DEFAULT CHARACTER SET utf8 ;
</code></pre>

blocks|key|1563871|text|使用bash：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1563872|if+[+"`mysql+-u'USER'+-p'PASSWORD'+-se'USE+$DATABASE_NAME;'+2>&1`"+==+""+];+then
++++echo+$DATABASE_NAME+exist
else
++++echo+$DATABASE_NAME+doesn't+exist
fi|code-block|syntax|javascript|1563873|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Using bash:

<pre><code>if [ "`mysql -u'USER' -p'PASSWORD' -se'USE $DATABASE_NAME;' 2&gt;&amp;1`" == "" ]; then
 echo $DATABASE_NAME exist
else
 echo $DATABASE_NAME doesn't exist
fi
</code></pre>

blocks|key|1563737|text|SELECT+IF('database_name'+IN(SELECT+SCHEMA_NAME+FROM+INFORMATION_SCHEMA.SCHEMATA),+1,+0)+AS+found;|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1563738|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>SELECT IF('database_name' IN(SELECT SCHEMA_NAME FROM INFORMATION_SCHEMA.SCHEMATA), 1, 0) AS found;
</code></pre>

blocks|key|1348629|text|使用此脚本，您可以获得是或否数据库存在，如果它不存在，它不会抛出异常。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1348630|SELECT+
++++IF(EXISTS(+SELECT+
++++++++++++++++SCHEMA_NAME
++++++++++++FROM
++++++++++++++++INFORMATION_SCHEMA.SCHEMATA
++++++++++++WHERE
++++++++++++++++SCHEMA_NAME+=+'DbName'),
++++++++'Yes',
++++++++'No')++as+exist|code-block|syntax|javascript|1348631|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

With this Script you can get Yes or No database exists, in case it does not exist it does not throw Exception.

<pre><code>SELECT 
 IF(EXISTS( SELECT 
 SCHEMA_NAME
 FROM
 INFORMATION_SCHEMA.SCHEMATA
 WHERE
 SCHEMA_NAME = 'DbName'),
 'Yes',
 'No') as exist
</code></pre>

blocks|key|1564113|text|下面是我在bash脚本中执行此操作的方法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1564114|#!/bin/sh

DATABASE_USER=*****
DATABASE_PWD=*****
DATABASE_NAME=my_database

if+mysql+-u$DATABASE_USER+-p$DATABASE_PWD+-e+"use+$DATABASE_NAME";
then
echo+"Database+$DATABASE_NAME+already+exists.+Exiting."
exit
else
echo+Create+database
mysql+-u$DATABASE_USER+-p$DATABASE_PWD+-e+"CREATE+DATABASE+$DATABASE_NAME"
fi|code-block|syntax|javascript|1564115|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Here's my way of doing it inside a bash script:

<pre><code>#!/bin/sh

DATABASE_USER=*****
DATABASE_PWD=*****
DATABASE_NAME=my_database

if mysql -u$DATABASE_USER -p$DATABASE_PWD -e "use $DATABASE_NAME";
then
echo "Database $DATABASE_NAME already exists. Exiting."
exit
else
echo Create database
mysql -u$DATABASE_USER -p$DATABASE_PWD -e "CREATE DATABASE $DATABASE_NAME"
fi
</code></pre>

blocks|key|1563525|text|Rails代码：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1563526|ruby-1.9.2-p290+:099+>+ActiveRecord::Base.connection.execute("USE+INFORMATION_SCHEMA")

ruby-1.9.2-p290+:099+>+ActiveRecord::Base.connection.execute("SELECT+SCHEMA_NAME+FROM+++++++++INFORMATION_SCHEMA.SCHEMATA+WHERE+SCHEMA_NAME+=+'entos_development'").to_a
SQL+(0.2ms)+SELECT+SCHEMA_NAME+FROM+INFORMATION_SCHEMA.SCHEMATA+WHERE+SCHEMA_NAME+=+++++++++++++++'entos_development'
=>+[["entos_development"]]+
ruby-1.9.2-p290+:100+>+ActiveRecord::Base.connection.execute("SELECT+SCHEMA_NAME+FROM++++++++++++++INFORMATION_SCHEMA.SCHEMATA+WHERE+SCHEMA_NAME+=+'entos_development1'").to_a
SQL+(0.3ms)+SELECT+SCHEMA_NAME+FROM+INFORMATION_SCHEMA.SCHEMATA+WHERE+SCHEMA_NAME+=++++++++++++'entos_development1'
=>+[]|code-block|syntax|javascript|1563527|=>+entos_development存在，entos_development1不存在|1563528|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Rails Code:

<pre><code>ruby-1.9.2-p290 :099 &gt; ActiveRecord::Base.connection.execute("USE INFORMATION_SCHEMA")

ruby-1.9.2-p290 :099 &gt; ActiveRecord::Base.connection.execute("SELECT SCHEMA_NAME FROM INFORMATION_SCHEMA.SCHEMATA WHERE SCHEMA_NAME = 'entos_development'").to_a
SQL (0.2ms) SELECT SCHEMA_NAME FROM INFORMATION_SCHEMA.SCHEMATA WHERE SCHEMA_NAME = 'entos_development'
=&gt; [["entos_development"]] 
ruby-1.9.2-p290 :100 &gt; ActiveRecord::Base.connection.execute("SELECT SCHEMA_NAME FROM INFORMATION_SCHEMA.SCHEMATA WHERE SCHEMA_NAME = 'entos_development1'").to_a
SQL (0.3ms) SELECT SCHEMA_NAME FROM INFORMATION_SCHEMA.SCHEMATA WHERE SCHEMA_NAME = 'entos_development1'
=&gt; []
</code></pre>

=> entos_development exist , entos_development1 not exist

blocks|key|1563725|text|IF+EXISTS+(SELECT+SCHEMA_NAME+FROM+INFORMATION_SCHEMA.SCHEMATA+WHERE+SCHEMA_NAME+=+N'YourDatabaseName')
BEGIN++++
++++--+Database+exists,+so+do+your+stuff+here.
END|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1563726|如果您使用的是MSSQL而不是MySQL，请参阅此answer+from+a+similar+thread。|unstyled|offset|length|1563727|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/679000/how-to-check-if-a-database-exists-in-sql-server^0|0|P|S|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|R|8|@]|9|@$G|S|H|T|1|U]]|A|$]]|$1|I|3|-4|5|F|7|V|8|@]|9|@]|A|$]]]|J|$K|$5|L|M|N|A|$O|P]]]]

<pre><code>IF EXISTS (SELECT SCHEMA_NAME FROM INFORMATION_SCHEMA.SCHEMATA WHERE SCHEMA_NAME = N'YourDatabaseName')
BEGIN 
 -- Database exists, so do your stuff here.
END
</code></pre>

If you are using MSSQL instead of MySQL, see this <a href="https://stackoverflow.com/questions/679000/how-to-check-if-a-database-exists-in-sql-server">answer from a similar thread</a>.

blocks|key|1348577|text|我简单地使用了以下查询：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1348578|"USE+'DBname'"|code-block|syntax|javascript|1348579|然后检查结果是否为FALSE。否则，可能会出现拒绝访问的错误，但我无法确定。因此，在涉及特权的情况下，可以使用：|1348580|"SHOW+DATABASES+LIKE+'DBname'"|1348581|正如前面提到的。|1348582|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

I am using simply the following query:

<pre><code>"USE 'DBname'"
</code></pre>

Then check if the result is FALSE.
Otherwise, there might be an access denied error, but I cannot know that.
So, in case of privileges involved, one can use:

<pre><code>"SHOW DATABASES LIKE 'DBname'"
</code></pre>

as already mentioned earlier.

blocks|key|1348715|text|另一个php解决方案，但使用PDO：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1348716|<?php
try+{
+++$pdo+=+new+PDO('mysql:host=localhost;dbname=dbname',+'root',+'password',+[+PDO::ATTR_ERRMODE+=>+PDO::ERRMODE_EXCEPTION+]);
+++echo+'table+dbname+exists...';
}
catch+(PDOException+$e)+{
+++die('dbname+not+found...');
}|code-block|syntax|javascript|1348717|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Another php solution, but with PDO:
<pre><code>&lt;?php
try {
 $pdo = new PDO('mysql:host=localhost;dbname=dbname', 'root', 'password', [ PDO::ATTR_ERRMODE =&gt; PDO::ERRMODE_EXCEPTION ]);
 echo 'table dbname exists...';
}
catch (PDOException $e) {
 die('dbname not found...');
}
</code></pre>

blocks|key|1348415|text|下面的解决方案对我很有效：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1348416|mysql+-u${MYSQL_USER}+-p${MYSQL_PASSWORD}+\
-s+-N+-e+"SELECT+SCHEMA_NAME+FROM+INFORMATION_SCHEMA.SCHEMATA+WHERE+SCHEMA_NAME='${MYSQL_DATABASE}'"|code-block|syntax|javascript|1348417|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Following solution worked for me:

<pre><code>mysql -u${MYSQL_USER} -p${MYSQL_PASSWORD} \
-s -N -e "SELECT SCHEMA_NAME FROM INFORMATION_SCHEMA.SCHEMATA WHERE SCHEMA_NAME='${MYSQL_DATABASE}'"
</code></pre>

blocks|key|1348768|text|当您没有足够的权限查看数据库时，使用INFORMATION_SCHEMA或show+databases是不可靠的。当你不能访问数据库时，它看起来就像不存在。然后，创建将失败。另一种更精确检查的方法是使用use命令的输出，尽管我不知道这种方法有多可靠(文本输出在未来的版本/其他语言中会发生变化...)因此，请注意。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1348769|CHECK=$(mysql+-sNe+"use+DB_NAME"+2>&1)
if+[+$?+-eq+0+];+then
++#+database+exists+and+is+accessible
elif+[+!+-z+"$(echo+$CHECK+%7C+grep+'Unknown+database')"+];+then
++#+database+does+not+exist
elif+[+!+-z+"$(echo+$CHECK+%7C+grep+'Access+denied')"+];+then
++#+cannot+tell+if+database+exists+(not+enough+permissions)"
else
++#+unexpected+output
fi|code-block|syntax|javascript|1348770|entityMap^0|I|I|11|E|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|S|8|@]|D|@]|E|$]]]|L|$]]

Using the <code>INFORMATION_SCHEMA</code> or <code>show databases</code> is not reliable when you do not have enough permissions to see the database. It will seem that the DB does not exist when you just don't have access to it. The creation would then fail afterwards. Another way to have a more precise check is to use the output of the use command, even though I do not know how solid this approach could be (text output change in future versions / other languages...) so be warned.
<pre><code>CHECK=$(mysql -sNe &quot;use DB_NAME&quot; 2&gt;&amp;1)
if [ $? -eq 0 ]; then
 # database exists and is accessible
elif [ ! -z &quot;$(echo $CHECK | grep 'Unknown database')&quot; ]; then
 # database does not exist
elif [ ! -z &quot;$(echo $CHECK | grep 'Access denied')&quot; ]; then
 # cannot tell if database exists (not enough permissions)&quot;
else
 # unexpected output
fi
</code></pre>

blocks|key|1348798|text|SELECT+COUNT(*)+FROM+INFORMATION_SCHEMA.SCHEMATA+WHERE+SCHEMA_NAME+=+'DbName'|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1348799|1-存在，0-不存在|unstyled|1348800|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>SELECT COUNT(*) FROM INFORMATION_SCHEMA.SCHEMATA WHERE SCHEMA_NAME = 'DbName'
</code></pre>
1 - exists, 0 - not

Is it possible to check if a (MySQL) database exists after having made a connection.

I know how to check if a table exists in a DB, but I need to check if the DB exists. If not I have to call another piece of code to create it and populate it.

I know this all sounds somewhat inelegant - this is a quick and dirty app.

How to check if mysql database exists

数据库

MySQL

是否可以在建立连接后检查(MySQL)数据库是否存在。我知道如何检查表是否存在于DB中，但我需要检查DB是否存在。如果不是，我必须调用另一段代码来创建它并填充它。我知道这一切听起来有点不雅观--这是一个又快又脏的应用程序。

问如何检查mysql数据库是否存在
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回答 20

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