$result2=mysql_query("exists adimin1");这样判断数据库是否存在好像是错的。本人初学php,求高手们指正啊!已经连上mysql,创建数据库已经成功。
$mysql_command1="create database admin1";
$result1=mysql_query($mysql_command1) or die("database creation failure:".mysql_errno().":".mysql_error());
if($result1)
echo "database created successfully!";
$mysql_command1="create database admin1";
$result1=mysql_query($mysql_command1) or die("database creation failure:".mysql_errno().":".mysql_error());
if($result1)
echo "database created successfully!";
if(mysql_select_db("database",$link)){
echo "库存在";
}
SELECT count(1) FROM `information_schema`.`SCHEMATA` where SCHEMA_NAME like '数据库名称'