是否有一种方法可以在MySQL中使用PHP获取表的列名?
当前回答
你可能还想检查mysql_fetch_array(),如下所示:
$rs = mysql_query($sql);
while ($row = mysql_fetch_array($rs)) {
//$row[0] = 'First Field';
//$row['first_field'] = 'First Field';
}
其他回答
如果你喜欢的话,还有这个:
mysql_query('SHOW COLUMNS FROM tableName');
这个怎么样:
SELECT @cCommand := GROUP_CONCAT( COLUMN_NAME ORDER BY column_name SEPARATOR ',\n')
FROM INFORMATION_SCHEMA.COLUMNS
WHERE TABLE_SCHEMA = 'my_database' AND TABLE_NAME = 'my_table';
SET @cCommand = CONCAT( 'SELECT ', @cCommand, ' from my_database.my_table;');
PREPARE xCommand from @cCommand;
EXECUTE xCommand;
这对我很有效。
$sql = "desc MyTableName";
$result = @mysql_query($sql);
while($row = @mysql_fetch_array($result)){
echo $row[0]."<br>";
}
下面的SQL语句几乎是等价的:
SELECT COLUMN_NAME
FROM INFORMATION_SCHEMA.COLUMNS
WHERE table_name = 'tbl_name'
[AND table_schema = 'db_name']
[AND column_name LIKE 'wild']
SHOW COLUMNS
FROM tbl_name
[FROM db_name]
[LIKE 'wild']
参考:INFORMATION_SCHEMA列
这个解决方案来自命令行mysql
mysql>USE information_schema;
在下面的查询中,只需将<——DATABASE_NAME——>更改为您的数据库,<——TABLENAME——>更改为您的表名,其中您只需要description语句的字段值
mysql> SELECT COLUMN_NAME FROM COLUMNS WHERE TABLE_SCHEMA = '<--DATABASE_NAME-->' AND TABLE_NAME='<--TABLENAME-->';