CodeIgniter-单击“在数据库中创建表”
我有一个输入字段和一个按钮。当我单击按钮时,必须在数据库中创建一个表,其名称应为输入字段中的输入。控制器:CodeIgniter-单击“在数据库中创建表”,codeigniter,Codeigniter,我有一个输入字段和一个按钮。当我单击按钮时,必须在数据库中创建一个表,其名称应为输入字段中的输入。控制器: function create() { $table = $this->input->post('table'); $this->M_users->create($table); } function create($table) { $sql = "CREATE TABLE ".$table." ( id INT(6) UNSIGN
function create()
{
$table = $this->input->post('table');
$this->M_users->create($table);
}
function create($table)
{
$sql = "CREATE TABLE ".$table." (
id INT(6) UNSIGNED AUTO_INCREMENT PRIMARY KEY,
firstname VARCHAR(30) NOT NULL,
lastname VARCHAR(30) NOT NULL,
email VARCHAR(50),
reg_date TIMESTAMP
)";
$query = $this->db->query($sql);
return $query;
}
<form method="post" action="<?php echo base_url('create');?>">
<input type="text" name="table">
<input type="submit" name="">
</form>
型号:
function create()
{
$table = $this->input->post('table');
$this->M_users->create($table);
}
function create($table)
{
$sql = "CREATE TABLE ".$table." (
id INT(6) UNSIGNED AUTO_INCREMENT PRIMARY KEY,
firstname VARCHAR(30) NOT NULL,
lastname VARCHAR(30) NOT NULL,
email VARCHAR(50),
reg_date TIMESTAMP
)";
$query = $this->db->query($sql);
return $query;
}
<form method="post" action="<?php echo base_url('create');?>">
<input type="text" name="table">
<input type="submit" name="">
</form>
查看:
function create()
{
$table = $this->input->post('table');
$this->M_users->create($table);
}
function create($table)
{
$sql = "CREATE TABLE ".$table." (
id INT(6) UNSIGNED AUTO_INCREMENT PRIMARY KEY,
firstname VARCHAR(30) NOT NULL,
lastname VARCHAR(30) NOT NULL,
email VARCHAR(50),
reg_date TIMESTAMP
)";
$query = $this->db->query($sql);
return $query;
}
<form method="post" action="<?php echo base_url('create');?>">
<input type="text" name="table">
<input type="submit" name="">
</form>
谢谢,我也可以使用迁移和dbforge来完成这项工作吗?我不确定,但我认为你可以。如果你已经完成了这项工作,请接受答案。@RedEye\u 0:是的,你可以使用codeigniter中的dbforge类来完成这项工作。@RedEye\u 0请参考此内容