[英]How to join multiple column values into one column while inserting
I have a table which looks like this: 我有一张桌子,看起来像这样:
ID | Name | Address
I have a form which looks like this: 我有一个看起来像这样的表格:
<input type="text" class="form-control" name="name"/>
<input type="text" class="form-control" name="address"/>
My PHP Code: 我的PHP代码:
$name = $_POST['name'];
$address = $_POST['address'];
$sql = "Insert into staff(client_name,address) VALUES ('$client_name','$address')";
$this->db->query($sql);
What i want to do is, join the values posted by user into one column.ie if user types JOHN in Name field and USA in address field. 我想要做的是,将用户发布的值加入一列。即,如果用户在“名称”字段中键入JOHN,在地址字段中键入USA。 I want to following in my database table 我想在我的数据库表中关注
ID | Name | Address
1 | JOHN USA | USA
使用这样的方法:
$sql = "Insert into staff(client_name,address) VALUES ('$client_name $address','$address')";
Simply concatenate your $_POST['name']
with $_POST['address']
in $client_name
before inserting to database. 在插入数据库之前, $client_name
将$client_name
的$_POST['name']
与$_POST['address']
起来。 Like this : 像这样 :
$client_name = $_POST['name'] .' '. $_POST['address'];
$address = $_POST['address'];
$sql = "Insert into staff(client_name,address) VALUES ('$client_name','$address')";
$this->db->query($sql);
$client_name = $_POST['name'];
$address = $_POST['address'];
$name = $client_name .' '. $address;
$sql = "Insert into staff(client_name,address) VALUES ('$name','$address')";
$this->db->query($sql);
try this 尝试这个
$name = $_POST['name'];
$address = $_POST['address'];
$client_name = $name.' | '.$address;
$sql = "Insert into staff(client_name,address) VALUES ('$client_name','$address')";
$this->db->query($sql);
Use json_encode to encode the content and then store it in database field and when you retrieve the value then use json_decode 使用json_encode对内容进行编码,然后将其存储在数据库字段中,当您检索值时,请使用json_decode
$form_data_json = json_encode( $_POST );
// store $form_data_json in database.
// When you want to get stored data, just fetch it from db. let it be stored in $fetch
$original_post_array = json_decode( $fetch, true );
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