How to convert sql query to codeigniter query
Can somebody help me convert this Sql Query
SELECT *
FROM customer c
LEFT JOIN customer_order co
ON c.customer_number = co.customer_number
AND co.order_status IN ('preparing', 'prepared')
WHERE c.customer_status='unpaid'
AND c.order_status = 'unserve'
AND co.cus_ord_no IS null
into Codeigniter query just like the image below for example
mysql codeigniter codeigniter-3
add a comment |
Can somebody help me convert this Sql Query
SELECT *
FROM customer c
LEFT JOIN customer_order co
ON c.customer_number = co.customer_number
AND co.order_status IN ('preparing', 'prepared')
WHERE c.customer_status='unpaid'
AND c.order_status = 'unserve'
AND co.cus_ord_no IS null
into Codeigniter query just like the image below for example
mysql codeigniter codeigniter-3
Possible duplicate of How to execute my SQL query in CodeIgniter
– sad saddest
Nov 21 '18 at 23:57
add a comment |
Can somebody help me convert this Sql Query
SELECT *
FROM customer c
LEFT JOIN customer_order co
ON c.customer_number = co.customer_number
AND co.order_status IN ('preparing', 'prepared')
WHERE c.customer_status='unpaid'
AND c.order_status = 'unserve'
AND co.cus_ord_no IS null
into Codeigniter query just like the image below for example
mysql codeigniter codeigniter-3
Can somebody help me convert this Sql Query
SELECT *
FROM customer c
LEFT JOIN customer_order co
ON c.customer_number = co.customer_number
AND co.order_status IN ('preparing', 'prepared')
WHERE c.customer_status='unpaid'
AND c.order_status = 'unserve'
AND co.cus_ord_no IS null
into Codeigniter query just like the image below for example
mysql codeigniter codeigniter-3
mysql codeigniter codeigniter-3
edited Nov 21 '18 at 19:49
ropenrom24
1068
1068
asked Nov 21 '18 at 16:45
Jeremy PaulJeremy Paul
10510
10510
Possible duplicate of How to execute my SQL query in CodeIgniter
– sad saddest
Nov 21 '18 at 23:57
add a comment |
Possible duplicate of How to execute my SQL query in CodeIgniter
– sad saddest
Nov 21 '18 at 23:57
Possible duplicate of How to execute my SQL query in CodeIgniter
– sad saddest
Nov 21 '18 at 23:57
Possible duplicate of How to execute my SQL query in CodeIgniter
– sad saddest
Nov 21 '18 at 23:57
add a comment |
2 Answers
2
active
oldest
votes
When query statements do not have clauses that need to change conditionally then using $this->db-query()
is the way to go.
$sql = "SELECT * FROM customer c LEFT JOIN customer_order co
ON c.customer_number=co.customer_number AND co.order_status IN ('preparing', 'prepared')
WHERE c.customer_status='unpaid' AND c.order_status='unserve' AND co.cus_ord_no IS null";
$query = $this->db->query($sql)->result();
echo json_encode($query);
It might be wise to include a check on the return from query()
though because if it fails (returns false) then the call to result()
will throw an exception. One way that can be handled is like this.
$query = $this->db->query($sql);
if($query !== FALSE)
{
echo json_encode($query->result());
return;
}
echo json_encode(); // respond with an empty array
Query Builder (QB) is a nice tool, but it is often overkill. It adds a lot of overhead to create a string that literally is passed to $db->query()
. If you know the string and it doesn't need to be restructured for some reason you don't need QB.
QB is most useful when you want to make changes to your query statement conditionally. Sorting might be one possible case.
if($order === 'desc'){
$this->db->order_by('somefield','DESC');
} else {
$this->db->order_by('somefield','ASC');
}
$results = $this->db
->where('other_field', "Foo")
->get('some_table')
->result();
So if the value of $order
is 'desc'
the query statement would be
SELECT * FROM some_table WHERE other_field = 'Foo' ORDER BY somefield 'DESC'
But if you insist on using Query Builder I believe this your answer
$query = $this->db
->join('customer_order co', "c.customer_number = co.customer_number AND co.order_status IN ('preparing', 'prepared')", 'left')
->where('c.customer_status','unpaid')
->where('c.order_status','unserve')
->where('co.cus_ord_no IS NULL')
->get('customer c');
//another variation on how to check that the query worked
$result = $query ? $query->result() : ;
echo json_encode($result);
thank you it works
– Jeremy Paul
Nov 22 '18 at 10:24
add a comment |
You can do
public function view_customers()
{
$sql = "SELECT * FROM customer c LEFT JOIN customer_order co ON c.customer_number = co.customer_number AND co.order_status IN ('preparing', 'prepared') WHERE c.customer_status='unpaid' AND c.order_status = 'unserve' AND co.cus_ord_no IS null";
return $this->db->query($sql)->result();
}
You can use row()
for one output to object
, or row_array()
if one output but array
. result()
is multiple objects
and result_array()
is multiple arrays
.
My way do usually is like this:
Controller:
public function view()
{
$this->load->model('My_Model');
$data = new stdclass;
$data->user_lists = $this->my_model->view_users(array('nationality'=>'AMERICAN'));
}
Model:
public function view_users($param = null) //no value passed
{
$condition = '1';
if (!empty($param)) { //Having this will trap if you input an array or not
foreach ($param as $key=>$val) {
$condition .= " AND {$key}='{$val}'"; //Use double quote so the data $key and $val will be read.
}
}
$sql = "SELECT * FROM users WHERE {$condition}"; //Use double quote so the data $condition will be read.
// Final out is this "SELECT * FROM users WHERE 1 AND nationality='AMERICAN'";
return $this->db->query($sql)->result();
}
this method also works for me thank you!
– Jeremy Paul
Nov 22 '18 at 10:25
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
When query statements do not have clauses that need to change conditionally then using $this->db-query()
is the way to go.
$sql = "SELECT * FROM customer c LEFT JOIN customer_order co
ON c.customer_number=co.customer_number AND co.order_status IN ('preparing', 'prepared')
WHERE c.customer_status='unpaid' AND c.order_status='unserve' AND co.cus_ord_no IS null";
$query = $this->db->query($sql)->result();
echo json_encode($query);
It might be wise to include a check on the return from query()
though because if it fails (returns false) then the call to result()
will throw an exception. One way that can be handled is like this.
$query = $this->db->query($sql);
if($query !== FALSE)
{
echo json_encode($query->result());
return;
}
echo json_encode(); // respond with an empty array
Query Builder (QB) is a nice tool, but it is often overkill. It adds a lot of overhead to create a string that literally is passed to $db->query()
. If you know the string and it doesn't need to be restructured for some reason you don't need QB.
QB is most useful when you want to make changes to your query statement conditionally. Sorting might be one possible case.
if($order === 'desc'){
$this->db->order_by('somefield','DESC');
} else {
$this->db->order_by('somefield','ASC');
}
$results = $this->db
->where('other_field', "Foo")
->get('some_table')
->result();
So if the value of $order
is 'desc'
the query statement would be
SELECT * FROM some_table WHERE other_field = 'Foo' ORDER BY somefield 'DESC'
But if you insist on using Query Builder I believe this your answer
$query = $this->db
->join('customer_order co', "c.customer_number = co.customer_number AND co.order_status IN ('preparing', 'prepared')", 'left')
->where('c.customer_status','unpaid')
->where('c.order_status','unserve')
->where('co.cus_ord_no IS NULL')
->get('customer c');
//another variation on how to check that the query worked
$result = $query ? $query->result() : ;
echo json_encode($result);
thank you it works
– Jeremy Paul
Nov 22 '18 at 10:24
add a comment |
When query statements do not have clauses that need to change conditionally then using $this->db-query()
is the way to go.
$sql = "SELECT * FROM customer c LEFT JOIN customer_order co
ON c.customer_number=co.customer_number AND co.order_status IN ('preparing', 'prepared')
WHERE c.customer_status='unpaid' AND c.order_status='unserve' AND co.cus_ord_no IS null";
$query = $this->db->query($sql)->result();
echo json_encode($query);
It might be wise to include a check on the return from query()
though because if it fails (returns false) then the call to result()
will throw an exception. One way that can be handled is like this.
$query = $this->db->query($sql);
if($query !== FALSE)
{
echo json_encode($query->result());
return;
}
echo json_encode(); // respond with an empty array
Query Builder (QB) is a nice tool, but it is often overkill. It adds a lot of overhead to create a string that literally is passed to $db->query()
. If you know the string and it doesn't need to be restructured for some reason you don't need QB.
QB is most useful when you want to make changes to your query statement conditionally. Sorting might be one possible case.
if($order === 'desc'){
$this->db->order_by('somefield','DESC');
} else {
$this->db->order_by('somefield','ASC');
}
$results = $this->db
->where('other_field', "Foo")
->get('some_table')
->result();
So if the value of $order
is 'desc'
the query statement would be
SELECT * FROM some_table WHERE other_field = 'Foo' ORDER BY somefield 'DESC'
But if you insist on using Query Builder I believe this your answer
$query = $this->db
->join('customer_order co', "c.customer_number = co.customer_number AND co.order_status IN ('preparing', 'prepared')", 'left')
->where('c.customer_status','unpaid')
->where('c.order_status','unserve')
->where('co.cus_ord_no IS NULL')
->get('customer c');
//another variation on how to check that the query worked
$result = $query ? $query->result() : ;
echo json_encode($result);
thank you it works
– Jeremy Paul
Nov 22 '18 at 10:24
add a comment |
When query statements do not have clauses that need to change conditionally then using $this->db-query()
is the way to go.
$sql = "SELECT * FROM customer c LEFT JOIN customer_order co
ON c.customer_number=co.customer_number AND co.order_status IN ('preparing', 'prepared')
WHERE c.customer_status='unpaid' AND c.order_status='unserve' AND co.cus_ord_no IS null";
$query = $this->db->query($sql)->result();
echo json_encode($query);
It might be wise to include a check on the return from query()
though because if it fails (returns false) then the call to result()
will throw an exception. One way that can be handled is like this.
$query = $this->db->query($sql);
if($query !== FALSE)
{
echo json_encode($query->result());
return;
}
echo json_encode(); // respond with an empty array
Query Builder (QB) is a nice tool, but it is often overkill. It adds a lot of overhead to create a string that literally is passed to $db->query()
. If you know the string and it doesn't need to be restructured for some reason you don't need QB.
QB is most useful when you want to make changes to your query statement conditionally. Sorting might be one possible case.
if($order === 'desc'){
$this->db->order_by('somefield','DESC');
} else {
$this->db->order_by('somefield','ASC');
}
$results = $this->db
->where('other_field', "Foo")
->get('some_table')
->result();
So if the value of $order
is 'desc'
the query statement would be
SELECT * FROM some_table WHERE other_field = 'Foo' ORDER BY somefield 'DESC'
But if you insist on using Query Builder I believe this your answer
$query = $this->db
->join('customer_order co', "c.customer_number = co.customer_number AND co.order_status IN ('preparing', 'prepared')", 'left')
->where('c.customer_status','unpaid')
->where('c.order_status','unserve')
->where('co.cus_ord_no IS NULL')
->get('customer c');
//another variation on how to check that the query worked
$result = $query ? $query->result() : ;
echo json_encode($result);
When query statements do not have clauses that need to change conditionally then using $this->db-query()
is the way to go.
$sql = "SELECT * FROM customer c LEFT JOIN customer_order co
ON c.customer_number=co.customer_number AND co.order_status IN ('preparing', 'prepared')
WHERE c.customer_status='unpaid' AND c.order_status='unserve' AND co.cus_ord_no IS null";
$query = $this->db->query($sql)->result();
echo json_encode($query);
It might be wise to include a check on the return from query()
though because if it fails (returns false) then the call to result()
will throw an exception. One way that can be handled is like this.
$query = $this->db->query($sql);
if($query !== FALSE)
{
echo json_encode($query->result());
return;
}
echo json_encode(); // respond with an empty array
Query Builder (QB) is a nice tool, but it is often overkill. It adds a lot of overhead to create a string that literally is passed to $db->query()
. If you know the string and it doesn't need to be restructured for some reason you don't need QB.
QB is most useful when you want to make changes to your query statement conditionally. Sorting might be one possible case.
if($order === 'desc'){
$this->db->order_by('somefield','DESC');
} else {
$this->db->order_by('somefield','ASC');
}
$results = $this->db
->where('other_field', "Foo")
->get('some_table')
->result();
So if the value of $order
is 'desc'
the query statement would be
SELECT * FROM some_table WHERE other_field = 'Foo' ORDER BY somefield 'DESC'
But if you insist on using Query Builder I believe this your answer
$query = $this->db
->join('customer_order co', "c.customer_number = co.customer_number AND co.order_status IN ('preparing', 'prepared')", 'left')
->where('c.customer_status','unpaid')
->where('c.order_status','unserve')
->where('co.cus_ord_no IS NULL')
->get('customer c');
//another variation on how to check that the query worked
$result = $query ? $query->result() : ;
echo json_encode($result);
edited Nov 21 '18 at 20:21
answered Nov 21 '18 at 19:58
DFriendDFriend
6,9871520
6,9871520
thank you it works
– Jeremy Paul
Nov 22 '18 at 10:24
add a comment |
thank you it works
– Jeremy Paul
Nov 22 '18 at 10:24
thank you it works
– Jeremy Paul
Nov 22 '18 at 10:24
thank you it works
– Jeremy Paul
Nov 22 '18 at 10:24
add a comment |
You can do
public function view_customers()
{
$sql = "SELECT * FROM customer c LEFT JOIN customer_order co ON c.customer_number = co.customer_number AND co.order_status IN ('preparing', 'prepared') WHERE c.customer_status='unpaid' AND c.order_status = 'unserve' AND co.cus_ord_no IS null";
return $this->db->query($sql)->result();
}
You can use row()
for one output to object
, or row_array()
if one output but array
. result()
is multiple objects
and result_array()
is multiple arrays
.
My way do usually is like this:
Controller:
public function view()
{
$this->load->model('My_Model');
$data = new stdclass;
$data->user_lists = $this->my_model->view_users(array('nationality'=>'AMERICAN'));
}
Model:
public function view_users($param = null) //no value passed
{
$condition = '1';
if (!empty($param)) { //Having this will trap if you input an array or not
foreach ($param as $key=>$val) {
$condition .= " AND {$key}='{$val}'"; //Use double quote so the data $key and $val will be read.
}
}
$sql = "SELECT * FROM users WHERE {$condition}"; //Use double quote so the data $condition will be read.
// Final out is this "SELECT * FROM users WHERE 1 AND nationality='AMERICAN'";
return $this->db->query($sql)->result();
}
this method also works for me thank you!
– Jeremy Paul
Nov 22 '18 at 10:25
add a comment |
You can do
public function view_customers()
{
$sql = "SELECT * FROM customer c LEFT JOIN customer_order co ON c.customer_number = co.customer_number AND co.order_status IN ('preparing', 'prepared') WHERE c.customer_status='unpaid' AND c.order_status = 'unserve' AND co.cus_ord_no IS null";
return $this->db->query($sql)->result();
}
You can use row()
for one output to object
, or row_array()
if one output but array
. result()
is multiple objects
and result_array()
is multiple arrays
.
My way do usually is like this:
Controller:
public function view()
{
$this->load->model('My_Model');
$data = new stdclass;
$data->user_lists = $this->my_model->view_users(array('nationality'=>'AMERICAN'));
}
Model:
public function view_users($param = null) //no value passed
{
$condition = '1';
if (!empty($param)) { //Having this will trap if you input an array or not
foreach ($param as $key=>$val) {
$condition .= " AND {$key}='{$val}'"; //Use double quote so the data $key and $val will be read.
}
}
$sql = "SELECT * FROM users WHERE {$condition}"; //Use double quote so the data $condition will be read.
// Final out is this "SELECT * FROM users WHERE 1 AND nationality='AMERICAN'";
return $this->db->query($sql)->result();
}
this method also works for me thank you!
– Jeremy Paul
Nov 22 '18 at 10:25
add a comment |
You can do
public function view_customers()
{
$sql = "SELECT * FROM customer c LEFT JOIN customer_order co ON c.customer_number = co.customer_number AND co.order_status IN ('preparing', 'prepared') WHERE c.customer_status='unpaid' AND c.order_status = 'unserve' AND co.cus_ord_no IS null";
return $this->db->query($sql)->result();
}
You can use row()
for one output to object
, or row_array()
if one output but array
. result()
is multiple objects
and result_array()
is multiple arrays
.
My way do usually is like this:
Controller:
public function view()
{
$this->load->model('My_Model');
$data = new stdclass;
$data->user_lists = $this->my_model->view_users(array('nationality'=>'AMERICAN'));
}
Model:
public function view_users($param = null) //no value passed
{
$condition = '1';
if (!empty($param)) { //Having this will trap if you input an array or not
foreach ($param as $key=>$val) {
$condition .= " AND {$key}='{$val}'"; //Use double quote so the data $key and $val will be read.
}
}
$sql = "SELECT * FROM users WHERE {$condition}"; //Use double quote so the data $condition will be read.
// Final out is this "SELECT * FROM users WHERE 1 AND nationality='AMERICAN'";
return $this->db->query($sql)->result();
}
You can do
public function view_customers()
{
$sql = "SELECT * FROM customer c LEFT JOIN customer_order co ON c.customer_number = co.customer_number AND co.order_status IN ('preparing', 'prepared') WHERE c.customer_status='unpaid' AND c.order_status = 'unserve' AND co.cus_ord_no IS null";
return $this->db->query($sql)->result();
}
You can use row()
for one output to object
, or row_array()
if one output but array
. result()
is multiple objects
and result_array()
is multiple arrays
.
My way do usually is like this:
Controller:
public function view()
{
$this->load->model('My_Model');
$data = new stdclass;
$data->user_lists = $this->my_model->view_users(array('nationality'=>'AMERICAN'));
}
Model:
public function view_users($param = null) //no value passed
{
$condition = '1';
if (!empty($param)) { //Having this will trap if you input an array or not
foreach ($param as $key=>$val) {
$condition .= " AND {$key}='{$val}'"; //Use double quote so the data $key and $val will be read.
}
}
$sql = "SELECT * FROM users WHERE {$condition}"; //Use double quote so the data $condition will be read.
// Final out is this "SELECT * FROM users WHERE 1 AND nationality='AMERICAN'";
return $this->db->query($sql)->result();
}
answered Nov 21 '18 at 23:51


GGwGGw
239213
239213
this method also works for me thank you!
– Jeremy Paul
Nov 22 '18 at 10:25
add a comment |
this method also works for me thank you!
– Jeremy Paul
Nov 22 '18 at 10:25
this method also works for me thank you!
– Jeremy Paul
Nov 22 '18 at 10:25
this method also works for me thank you!
– Jeremy Paul
Nov 22 '18 at 10:25
add a comment |
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Possible duplicate of How to execute my SQL query in CodeIgniter
– sad saddest
Nov 21 '18 at 23:57