CODEWITHSUNDEEP
phpcodeignitercodeigniter-3
Shahzaib Ahsan
Shahzaib Ahsan

Reputation: 31

Storing Same Record my on Submit Button

Hello I'm looking to get the record and save it to my data base but all the time I'm getting the same sql statement for all the select options

Course ID INSERT INTO `tbl_assign_course` (`course_id_1`, course_id_2`, `course_id_3`) VALUES ('7', '7', '7')

Here is the Options list to choose:

How I can fix this logical error I will appreciate any kind of help 

<!DOCTYPE html>
<html>
<head>
    <title> The University</title>
    <link rel="stylesheet" type="text/css" href="<?php echo base_url(); ?>css/style.css">
</head>
<body>
<div class="topnav">
    <?php $this->load->view('header'); ?>
</div>
<?php
print_r($result);
?>

<form method="post" action="<?php echo(base_url() . "University/std_course_assign/" . $result[0]['course_id'] . "/" . $result[0]['course_i'] . "/" . $result[0]['course_id']); ?>">
    <table>
        <tr>
            <td>
                <select name='course_id_1'>
                    <option value="">--- Select Course ---</option>
                    <?php
                    for ($i = 0; $i < count($result); $i++) { ?>
                        <?php echo "<option value=" . $result[$i]['course_id'] . ">" . $result[$i]['course_name'] . "</option>"; ?>
                        <?php
                    }
                    ?>
            </td>
            <td>
                <select name='course_id_2'>
                    <option value="">--- Select Course ---</option>
                    <?php
                    for ($i = 0; $i < count($result); $i++) { ?>
                        <?php echo "<option value=" . $result[$i]['course_id'] . ">" . $result[$i]['course_name'] . "</option>"; ?>
                        <?php
                    }
                    ?>
            </td>
            <td>
                <select name='course_id_3'>
                    <option value="">--- Select Course ---</option>
                    <?php
                    for ($i = 0; $i < count($result); $i++) { ?>
                        <?php echo "<option value=" . $result[$i]['course_id'] . ">" . $result[$i]['course_name'] . "</option>"; ?>
                        <?php
                    }
                    ?>
                    <input type="hidden" name="std_id" value="<?= $std_id ?>">
                    <input type="Submit">

            </td>
        </tr>
    </table>
</form>
</body>
</html>

Model Function

public function std_course_assign($course_id) {
    $data = array('std_id' => $this->input->post('std_id'),
                  'course_id_1' => $this->input->post('course_id'),
                  'course_id_2' => $this->input->post('course_id'),
                  'course_id_3' => $this->input->post('course_id'),
        );
    $this->db->insert('tbl_assign_course', $data);
    $this->db->where('std_id', $this->input->post('std_id'));
    print_r($data);
}

Upvotes: 0

Views: 27

Answers (2)

helpdoc
helpdoc

Reputation: 1990

I think You are using WHERE condition with INSERT query, use correct INSERT query for desired result.

and use correct attribute name in insert query

Upvotes: 0

Anand Pandey
Anand Pandey

Reputation: 2025

Your insert query is wrong, change it to 'course_id' to 'course_id_1' 'course_id_2' and so on:

public function std_course_assign($course_id) { 
    $data= array( 
        'std_id'=>$this->input->post('std_id'), 
        'course_id_1'=>$this->input->post('course_id_1'), 
        'course_id_2'=>$this->input->post('course_id_2'), 
        'course_id_3'=>$this->input->post('course_id_3')
    ); 
    $this->db->insert('tbl_assign_course', $data); 
    $this->db->where('std_id',$this->input->post('std_id')); 
    print_r($data); 
}

Upvotes: 1

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