duanjuan1103 2015-06-18 05:58
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如何在db中插入一个值

in my view i have one dropdown menu and one textbox..in my dropdown list contain 3 values daily salary,weekly salary and monthly salary..And the text field for amount..

in my db contain 3 columns for daily,weekly and monthly if i select a value in my dropdown then enter an amount in my text field..that amount is entered into the particular column like daily,weekly and monthly

But shows a database error

view

<label>Mode of Salary</label>
<select name="salary_mode">
     <option disabled="disabled" selected>Select</option>
     <option value="1">Daily </option>
     <option value="2">Weekly</option>
     <option value="3">Monthly</option>
</select></br></br>
<input id="" name="amount_received" type="text" placeholder="Amount Received"  class="form-control">

controlller

for($i=0;$i<count($select_employee);$i++)
{
    $salary_mode=$this->input->post('salary_mode');
    // echo $salary_mode;
    $amount_received=$this->input->post('amount_received');
    // echo $amount_received;
    if($salary_mode==1)
    {
        $daily_salary=$amount_received;
        $weekly_salary=0;
        $monthly_salary=0;
    }
    else if($salary_mode==2)
    {
        $daily_salary=0;
        $weekly_salary=$amount_received;
        $monthly_salary=0;

    }
    else if($salary_mode==3)
    {
        $daily_salary=0;
        $weekly_salary=0;
        $monthly_salary=$amount_received;
    }
    $employee_id =TRIM($select_employee[$i]);
    $issued_date=$this->input->post('issued_date');

    $id=$this->session->userdata('userid');

    $salary_data=array(
        'employee_id'=>$employee_id,
        'user_id'=>$id,
        'daily_salary'=>$daily_salary,
        'weekly_salary'=>$weekly_salary,
        'monthly_salary'=>$monthly_salary,
        'issued_date'=>$issued_date,
    );
    $this->home_model->insert_salary($salary_data);
}

model

function insert_salary($salary_data)
{
    $this->db->insert('track_salary',$salary_data);
}
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1条回答 默认 最新

  • dongsechuan0535 2015-06-18 06:14
    关注

    try to set all three daily_salary, monthly_salary and weekly_salary =0 before the for loop, because you are using if else so its obvious when 1 condition is true the other two variables will not be set.

    $daily_salary=0;
    $weekly_salary=0;
    $monthly_salary=0;
    
        for($i=0;$i<count($select_employee);$i++)
            {    
                $salary_mode=$this->input->post('salary_mode');
                // echo $salary_mode;
                 $amount_received=$this->input->post('amount_received'); 
                // echo $amount_received;
                  if($salary_mode==1)
                 {
                   $daily_salary=$amount_received;  
                 }
                 else if($salary_mode==2)
                 {
                    $weekly_salary=$amount_received; 
                 }
                 else if($salary_mode==3)
                 {
                     $monthly_salary=$amount_received;
                 }
                 $employee_id =TRIM($select_employee[$i]);
                 $issued_date=$this->input->post('issued_date');
    
                 $id=$this->session->userdata('userid');
    
                $salary_data=array(
                   'employee_id'=>$employee_id,
                   'user_id'=>$id,
                   'daily_salary'=>$daily_salary,
                   'weekly_salary'=>$weekly_salary,
                   'monthly_salary'=>$monthly_salary,
                   'issued_date'=>$issued_date,
                );
                $this->home_model->insert_salary($salary_data);
            }
          }
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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