使用 jquery 和 php codeigniter 插入多行药物

Question

如果我想添加单击按钮而不是打开下一个药物输入字段而不是填充，我想添加一次插入一行的药物，如果一次 5 次药物输入比 5 次插入数据库，则插入下一行。但这里只有一个将数据插入数据库的时间复选框不起作用只接受第一个值

查看页面

我需要数据库

name      quantity     days     take medician
----------------------------------------------
Amit       1            2        morning,evening
-----------------------------------------------
Amitabh    2            3        evening,night
-----------------------------------------------
Amitabh    3            4        afternoon,night
kumar 
gupta

插入当前数据库

name      quantity     days     take medician
----------------------------------------------
Amit       1            2        morning
-----------------------------------------------

html查看页面

 <tbody>
        <?php foreach ($app_booking as $row){ ?>
        <input type="hidden" name="doctorname" id="doctorname" value="<?php echo $row["doctor_name"];?>">
        <input type="hidden" id="doctorid" name="doctorid" value="<?php echo $row["doctor_id"];?>">
        <input type="hidden" id="appid"  name="appid" value="<?php echo $row["appointment_id"];?>">
        <input type="hidden" id="uid" name="uid" value="<?php echo $row["user_id"];?>">
    <?php };?>
        <tr>
        <td><input class="form-control" type="text" name="name" id="name" required=""></td>
        <td><input class="form-control" type="text" id="quantity" name="quantity" required=""></td>
        <td><input class="form-control" type="text" name="days" id="days"  required=""></td>
        <td>
            <div class="form-check form-check-inline">
             <label class="form-check-label">                                               
            <input class="form-check-input checkbox_value" type="checkbox" name="take_medicine[]" id="morning" value="Morning"> Morning
            </label>
        </div>
        <div class="form-check form-check-inline">
            <label class="form-check-label">
            <input class="form-check-input checkbox_value" type="checkbox" name="take_medicine[]" id="afternoon" value="Afternoon"> Afternoon
            </label>
            </div>
        <div class="form-check form-check-inline">
        <label class="form-check-label">
            <input class="form-check-input checkbox_value" type="checkbox" name="take_medicine[]" id="evening" value="Evening"> Evening
            </label>
        </div>
        <div class="form-check form-check-inline">
        <label class="form-check-label">
        <input class="form-check-input checkbox_value" type="checkbox" name="take_medicine[]" value="Night" id="night"> Night
        </label>
        </div>
        </td>
        <td>
        <a href="#"  name="add" id="btn1" value="Add" class="btn bg-success-light"><i class="fas fa-plus-circle"></i> Add Item</a>
        </td>
        <button type="button" onclick="save_medical_records();" class="btn btn-primary submit-btn">Save</button>
            </tr>
        </tbody>

ajax 代码

 function save_medical_records()
           {
               
               var doctorname           =   $("input[name='doctorname']").val();
               var doctorid             =   $("input[name='doctorid']").val();  
               var appid                =   $("input[name='appid']").val();
               var uid                  =   $("input[name='uid']").val();
               var name                 =   $("input[name='name']").val();             
               var quantity             =   $("input[name='quantity']").val();
               var days                 =   $("input[name='days']").val();
               
                
                const take_meds = $("[name='take_medicine[]']:checked").map(function() { return this.value}).get()
                 console.log(take_meds);
                
               
                   $.ajax({
                       url:"<?php echo base_url() ?>add-prescription",
                       data:"doctorname="+doctorname+"&doctorid="+doctorid+"&appid="+appid+"&uid="+uid+"&name="+name+"&quantity="+quantity+"&days="+days+"&take_meds="+take_meds,
                       //data: $("[name=doctorname]").serialize(),$("[name=doctorid]").serialize(),$("[name=appid]").serialize(),$("[name=uid]").serialize(),$("[name=name]").serialize(),$("[name=quantity]").serialize(),$("[name=days]").serialize(),$("[name=take_meds]").serialize(),
                       
                       
                       
                       type:"post",
                       success:function(response){ 
                           if(response==1)
                              
                              $("#msg").css("display","block"); 
                           
                       }
                       });
               }

            
    

控制器

public function add_prescription()
    {
        $result = $this->doctor_health_model->add_prescription();
        echo $result;

    }

模型

public function add_prescription()
    {
        $db2 = $this->load->database('dpr',TRUE);
        $medicine       = $this->input->post('name');
        $uid            = $this->input->post('uid');
        $appid          = $this->input->post('appid');
        $doctor_id      = $this->input->post('doctorid');
        $doctor_name    = $this->input->post('doctorname');
        $quantity       = $this->input->post('quantity');
        $days           = $this->input->post('days');
        $take_medicine  = $this->input->post('take_medicine');  
        $today_date     = date("Y-m-d");
        
     $insert =$db2->query('INSERT INTO dpr_save_farmacytest (medicine,user_id,appointment_id,doctor_id,doctor_name,quantity,days,take_medicine_time,created_date) 
            VALUES ("'.$medicine.'","'.$uid.'","'.$appid.'","'.$doctor_id.'","'.$doctor_name.'","'.$quantity.'","'.$days.'","'.$take_medicine.'","'.$today_date.'")');
            //echo $db2->last_query();
            return $insert;

        }
    }

Answer 1

除了 SQL 注入问题、在一个表字段中存储多个值等问题之外，您的直接问题由以下人员回答：

这就是 jQuery 的工作原理 - 这将返回一个值，从第一个具有相同名称的元素开始

var take_medicine =   $("input[name='take_medicine[]']").val();

您需要获取每个选中的复选框 - 例如

const take_meds = $("[name='take_medicine[]']:checked")
       .map(function() { return this.value}).get();
console.log(take_meds);

<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<div class="form-check form-check-inline">
  <label class="form-check-label"><input class="form-check-input checkbox_value" checked type="checkbox" name="take_medicine[]" id="morning" value="Morning"> Morning</label>
</div>
<div class="form-check form-check-inline">
  <label class="form-check-label"><input class="form-check-input checkbox_value" type="checkbox" name="take_medicine[]" id="afternoon" value="Afternoon"> Afternoon</label>
</div>
<div class="form-check form-check-inline">
  <label class="form-check-label"><input class="form-check-input checkbox_value" checked type="checkbox" name="take_medicine[]" id="evening" value="Evening"> Evening</label>
</div>

如果您使用serialize，您可能会过上更简单的生活

如果您有 <form id="myForm" - 您将整个 AJAX 代码更改为

$("#myForm").on("submit", function(e) {
  e.preventDefault();
  $.ajax({
    url: "<?php echo base_url() ?>add-prescription",
    data: $(this).serialize(),
    type: "post",
    success: function(response) {
      if (response == 1) $("#msg").css("display", "block");
    },
    error: function(jxhr) {
      console.log("Error", jxhr)
    }
  })
});