将表单值插入到数据库中

Question

如何在数据库中插入此表单的值？

<tr> 
    <td>1. Do you feel well and healthy?</label></td> 
    <input type="hidden" name="question" value="1. Do you feel well and healthy?"> 
    <td><input type="radio" name="answer" value="Yes" checked></td> 
    <td><input type="radio" name="answer" value="No"></td> 
    <td><input type="text" name="remarks"></td> 
</tr> 
<tr> 
    <td><ul><li>Have any flu or cold?</li></ul></td> 
    <input type="hidden" name="question" value="Have any flu or cold?"> 
    <td><input type="radio" name="answer" value="Yes" checked> </td> 
    <td><input type="radio" name="answer" value="No"></td> 
    <td><input type="text" name="remarks"></td> 
</tr> 

因为当我对这个数据插入到数据库中只插入最后一个值是2的表行中的“有烟” ......

$medical = new MedicalHistory; 
$medical->username = $value; 
$medical->question = $data['question']; 
$medical->answer = $data['answer']; 
$medical->remarks = $data['remarks1']; 
$medical->save(); 

我使用laravel作为一个框架

Answer 1

你的形式应该是这样的：

<tr> 
    <td>1. Do you feel well and healthy?</label></td> 
    <input type="hidden" name="question" value="1. Do you feel well and healthy?"> 
    <td><input type="radio" name="answer" value="Yes" checked></td> 
    <td><input type="radio" name="answer" value="No"></td> 
    <td><input type="text" name="remarks"></td> 
</tr> 
<tr> 
    <td><ul><li>Have any flu or cold?</li></ul></td> 
    <input type="hidden" name="question2" value="Have any flu or cold?"> 
    <td><input type="radio" name="answer2" value="Yes" checked> </td> 
    <td><input type="radio" name="answer2" value="No"></td> 
    <td><input type="text" name="remarks2"></td> 
</tr> 
<?php 
$medical = new MedicalHistory; 
$medical->username = $value; 
$medical->question = $data['question']; 
$medical->answer = $data['answer']; 
$medical->remarks = $data['remarks1']; 
$medical->save(); 

$medical = new MedicalHistory; 
$medical->username = $value; 
$medical->question = $data['question2']; 
$medical->answer = $data['answer2']; 
$medical->remarks = $data['remarks2']; 
$medical->save(); 
?> 

其实你的表单元素越来越OVERR idden。

感谢 阿米特

Answer 2

原因仅被插入的第二种形式是因为你重用你的领域相同的名称。

一个解决方案将给你的所有领域提供独特的名称（单选按钮对除外）。

更好的解决方案将使得阵列投入，我已经做了一个快速搜索你以及与此想出了：How to get form input array into PHP array

您应该修改代码以数组变量，因此，如果您想保存多个问题，在两个问题字段中使用名称'question []'而不是'question'。

前端表例如：

<tr> 
    <td>1. Do you feel well and healthy?</label></td> 
    <input type="hidden" name="question[]" value="1. Do you feel well and healthy?"> 
    <td><input type="radio" name="answer[0]" value="Yes" checked></td> 
    <td><input type="radio" name="answer[0]" value="No"></td> 
    <td><input type="text" name="remarks[]"></td> 
</tr> 
<tr> 
    <td><ul><li>Have any flu or cold?</li></ul></td> 
    <input type="hidden" name="question[]" value="Have any flu or cold?"> 
    <td><input type="radio" name="answer[1]" value="Yes" checked> </td> 
    <td><input type="radio" name="answer[1]" value="No"></td> 
    <td><input type="text" name="remarks[]"></td> 
</tr> 

以处理PHP逻辑，你应该做这样的事情你的数据的新格式：

$question = $_POST['question']; 
$answer = $_POST['answer']; 
$remark = $_POST['remarks']; 

foreach($question as $key => $q) { 
    print "The question is ".$q.", answer is ".$answer[$key]. 
      ", and remarks are ".$remark[$key].". Thank you\n"; 
} 

既然你要保存的数据，你应该这样做：

$value = 'idk_where_you_assign_names'; 
$question = $_POST['question']; 
$answer = $_POST['answer']; 
$remark = $_POST['remarks']; 

foreach($question as $key => $q) { 
    $medical = new MedicalHistory; 
    $medical->username = $value; 
    $medical->question = $q; 
    $medical->answer = $answer[$key]; 
    $medical->remarks = $remark[$key]; 
    $medical->save(); 
} 

经过我的编辑它现在应该工作。