Question

我有maiden name文本框字段，仅当性别为女性并且用户状态不是单一时才会启用，但只要用户注册了单个状态，就会出现警告，但是仍然保存数据。为什么以及如何删除该警告消息，因为它在消息回显时可见。

我的HTML代码：

<label> Maiden Name</label>
     <input type="text" class="form-control" name="maiden" id="mySelect" value="" disabled=true placeholder="Maiden Nname" onKeyPress="return ValidateAlpha(event);" onblur="toUpper(this.value, this.id);" >

<label>Gender</label>
       <input type=radio name="gender" id="gender" value='Male'>&nbsp;Male &nbsp;
       <input type=radio name="gender" id="gender" value='Female'>&nbsp;Female 

<label>Status</label>
    <select name="status" id="status" class="form-control">
        <option value="Single">Single</option>
        <option value="Married">Married</option>
        <option value="Widow">Widow</option>
    </select>

这是我的javascript：

<script>
    //whenever an item with the name attribute set to gender or status is changed:
    $(document).on("change", "[name=gender],[name=status]", function(){
        //gather the checked gender
        var gender = $('[name=gender]:checked').val();
      //gather the selected status
      var status = $('[name=status] option:selected').val();
      //if the user checked female and selected single
      if(gender == 'Female' && status != 'Single'){
        //enable the maidenname input
        $('[name=maiden]').prop('disabled', false);
      }
     else{
        //disable the maidenname input
        $('[name=maiden]').prop('disabled', true);
      }
    });
</script>

我的php

<?php
include('dbcontroller.php');
if(isset($_POST['submit']))
{
    $gender = $conn->real_escape_string(trim($_POST['gender']));
    $status = $conn->real_escape_string(trim($_POST['status']));
    $maiden = $conn->real_escape_string(trim($_POST['maiden']));

    $query = "INSERT INTO requests(gender, status, maiden) VALUES('$gender','$status','$maiden')";
 }   ?>

Answer 1

如果没有您的代码，很难具体说明，通常您也会发布实际错误，但听起来您需要......

 if(empty($_POST['maidenname'])) {$_POST['maidenname'] ='';}

每当我注册但仍然保存数据时，为什么我会有一个未定义的变量警告？

1 个答案: