错误，一些记录输入数据库而一些记录没有 [英] Error , some records entered into the database and some's not
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问题描述
我正在尝试将数据添加到数据库中，只有两个记录正确输入（名字，成员类型），ID总是0
这里是错误：
注意：未定义的索引：C中的memberid：\ xampp \\\ htdocs \ ...
注意：未定义的索引：C中的姓氏：\ xampp \\\ htdocs \ ..
注意：未定义的索引：C中的phonenumber：\ xampp \\\ htdocs \ ...
注意：未定义的索引：C：\ xampp \htdocs \中的电子邮件...
注意：未定义的索引：C中的用户名：\ xampp \htdocs \ ...
注意：未定义的索引：C中的密码：\ xampp \htdocs \ ...
注意：未定义的索引：C部门：\ xampp \\\ htdocs \ ...
记录添加成功。
我尝试了什么：
这里是代码
对于HTML
< pre > < 表格 action = test.php 方法 = 发布 >
< div class = form-group >
< label for = element-1 class = control-标签 > 名字< / label >
< 输入 type = text id = element-1 占位符 = 在此输入您的名字 class = 表单控制 ， < span class =code-attribute> name = firstname >
< / div >
< div class = form-group >
< label for = ele ment-2 class = control-label > 姓氏< ; / label >
< 输入 type = text id = element-2 占位符 = 在此输入您的姓氏 class = form-control ，name = 姓氏 >
< / div >
< div class = form-group >
< 标签 = element-2 class = control-label > 会员ID < / label >
< input 类型 = text id = element-2 占位符 = 在此输入您的ID class = form-control ，名称 = member_id >
< / div >
< div class = form-group >
< label for = element-2 class = control-label > 用户名< / label >
< 输入 type = text id = element-2 占位符 = 输入您的用户名这里 class = 表单控件 ，名称 = 用户名 >
< / div >
< div class = form-group >
< label = element-5 class = control-label > Passw ord < / label >
< 输入 type = password id = element-5 占位符 = 在此输入您的密码 class = form-control < span class =code-attribute>，name = 密码 >
< / div >
< div class = form-group >
< label for = element-5 class = control-label > 确认密码< / label >
< 输入 type = 密码 id = element-5 占位符 = 在此确认您的密码 class = 表单控件 ，名称 = confirmmpassword >
< / div >
< div class = form-group >
< label = element-5 类 = control-label > 电子邮件< / label >
< 输入 type = < span class =code-keyword> text id = element-5 占位符 = 输入您的电子邮件这里 class = 表单控制 ，名称 = email >
< / div >
< span class =code-keyword>< div class = tyled-select blue semi-squar >
< label = element-2 class = control-label > 会员类型< / label >
< 选择 名称 = membertype >
< 选项 value = 已选择 data-default > 选择成员输入< / option >
< 选项 value = 教授 > 教授< / option >
< 选项 < span class =code-attribute> value = assistent > 辅助< / option >
< 选项 value = student > 学生< / option >
< / select >
< / div >
< div class = form-group >
< label = element-2 class = < span class =code-keyword> control-label > 电话号码< / label >
< 输入 类型 = text id = element-2 占位符 = 在此输入您的电话号码 class = form-control ，名称 = phonenumber >
< / div >
< div class = 表单组 >
< label for = element-2 class = control-label > 部门< / label >
< 输入 类型 = text id = element-2 占位符 = 在此处输入您的部门 class = 表单控件 ，名称 = 部门 >
< / div >
< 按钮 class = btn btn-primary type = 提交 < span class =code-attribute> name = submit > 添加< / button >
< / form > 这是我的PHP代码
if（isset（ $ _POST [' submit']））
{
// 逃避用户输入以确保安全
$ id = mysqli_real_escape_string（$ link，$ _REQUEST [' member_id']）;
$ first_name = mysqli_real_escape_string（$ link，$ _REQUEST [' 姓名']）;
$ last_name = mysqli_real_escape_string（$ link，$ _REQUEST [' 名字']）;
$ phone = mysqli_real_escape_string（$ link，$ _ REQUEST [' phonenumber']）;
$ email = mysqli_real_escape_string（$ link，$ _ REQUEST [' email']）;
$ username = mysqli_real_escape_string（$ link，$ _ REQUEST [' username']）;
$ password = mysqli_real_escape_string（$ link，$ _ REQUEST [' password']）;
$ department = mysqli_real_escape_string（$ link，$ _ REQUEST [' department']）;
if （isset（$ _ POST [' membertype']））{
$ membertype = $ _POST [' membertype'];
}
// 尝试插入查询执行
$ sql = INSERT INTO`member`（` ID`，`first_name`，`last_name`，`member_type`，`password`，`email`，`phoneNo`，`department`，`user_name`）
VALUES（'$ id'，'$ first_name' ，'$ last_name'，'$ membertype'，'$ password'，'$ email'，'$ phone'，'$ department'，'$ username'）;
if（mysqli_query（$ link，$ sql））{
echo 记录添加成功。;
} else {
echo 错误：无法执行$ sql。。 mysqli_error（$链接）;
}
}
解决方案
_POST [' submit']））
{
// < span class =code-comment>逃避安全的用户输入
id = mysqli_real_escape_string （
链路，
I am trying to add data to the database only two records entered correctly (first name, member type), and the ID always be 0 here's the error: Notice: Undefined index: memberid in C:\xampp\htdocs\... Notice: Undefined index: lastname in C:\xampp\htdocs\.. Notice: Undefined index: phonenumber in C:\xampp\htdocs\... Notice: Undefined index: email in C:\xampp\htdocs\... Notice: Undefined index: username in C:\xampp\htdocs\... Notice: Undefined index: password in C:\xampp\htdocs\... Notice: Undefined index: department in C:\xampp\htdocs\... Records added successfully.
What I have tried:
here's the code
For HTML
<pre><form action="test.php" method="post">
<div class="form-group">
<label for="element-1" class="control-label">First Name</label>
<input type="text" id="element-1" placeholder="Enter Your First Name Here" class="form-control" , name="firstname">
</div>
<div class="form-group">
<label for="element-2" class="control-label">Last Name</label>
<input type="text" id="element-2" placeholder="Enter Your Last Name Here" class="form-control",name="lastname">
</div>
<div class="form-group">
<label for="element-2" class="control-label">Member ID</label>
<input type="text" id="element-2" placeholder="Enter Your ID Here" class="form-control" ,name="member_id">
</div>
<div class="form-group">
<label for="element-2" class="control-label">User Name</label>
<input type="text" id="element-2" placeholder="Enter Your User Name Here" class="form-control",name="username">
</div>
<div class="form-group">
<label for="element-5" class="control-label"> Password</label>
<input type="password" id="element-5" placeholder="Enter Your Password Here" class="form-control",name="password">
</div>
<div class="form-group">
<label for="element-5" class="control-label">confirm Password</label>
<input type="password" id="element-5" placeholder="confirm Your Password Here" class="form-control",name="confirmpassword">
</div>
<div class="form-group">
<label for="element-5" class="control-label"> Email</label>
<input type="text" id="element-5" placeholder="Enter Your Email Here" class="form-control",name="email">
</div>
<div class="tyled-select blue semi-squar">
<label for="element-2" class="control-label">Member Type</label>
<select name="membertype">
<option value="" selected data-default>select member type</option>
<option value="professor">Professor</option>
<option value="assistent">Assistent</option>
<option value="student">Student</option>
</select>
</div>
<div class="form-group">
<label for="element-2" class="control-label">Phone Number</label>
<input type="text" id="element-2" placeholder="Enter Your Phone Number Here" class="form-control",name="phonenumber">
</div>
<div class="form-group">
<label for="element-2" class="control-label">department</label>
<input type="text" id="element-2" placeholder="Enter Your Department Here" class="form-control",name="department">
</div>
<button class="btn btn-primary" type="submit" name="submit">Add</button>
</form> and here is my php code
if(isset($_POST['submit']))
{
// Escape user inputs for security
$id = mysqli_real_escape_string($link, $_REQUEST['member_id']);
$first_name = mysqli_real_escape_string($link, $_REQUEST['firstname']);
$last_name = mysqli_real_escape_string($link, $_REQUEST['lastname']);
$phone= mysqli_real_escape_string($link,$_REQUEST['phonenumber']);
$email= mysqli_real_escape_string($link,$_REQUEST['email']);
$username= mysqli_real_escape_string($link,$_REQUEST['username']);
$password= mysqli_real_escape_string($link,$_REQUEST['password']);
$department= mysqli_real_escape_string($link,$_REQUEST['department']);
if (isset($_POST['membertype']) ){
$membertype= $_POST['membertype'];
}
// attempt insert query execution
$sql = "INSERT INTO `member` (`ID`, `first_name`, `last_name`, `member_type`, `password`, `email`, `phoneNo`, `department`, `user_name`)
VALUES ('$id', '$first_name', '$last_name','$membertype','$password','$email','$phone','$department','$username')";
if(mysqli_query($link, $sql)){
echo "Records added successfully.";
} else{
echo "ERROR: Could not able to execute $sql. " . mysqli_error($link);
}
} 解决方案
_POST['submit'])) { // Escape user inputs for security
id = mysqli_real_escape_string(
link,
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