各位大佬!!为什么我将网页上输入的form表单的内容利用PHP接收然后传送到数据时一直添加失败!我哭了我,改了前面许多错之后,最后报一个我填写的内容的周围有错误,错误提示是:添加数据失败You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '女','513023200002210222','1235654@ff.com','12345468768')' at line 1
html部分代码如下
<div class="form-group">
<label for="idcard" class="col-sm-2 control-label">身份证号码</label>
<div class="col-sm-6">
<input type="text" class="form-control" name ="number" required="required" placeholder="请输入您的身份证号码">
<span id="idspan" style="color: red">
<div class="form-group">
<label for="email" class="col-sm-2 control-label">邮箱</label>
<div class="col-sm-6">
<input type="email" class="form-control" name="email" id="email" required="required" class="form-control"onclick="check(form)"placeholder="请输入您的邮箱">
<span id="emailspan" style="color: red">
<div class="form-group">
<label for="pnumber" class="col-sm-2 control-label">手机号</label>
<div class="col-sm-6">
<input type="tel" class="form-control" name="pnumber"required="required" class="form-control"onclick="checkemail()" placeholder="请输入您的手机号">
<div class="form-group">
<div class="col-sm-offset-2 col-sm-10">
<button type="submit" class="btn btn-default">提交</button>
</form>
php部分代码如下
<?php
$servername = "localhost";
$username = "root";
$password = "123456";
$dbname = "news";
$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_error) {
die("连接失败: " . $conn->connect_error);
}
mysqli_select_db($conn,"news");
$sql = "INSERT INTO news_tb3 (name,sex,num,email,pnumber) VALUES('$_POST[name],'$_POST[sex]','$_POST[number]','$_POST[email]','$_POST[pnumber]')";
if(mysqli_query($conn,$sql))
echo"添加数据成功";
else
echo"添加数据失败".mysqli_error(($conn));
mysqli_close($conn)
?>