weixin_50966102 2022-08-23 16:34 采纳率: 50%
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php提交数据给数据库失败

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)
?>

2条回答 默认 最新

  • 三只小菜猿 PHP领域新星创作者 2022-08-23 16:59
    关注

    sql语句从女那个地方开始报错,你看看数据库性别字段存得啥,你这里传得字符串是不是数据库是整型还是什么

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论
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