php mysqli synax错误
我的代码怎么了?我确信php mysqli synax错误,php,mysql,Php,Mysql,我的代码怎么了?我确信$\u POST['item']具有有效值 <?php $data = $_POST['item']; $conn = mysqli_connect("localhost","root",""); mysqli_select_db($conn, "ajaxexample"); $q = INSERT INTO user (userList) VALUES ('$data'); if(mysqli_query($conn, $q)){ echo 1; } ?
$\u POST['item']
具有有效值
<?php
$data = $_POST['item'];
$conn = mysqli_connect("localhost","root","");
mysqli_select_db($conn, "ajaxexample");
$q = INSERT INTO user (userList) VALUES ('$data');
if(mysqli_query($conn, $q)){
echo 1;
}
?>
put
插入用户(userList)值(“$data”)代码>用双引号括起来
例如:
PHP字符串文本需要加引号
要通过仅更改一行来解决此问题,请执行以下操作:
$q = "INSERT INTO user (userList) VALUES ('" . mysqli_real_escape_string($data . "')";
不是mysqli\u select\u db
您看起来好像缺少了一些引号。与其“确定”为什么不回显$data的值并检查。使用mysqli\u real\u escape\u string($\u POST['item'])
对于securityshow表名和结构就是这样:我创建了一个类似的表……我只发现了“双引号”问题。它应该是双引号而不是单引号。
$q = "INSERT INTO user (userList) VALUES ('" . mysqli_real_escape_string($data . "')";
<?php
$data = $_POST['item'];
$conn = mysqli_connect("localhost","root","", "ajaxexample");
$q = INSERT INTO user (userList) VALUES ('$data');
if(mysqli_query($conn, $q)){
echo 1;
}
?>
Not mysqli_select_db