通过PHP将数据插入MySQL表时出错

Question

我有一个带有表“ AllParties”的MySQL数据库。 当我使用PHP将数据插入该表时（如下所示），我没有收到错误消息，但是该数据未插入该表中。 我尝试从MySQL服务器获取错误报告，但模棱两可，它说MySQL语法中有错误。 请注意，我在Mac CodeRunner中运行我的代码，可能这是问题所在吗？ 另外，$ con是成功的连接。

<?php
$con=mysqli_connect("sql3.freemysqlhosting.net","*********","********","*********");

$name = "Will's Party";
$date = 'October 1st, 2013';
$housenum = '333 East Street';
$city = "Golden Gate";
$state = "California"; 
$time = "7:00";
$tag = "Serra";

mysqli_select_db('AllParties', $con);

$alpha = mysqli_query($con, 'INSERT INTO AllParties(Party Name, Date, House number and       street name, City, State, Time, Tag)  VALUES("$name","$date","$housenum","$city","$state","$time","$tag")');
?>

Answer 1

任何带有空格的列名以及类似的命名列都是使开发人员非常沮丧的好方法，需要转义：

INSERT INTO AllParties (`Party Name`, ...)

反引号（不是常规引号）用于转义数据库，表和列的名称。 正则引号'和"用于字符串。

Answer 2

<?php
$con=mysqli_connect("sql3.freemysqlhosting.net","*********","********","*********");

$name = 'Will\'s Party';
$date = 'October 1st, 2013';
$housenum = '333 East Street';
$city = 'Golden Gate';
$state = 'California'; 
$time = '7:00';
$tag = 'Serra';

mysqli_select_db('AllParties', $con);

$alpha = mysqli_query($con, "INSERT INTO AllParties(`Party Name`, `Date`, `House   number and street name`, `City`, `State`, `Time`, `Tag`)    VALUES('{$name}','{$date}','{$housenum}','{$city}','{$state}','{$time}','{$tag}')");
?>

{$ x}将评估x字符串