Php SQL创建表语法错误

我试图创建500个表，所有这些表都命名为[1-500]，并且只有一个时间戳和一个对应的值totalviews

<?php
$i = 1;
while ($i<500) {
    $con=mysqli_connect("localhost","user","pass","database");
    if (mysqli_connect_errno($con))
      {
      echo "Failed to connect to MySQL: " . mysqli_connect_error();
      }
    $sql = "CREATE TABLE \"".$i."\" ( 
    'TheTime' INT UNSIGNED PRIMARY KEY, 
    'TotalViews' INT UNSIGNED
    )";
    if (!mysqli_query($con, $sql))
      {
      die('Error: ' . mysqli_error($con));
      };
    echo "1 table created added: ".$i;
    mysqli_close($con);
$i++;
    }
?>

但是我得到了一个错误：错误：您的SQL语法有一个错误；检查与您的MySQL服务器版本对应的手册，以了解在第1行的“1”时间“INT UNSIGNED”主键、“TotalViews”INT UNSIGNED”附近使用的正确语法

---

我做错了什么？我应该以不同的方式完成这一点吗？

列名不能用单引号括起来，您可以使用反勾号

所以

应该是

`TheTime` 

及

到

---

表名：100，比如说，应该是反勾号

我是pendatic，把所有的名字都放在后面，但除了保留的单词和不寻常的名字之外，它们不是必需的

下面是测试代码：

<?php
$i = 1;
while ($i<5) {
    $con=mysqli_connect("localhost","test","test","testmysql");
    if (mysqli_connect_errno($con))
      {
      echo "Failed to connect to MySQL: " . mysqli_connect_error();
      }
    $sql = 'CREATE TABLE `'.$i.'` ('
    .'`TheTime` INT UNSIGNED PRIMARY KEY,'
    .'`TotalViews` INT UNSIGNED)';
    if (!mysqli_query($con, $sql))
      {
      die('Error: ' . mysqli_error($con));
      };
    echo "1 table created added: ".$i;
    mysqli_close($con);
$i++;
    }
?>

---

首先，您应该在开始while循环之前移动连接。 对于DB名称，请使用“代替”，对于字段，请使用“代替”符号：

<?php
$con = mysqli_connect("localhost", "user", "pass", "database");
if (mysqli_connect_errno($con)) {
    echo "Failed to connect to MySQL: " . mysqli_connect_error();

    die();
}

$i = 1;    
while ($i < 500) {
    $sql = "CREATE TABLE `" . $i . "` (
    `TheTime` INT UNSIGNED PRIMARY KEY,
    `TotalViews` INT UNSIGNED
    )";

    if (!mysqli_query($con, $sql)) {
        die('Error: ' . mysqli_error($con));
    }

    echo "1 table created added: " . $i;
    $i++;
}

mysqli_close($con);

---

PS：我建议您对mysqli使用面向对象的接口，它更简单。详细信息：

谢谢，但我现在仍然得到错误：您的SQL语法有错误；检查与您的MySQL服务器版本对应的手册，了解在第1hmm行使用接近“1 TheTime INT UNSIGNED PRIMARY KEY，TotalViews INT UNSIGNED”的正确语法。表名为1,2…等，因此您需要将它们与列名一样包含在backtics中。不要将整数用作列/表标识符。另外，创建500张完全相同的表格听起来是一个非常糟糕和愚蠢的想法。不知道为什么答案被否决了？不过，谢谢你接受。

`TotalViews` 

<?php
$i = 1;
while ($i<5) {
    $con=mysqli_connect("localhost","test","test","testmysql");
    if (mysqli_connect_errno($con))
      {
      echo "Failed to connect to MySQL: " . mysqli_connect_error();
      }
    $sql = 'CREATE TABLE `'.$i.'` ('
    .'`TheTime` INT UNSIGNED PRIMARY KEY,'
    .'`TotalViews` INT UNSIGNED)';
    if (!mysqli_query($con, $sql))
      {
      die('Error: ' . mysqli_error($con));
      };
    echo "1 table created added: ".$i;
    mysqli_close($con);
$i++;
    }
?>

<?php
$con = mysqli_connect("localhost", "user", "pass", "database");
if (mysqli_connect_errno($con)) {
    echo "Failed to connect to MySQL: " . mysqli_connect_error();

    die();
}

$i = 1;    
while ($i < 500) {
    $sql = "CREATE TABLE `" . $i . "` (
    `TheTime` INT UNSIGNED PRIMARY KEY,
    `TotalViews` INT UNSIGNED
    )";

    if (!mysqli_query($con, $sql)) {
        die('Error: ' . mysqli_error($con));
    }

    echo "1 table created added: " . $i;
    $i++;
}

mysqli_close($con);