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使用表 2 值添加到表 1 时出现 PHP SQL 语法错误

转载 作者:行者123 更新时间:2023-11-29 12:07:21 26 4
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运行此 php 代码时出现错误,该代码将根据表 2 值向表 1 添加值,其中表 3 值等于 0

我的 table

------------------------------
id| table 1| table 2 | table 3|
-------------------------------
01| 100 | 10 | 0 |

PHP 代码

<?php
$servername = "localhost";
$username = "myuser";
$password = "mypass";
$dbname = "lol";

// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}

$sql = "UPDATE table table1 = table1 + table2 WHERE table 3 = 0";



if (mysqli_query($conn, $sql)) {
echo "Record updated successfully";
} else {
echo "Error updating record: " . mysqli_error($conn);
}

mysqli_close($conn);
?>

它说我的sql语法有错误

Error updating record: 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 '= table1 + table2 WHERE table3 = 0' at line 1

如何使用正确的 sql 语法实现此目的?

最佳答案

您忘记了 SET 指令,并且您的列名称与查询中的名称不匹配,为此您必须使用反引号 -

 $sql = "UPDATE `table` SET `table 1` = `table 1` + `table 2` WHERE `table 3` = 0";

你真的应该了解prepared statements

关于使用表 2 值添加到表 1 时出现 PHP SQL 语法错误,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/31253718/

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