PHP ~ 列数与第 1 行的值数不匹配

Question

我正在尝试插入两个表，但出现此错误

Error: INSERT INTO provide_help (amount) VALUES ( 40,000.00) Column count doesn't match value count at row 1`

下面是我的插入代码

<?php

    session_start(); {



    //Include database connection details
    include('../../dbconnect.php');


$amount =  strip_tags($_POST['cat']);
$field1amount = $_POST['cat'];
$field2amount = $field1amount + ($field1amount*0.5);



$sql = "INSERT INTO provide_help (amount) VALUES ( $field1amount)";
if (mysqli_query($conn, $sql)) 


$sql = "INSERT INTO gh (ph_id, amount) VALUES (LAST_INSERT_ID(), $field2amount)";
if (mysqli_query($conn, $sql)) 

 {
    $_SESSION['ph'] ="<center><div class='alert alert-success' role='alert'>Request Accepted.</div></center>";
   header("location: PH.php");
} else {
    echo "Error: " . $sql . "<br>" . mysqli_error($conn);
}

mysqli_close($conn);
}
?>

但是当我做这样的事情时它会起作用

$sql = "INSERT INTO provide_help (amount) VALUES ( $field2amount)";

我只是将 $field1amount 更改为 $field2amount

但我不想那样做，我还想获取 $field1amount 的值并插入它

请提供任何帮助，谢谢

Answer 1

问题是因为您传递的数字中有一个逗号，而不是字符串。您需要传入 "40,000.00" 或 40000.00。 MySQL 将其解释为两个值:40 和 000.00。

使用准备好的语句将缓解这种情况(以及您的安全问题)，因为绑定(bind)会将 40,000.00 解释为字符串。让您入门的一个非常基本的示例是:

$sql = "INSERT INTO provide_help (amount) VALUES (?)";
$stmt = $mysqli->prepare($sql);

/*
    - the "s" below means string
    - NOTE you should still validate the $_POST value,
      don't just accept whatever is sent through your form - 
      make sure it matches the format you're expecting at least
      or you'll have data validation issues later on
*/
$stmt->bindParam("s", $field1amount);
$stmt->execute($fieldAmount1);
$result = $res->fetch_assoc();