php - 'Bob' 中的未知列 'field list' ... php 或 mysql 问题

Question

我有一个奇怪的问题，我希望它是我忽略的愚蠢问题。

我收到一个错误，通常在 mysql 表中缺少列或拼写错误的 php 变量时发生:“‘field list’ 中的未知列‘Bob’”...

除了“Bob”不是列的名称，它是我试图分配给该行的 VALUE。为什么 php 会混淆这两者？？

这是我认为是错误位置的 php 函数:

function recordGuest($id,$fname,$lname,$dinner){
    $conn = connect("wedding");
    $guest_query = "INSERT INTO guest (fname,lname,person_id)
                    VALUES (".$fname.",".$lname.",".$id.");";
    mysql_query($guest_query,$conn) or die(mysql_error());

    $guest_id_query = "SELECT id FROM guest WHERE person_id = ".$id.";";
    $guest_id_result = mysql_query($guest_id_query,$conn) or die(mysql_error());
    $guest_id = "";
    while($row = mysql_fetch_array($guest_id_result)){
        $guest_id = $row["id"];
    }

    $guest_dinner_query = " INSERT INTO guest_dinner (dinner_id,guest_id)
                            VALUES (".$dinner.",".$guest_id.");";
    mysql_query($guest_dinner_query,$conn) or die(mysql_error());
}

下面是处理表单并执行上述函数的php代码:

<?php
include("functions.php");

$code = $_POST["code"];
$type = $_POST["type"];
$people = getPeople($code);
$ids = $people["ids"];
$email= "";

for($i = 0; $i < count($ids); $i++){
    $response = $_POST["response_".$i];
    $dinner = $_POST["dinner_".$i];
    recordResponse($ids[$i],$response);
    if($dinner != "null"){
        recordDinner($ids[$i],$dinner);
    }
    if($type == 3){
        $guest_responses = $_POST["guest_response_".$i];
        $guest_fname = $_POST["guest_fname_".$i];
        $guest_lname = $_POST["guest_lname_".$i];
        $guest_dinner_response = $_POST["guest_dinner_response_".$i];
        if($guest_dinner_response != "null"){
            recordGuest($ids[$i],$guest_fname,$guest_lname,$guest_dinner_response);
        }
    }
}
?>

这是我的“访客”mysql 表的样子:

guest
    id int auto inc (primary key)
    fname varchar
    lname varchar
    person_id int

如有任何帮助，我们将不胜感激。

Answer 1

在尝试插入字符串值之前，用单引号将字符串括起来，即 'Bob'