用变量失败插入语句

Question

新的我遇到了这个问题。

它似乎无法从我上次语句中的两个第一个语句中使用我的ID作为变量资源，因此SQLcharacter语句失败。

我做错了什么？

$sqlimg = ("INSERT INTO cimages(image) VALUES(?)");
$stmtimg = $conn->prepare($sqlimg);
$stmtimg->bind_param('s', $image);
$stmtimg->execute();
$img_id = $stmtimg->insert_id;
// I insert the picture first, and retrieve it's ID
$sqlstats = ("INSERT INTO cstats(Strength, Dexterity, Constitution, 
Intelligence, Wisdom, Charisma, Aligment) VALUES(?, ?, ?, ?, ?, ?, ?)");
$stmtstats = $conn->prepare($sqlstats);
$stmtstats->bind_param("iiiiiis", $strength, $dexterity, $constitution, 
$intelligence, $wisdom, $charisma, $aligment);
$stmtstats->execute();
$stats_id = $stmtstats->insert_id;
// I insert the characters stats, and retrieve it's ID
// Last I insert The user_id and img_id and stats_id
$user_id = mysqli_real_escape_string($conn, $_POST['user_id']);
// I've used the session id to get the user_id already 

$sqlcharacter = ("INSERT INTO characters(Cname, Clast, Crace, house, 
location, Bgstory, user_id, img_id, stats_id) VALUES(?, ?, ?, ?, ?, ?, ?, 
$img_id, $stats_id)");
$stmtChar = $conn->prepare($sqlcharacter);
$stmtChar->bind_param('ssssssiii', $Cname, $Clast, $Crace, $house, 
$location, $Bgstory, $user_id, $img_id, $stats_id);
$stmtChar->execute();

Answer 1

$sqlcharacter字符串看起来好像您有两个变量$img_id和$stats_id而不是?，所以我认为这就是为什么它不绑定这些值的原因。

尝试更改以下内容：

"INSERT INTO characters(Cname, Clast, Crace, house, 
 location, Bgstory, user_id, img_id, stats_id) VALUES(?, ?, ?, ?, ?, ?, ?, 
 $img_id, $stats_id)"

：

"INSERT INTO characters(Cname, Clast, Crace, house, 
 location, Bgstory, user_id, img_id, stats_id) VALUES(?, ?, ?, ?, ?, ?, ?, 
 ?, ?)"