插入语句，变量失败

问题描述：

new here I've come across this problem.

It doesn't seem to be able to use my id's from the two first statements in my last statements as a variable resource, so the sqlcharacter statement fails.

What do i do wrong?

$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();

new here 我遇到过这个问题。 p>

它似乎无法将我上一个语句中的两个第一个语句中的id作为变量资源使用，因此sqlcharacter语句失败。 p> \ n

我做错了什么？ p>

  $ sqlimg =（“INSERT INTO cimages（image）VALUES（？）”）; 
 $ stmtimg = $ conn-＆gt; prepare（$ sqlimg）; 
  $ stmtimg-＆gt; bind_param（'s'，$ image）; 
 $ stmtimg-＆gt; execute（）; 
 $ img_id = $ stmtimg-＆gt; insert_id; 
 
 //我先插入图片， 并检索它的ID 
 
 $ sqlstats =（“INSERT INTO cstats（力量，敏捷，体质，
智能，智慧，魅力，讽刺）值（？，？，？，？，？，？，？）”）  ; 
 $ stmtstats = $ conn-＆gt; prepare（$ sqlstats）; 
 $ stmtstats-＆gt; bind_param（“iiiiiis”，$ strength，$ dexterity，$ constitution，
 $ intelligence，$ wisdom，$ charisma，  $ aligment）; 
 $ stmtstats-＆gt; execute（）; 
 $ stats_id = $ stmtstats-＆gt; insert_id; 
 
 //我插入字符统计信息，并检索它的ID 
 //最后我插入 user_id和img_id和stats_id 
 $ user_id = mysqli_real_escape_string（$ conn，$ _POST ['user_id']）; 
 //我已经使用会话ID来获取user_id 
 
 
 $ sqlcharacter =  （“INSERT INTO字符（Cname，Clast，Crace，house，
location，Bgstory，user_id，img_id，stats_id）  ）VALUES（？，？，？，？，？，？，？，
 $ img_id，$ stats_id）“）; 
 $ stmtChar = $ conn-＆gt; prepare（$ sqlcharacter）; 
 $ stmtChar-＆gt  ; bind_param（'ssssssiii'，$ Cname，$ Clast，$ Crace，$ house，
 $ location，$ Bgstory，$ user_id，$ img_id，$ stats_id）; 
 $ stmtChar-＆gt; execute（）; 
   code>  pre> 
  div>

答

The $sqlcharacter string looks like you've got two variables $img_id and $stats_id in there instead of ?, so I think that's why it's not binding those values.

Try changing this:

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

To this:

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

插入语句，变量失败

相关推荐