0

我即将制作一个复选框，将其存储在我检查的位置，但出现错误：

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 ' Array, 5' at line 1

我现在更新了我的代码，我得到一个新的错误：

“无法添加或更新子行：外键约束失败 ( dashboard. das_custermers_employees, CONSTRAINT das_custermers_employees_ibfk_5FOREIGN KEY ( das_employess_id) REFERENCESdas_employer_roles ( id))”

希望有人可以提供帮助...我已将所有代码放入 så 希望有人能看到问题所在

我的 HTML

    <div class="contact_form_div">
            <form method="post">
                <?php
                global $wpdb;                        

if (isset($_POST['add_contact_btn'])) {

    mysql_query("DELETE FROM das_custermers_employees WHERE das_custermers_id = $user_custermers_id") or die(mysql_error());


    if (!empty($_POST['role_checkbox']) && is_array($_POST['role_checkbox'])) {
    foreach ($_POST['role_checkbox'] as $employess_roles) {
        foreach ($employess_roles as $em_id ){
            $employess_hidden = $_POST["employess_hidden_$em_id"];
        mysql_query("INSERT INTO das_custermers_employees(das_employess_id, employees_role_id, das_custermers_id) VALUES ($employess_hidden, $em_id, $user_custermers_id) ") or die(mysql_error());


        }
    }
    }
}
                        $show_employess = mysql_query("SELECT * FROM das_employess") or die(mysql_error());

                while ($my_employess = mysql_fetch_array($show_employess)) {

                    ?>                        
                        <b><? echo " " . $my_employess['first_name'] . " ";
                            echo $my_employess['last_name'] . " "; ?></b>
                        <?
                        $show_role = mysql_query("SELECT * FROM das_employer_roles") or die(mysql_error());

                        while ($my_role = mysql_fetch_array($show_role)) {

                            ?>
                            <input type="hidden" name="employess_hidden_<? echo $my_role['id']; ?>" value="<? echo $my_employess['id']; ?>" />
                            <em><p class="role_checkbox"><input type="checkbox" name="role_checkbox[][<? echo $my_role['id']; ?>]" <? //if (mysql_num_rows($role_query) == 1) echo 'checked="checked"'; ?> value="<? echo $my_role['id']; ?>"/><? echo " " . $my_role['role']; ?> </p></em>
                                <? } ?>                        
    <?
}

?>
                    <input value="Gem" name="add_contact_btn" type="submit" >
                    </form>
            </div>

这是表架构： 这是数据库架构：

4

2 回答 2

0

我认为：

mysql_query("INSERT INTO das_custermers_employees(das_employess_id, employees_role_id, das_custermers_id) VALUES ($employess_hidden, $employess_roles, $user_custermers_id ") or die(mysql_error());

应该：

mysql_query("INSERT INTO das_custermers_employees(das_employess_id, employees_role_id, das_custermers_id) VALUES ($employess_hidden, $em_id, $user_custermers_id ") or die(mysql_error());

您试图将数组插入到INSERT语句中，而不是foreach循环中的元素。

另外，改变：

$employess_hidden = $_POST["employess_id_$em_id"];

至：

$employess_hidden = $_POST["employess_hidden_$em_id"];

您的表单中没有employess_id_XXX输入。

于 2013-04-04T06:27:20.547 回答
0

你错过了什么..在下面检查...

mysql_query("INSERT INTO das_custermers_employees(das_employess_id, employees_role_id, das_custermers_id) VALUES ($employess_hidden, $em_id, $user_custermers_id)") or die(mysql_error());
于 2013-04-04T06:24:43.457 回答