Question

我是一个但未使用Mysqli，我遇到以下代码的问题..

$mysql = new mysqli(DB_SERVER, DB_USER, DB_PASSWORD, DB_NAME) or die('There was a problem connecting to the database');
     if (mysqli_connect_errno()) {
        printf("DB error: %s", mysqli_connect_error());
        exit();
     }

     $query = "INSERT INTO employee(id, name, age, address, phone, email, department,
         designation, joindt, terminate, salary, deduction, tds, pf)
         VALUES (:id, :name, :age, :address, :phone, :email, :department,
         :designation, :joindt, :terminate, :salary, :deduction, :tds, :pf)";

     $ins = $mysql->prepare($query);
     if(!ins){
            echo "prepare failed\n";
      echo "error: ", $mysql->error, "\n";
      echo "OBJECT NOT CREATED";
      return;

     }

运行此代码后，我的浏览器出现以下错误：

（！）致命错误：未捕获异常'mysqli_sql_exception'   消息'你的错误   SQL语法;检查手册   对应于您的MySQL服务器   用于正确语法的版本   靠近'in   C：\ wamp \ www \ payroll \ new_backend.php on   第40行

（！）mysqli_sql_exception：你有   SQL语法中的错误;检查   与您的MySQL对应的手册   用于正确语法的服务器版本   使用附近'加入，终止，工资，   演绎，tds，pf）VALUES（：id，   ：name，：age，'在C：\ wamp的第2行   \ www \ payroll \ new_backend.php就行了   40

我猜PREPARE语句没有创建$ ins对象。有什么帮助吗？

Answer 1

join是mySQL中的保留字。你需要将它包含在反引号中：

`join`

或 - 更好 - 更改列的名称。

此外，看起来mysqli不支持PDO风格的:fieldname绑定。查看manual on prepare().

中的示例

我现在无法测试，但正确的语法应该是这样的（缩写）：

$id = 10;
$name = "John Doe";

$query = "INSERT INTO employee(id, name) values (?, ?)";
$query->bind_param("i", $id);
$query->bind_param("s", $name);

Answer 2

查看mysqli :: prepare（）的手册页：

http://es.php.net/mysqli_prepare

准备好的语句希望?成为占位符，而不是:foo。

Answer 3

这是我在Pekka的帮助下到达的解决方案.. :)

 $mysql = new mysqli(DB_SERVER, DB_USER, DB_PASSWORD, DB_NAME) or die('There was a problem connecting to the database');
     if (mysqli_connect_errno()) {
        printf("DB error: %s", mysqli_connect_error());
        exit();
     }

     $query = "INSERT INTO employee(name, age, address, phone, email, department,
         designation, joindt, terminate, salary, deduction, tds, pf)
         VALUES (?,?,?,?,?,?,?,?,?,?,?,?,?)";

     $ins = $mysql->prepare($query);
     if(!$ins){
            echo "prepare failed\n";
      echo "error: ", $mysql->error, "\n";
      return;

     }

     $ins->bind_param('sisisssiiiiii', $data['name'], $data['age'], $data['address'],
             $data['phone'], $data['email'], $data['department'], $data['designation'],
             $data['joindate'], $data['terminationdate'], $data['salary'], $data['leave_deduction'], $data['tds'], $data['pf']);

     $ins->execute();

Mysqli“准备语句”不是创建对象

3 个答案: